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S.No Question Option A Option B Option C Option D Answer Solution Comments Status Action
1 If a draft tube is used with a Francis turbine (installed above tail race level), the pressure at the runner outlet is equal to atmospheric pressure is above atmospheric pressure is below atmosphere pressure depends upon turbine speed c Draft tube is a diverging tube filled at the exit of runner of turbine and used to utilize the kinetic energy available with water at exit of runner.
This draft tube at the end of the turbine increases the pressure of the exiting fluid at the expense of its velocity. This means that the turbine can reduce pressure to a higher extent without fear of back flow from the tail race.
Also, if the pressure of the fluid in the tail race is higher that at the exit of the turbine, a back flow of liquid into the turbine can result in significant damage. By placing a
draft tube (also called as diffuser tube or pipe) at the exit of the turbine, the turbine pressure head is increased by decreasing the exit velocity and both the overall efficiency and output of the turbine can be improved. The draft tube works by converting some of kinetic energy at the exit of the turbine runner into the useful pressure energy. It is located just under the runner and allowed to decelerate the flow velocity exiting the runner, thereby converting the excess of kinetic energy into static pressure.
Pressure at inlet of draft is below the atmospheric exit of draft tube will be equal to the atmosphere.		 Comments Active
2 Turbine gives best performance (i.e. work at peak efficiency) when they are operated at full or design load.
The performance of many turbines deteriorates considerably at part loads. Which of the following turbines is best suited for operation at part loads. ?		 Pelton Turbine Francis Turbine Propeller Turbine Kaplan Turbine d Part load operations:
It is the overload condition in which efficiency reduces. This is uneconomical for hydraulic power plant.
* Kaplan turbines has adjustable or movable blades hence for a required discharge the movable blades can be adjusted so that the losses due to flow separation can be
minimized Hence part load efficiency is considerably higher for the Kaplan turbine as it can maintain high efficiency over a wide range of operating condition.		 Comments Active
3 Which of the following statements is not correct about MHD power generation ? Lesser thermal pollution No moving parts High operation efficiency No direct conversion of heat into electrical energy d The advantage of MHD generator include the following
* MHD generators converts heat or thermal energy directly into electrical energy.
* It has no moving parts so mechanical loss would be minimal.
* Highly efficient has higher operational efficiency more than conventional generator, therefore, the overall cost of an MHD plant is less compared to conventional steam plants.
* Operational and maintenance cost are less
* It works on any type of fuel and has better fuel.
Disadvantages
* Aids in the high amount of losses that include fluid friction and heat transfer losses.
* Needs large magnets leading to a higher costs in implementing MHD generators.
* High operating temperatures in the range of 200áµ’ K to 2400áµ’K will corrode to components sooner.		 Comments Active
4 Which of the following statements is not true for gas turbines ? Low full load thermal efficiency Costly machines Self-starting unit Slow in its response to acceleration d As we know that these turbines are used in jet propulsion engine where acceleration is an important parameter. Comments Active
5 Francis turbine is a tangential flow reaction turbine axial flow reaction turbine radial flow reaction turbine mixed flow reaction turbine d In a Francis turbine, runner blades are divided into 2 parts. The lower half is made in the shape of a small bucket so that it uses the impulse action of water to rotate the turbine. The upper part of the blades uses the reaction force of water flowing through it. These two forces together make the runner rotates. It is an inward – flow reaction turbine that combines radial and axial flow concepts. It is an inward flow reaction turbine that combines radial and axial flow concepts. Comments Active
6 De-Laval turbine is pressure compounded impulse turbine simple single wheel impulse turbine velocity compounded impulse turbine simple single wheel reaction turbine b A simple impulse turbine is also called de – laval turbine since the expansion of the steam takes place in one set of nozzles. The expansion of steam from its initial pressure (steam chest pressure) to final pressure (condenser pressure) takes place in one set of nozzles. Due to high drop in pressure in the steam in the velocity of steam in the nozzle increases. The pressure of the steam when it moves over the blades remains constant but the velocity decreases. Comments Active
7 If maximum surface temperature of sea is 30 oC and temperature in depth is 4 oC, how much can be the maximum thermal efficiency of Ocean Thermal Conversion (OTEC) system ? 8.58% 13.3% 86.7% none of the above a \( T_{H}=30℃=303K\)
\(T_{L}=4℃=277K\)
Efficiency
\(η=(1-\frac{T_{L}}{T_{H}})100%\)
\(=(1-\frac{277}{303})100%\)
\(=8.58%\)		 Comments Active
8 The compression ratio for a practical diesel engine usually lies in the range. 3 – 5 6 – 8 10 – 15 16 – 22 d For petrol – (6 – 8)
For diesel – (16 – 22)		 Comments Active
9 Cavitation depends upon vapour pressure which is function of temperature absolute pressure or barometric pressure suction pressure (Hs) which is height of runner outlet above tail race level. all the above d Cavitation, formation of vapour bubbles within a liquid at low – pressure regions that occur in places where the liquid has been accelerated to high velocities, as in the operation of centrifugal pumps water turbines and marine propellers.
Cavitation is a phenomenon in which the static pressure of a liquid reduces to below the liquid’s vapour pressure, leading to the formation of small vapour – filled cavities in the liquid.		 Comments Active
10 Why intercooling in multistage compressors is done ? To minimize the work of compression To cool the air delivery To cool the air during compression None of these a An increase in pressure ratio in a single stage reciprocating compressor causes an increase in temperature a decrease in volumetric efficiency and an increase in work input.
So for the same higher pressure ratio, multi – stage compression is efficient.
Intercooling
* In multi – stage compression with intercooling, where the gas is compressed in stages and cooled between each stage by passing it through a heat exchanger called an intercooler.
* P – V and T – S diagram of the compression with intercooling is shown in the figure below P – V and T – S diagrams for a two stage steady flow compression process.
* It can be seen that interceding is done at constant pressure and is represented by a horizontal line on the P – V diagram.
* This reduction in temperature means a reduction in internal energy at the delivered air and since this energy must have come from the input energy required to drive the machine, this results in a decrease in input work requirement for a given mass of delivered air.
* The low pressure ratio in a cylinder improves volumetric efficiency.		 Comments Active
11 Bleeding in turbine means : leakage of steam steam doing no useful work removal of condensed steam extracting steam for preheating feed water d Bleed is amount of steam output from turbine through pipe and exit from final state of turbine. This bleed enter to feed water heater (low and high) and dearotor to increase unit efficiency. Comments Active
12 For a single stage impulse turbine having nozzle angle , maximum blade efficiency under ideal conditions is given by cos /2 cos2/2 cos  cos2 d \(V_{ω_{1}}+V_{ω_{2}}=AC+AF=FC=FB+BC\)
\(=Vr_{2}cosβ_{2}+Vr_{1}cosβ_{1}\)
\(=V_{r_{1}cos}β_{1}(1+\frac{Vr_{2}cosβ_{2}}{Vr_{1}cosβ_{1}})\)
\(=Vr_{1}cosβ_{1}[1+K_{B}\)]
\(Vω_{1}-u[1+K_{B}\)]
\(=V_{1}cosα_{1}-u\)[1+K_{β}]]
Now blade efficiency
\(η_{b}=\frac{2(Vω_{1}+Vω_{2})u}{V12}\)
Speed ratio
\(ρ=\frac{u}{v}\)
\(η_{β}=\frac{(V_{1}cosα_{1}-u)(1+K_{β})u}{V12}\)
\(=2[cosα_{1}-ρ\)(1+K_{β})ρ]
\(=2[ρcosα_{1}-ρ^{2}\)[1+K_{β}]]
Now blade efficiency
\(η_{b}=\frac{2(Vω_{1}+Vω_{2})u}{V12}\)
Speed ratio
\(ρ=\frac{u}{V_{1}}\)
\(η_{β}=\frac{(V_{1}cosα_{1}-u)[1+K_{β}\)u}{V12}]
\(=2[cosα_{1}-ρ\)(1+K_{β})ρ]
\(=2[ρcosα_{1}-ρ^{2}\)[1+K_{β}]]
For maximum efficiency
\(\frac{dη_{β}}{dρ}=0⇒2[cosα_{1}-2ρ\)(1+K_{β})]
\(∴ρ= \frac{cosα_{1}}{2}\)
And \(η_{bmax}=2(\frac{cosα_{1}}{2}.cosα_{1}-\frac{cosα12}{4})(1+K_{β})\)
\(=\frac{2cos^{2}α_{1}}{4}.(1+1)\)
\(=cos^{2}α_{1}\)
Take \(K_{β}=1\)
(blade friction factor)		 Comments Active
13 Specific speed of a turbine is defined as the speed of the turbine which produces unit power at unit discharge produces unit power at unit head delivers unit discharge at unit head delivers unit discharge at unit power b The specific speed value of a turbine is the speed of a geometrically similar turbine which would produce unit power (one kW) under unit head (one meter).
\(N_{s}=\frac{NP}{H^{5/4}}\)		 Comments Active
14 The degree of reaction for a turbomachinary in which heat drop in moving blades is 8 kJ/kg and in fixed blades 12 kJ/kg, would be 66.6% 150% 40% 166.6% c Degree of reaction
\(R=\frac{Enthalpy drop in moving blade}{Total enthalpy drop}\)
\(=(\frac{8}{12+8})100=40%\)		 Comments Active
15 Piston rings are generally made of following material. cast iron mild steel aluminium carbon steel a Due to high compressive strength and self-lubricating property, CI is best suited material. Comments Active
16 In parson’s steam turbine, steam expands in nozzles only blades only partly in nozzles and partly in blades none of the above c The Parson’s turbine is composed of moving blades (nozzles) alternating with fixed nozzles. In the reaction turbine, the steam is expanded in fixed nozzles and also in the moving nozzles. In other words, the steam is continuously expanding as it flows over the blades. There is pressure an velocity loss in the moving blades. The moving blades have a converging steam nozzle. Hence when the steam passes over the fixed blades, it expands with decrease in steam pressure and increases in kinetic energy. Comments Active
17 for high boiler efficiency the feed water is heated by recuperator convective heater super heater economizer d It improves efficiency by preheating economizers are heat exchanger that transfer the heat in the flue gas to another medium, generally to the boiler feed water although other streams are some times used such as make – up water. It is mechanical devices intended to reduce energy consumption, or to perform useful function such as preheating a flui(d) It is an accessories. Comments Active
18 Work ratio is a guide in the determination of the size of the gas turbine overall efficiency of the turbine mechanical efficiency of the turbine compressor efficiency a Work ratio is just the fraction or percentage of the work that the turbine produces and that is consumed by the pump. Work ratio is defined as the ratio of the network to the turbine work developed (W). It is denoted by WR \((W_{t}-W_{c})\)
\(WR=\frac{Net work}{Turbine work}=\frac{W_{t}-W_{c}}{W_{t}}\)
\(=-\frac{W_{c}}{W_{t}}=1-\frac{C_{P}(T_{2}-T_{1})}{C_{P}(T_{3}-T_{4})}\)
\(=1-\frac{T_{1}(\frac{T_{2}}{T_{2}}-1)}{T_{3}(1-\frac{T_{2}}{T_{3}})}\)
But \(\frac{T_{2}}{T_{1}}=rP\frac{γ-1}{γ},\frac{T_{3}}{T_{4}}=rP\frac{γ-1}{γ}\)
Substituting these ratios in WR
\(=1-\frac{T_{1}(rP\frac{γ-1}{γ}-1)}{T_{3}(1-\frac{1}{rP\frac{γ-1}{γ}}) }\)
\(=1-\frac{T_{1}}{T_{3}}[\frac{rP\frac{γ-1}{γ}(rP\frac{γ-1}{γ}-1)}{(rP\frac{γ-1}{γ}-1)}\)
it decides size of turbine. \(WR=1-\frac{T_{1}}{T_{3}}×rP\frac{γ-1}{γ}\)
From this reaction we note that for the given temperatures and , WR is maximum when is minimum. \(T_{1}\) \(T_{3}\) \(rP\frac{γ-1}{γ}\)
Similarly for the given pressure ratio again WR is maximum when is minimum, i.e., si minimum and is maximum. Maximum is according to metallurgical consideration and varies between 800℃ to 1000℃. \(r_{P},\) \(\frac{T_{1}}{T_{3}}\) \(T_{1}\) \(T_{3}\) \(T_{3}\)
Similarly will be depending on the atmospheric conditions and may be taken as varying between 15℃ and 30℃. \(T_{1}\)		 Comments Active
19 When an impeller has backward curved vanes in a centrifugal blower, then with an increase in flow rate, Euler head H increases decreases remains constant none of the above b With a backward curved impeller the air exits in a radial direction whereas with a forward curved the air exits tangentially from the circumference of the fan.
Euler’s head is difference of product of whirl velocity and peripheral velocity of outlet and inlet.		 Comments Active
20 In centrifugal blowers/compressors, the ratio of outlet whirl velocity to the blade velocity is known as work factor slip factor degree of reaction pressure coefficient b Slip is the internal back flow of fluid in a pump returning from the discharge side to the suction side. In a centrifugal pump, slip is the loss of efficiency from a theoretical ideal arising from design constraints and turbulence at the impeller vanes. Slip is the loss of pumping capacity due to fluid leaking back of through a pump from the discharge side to the inlet side. It can affect the efficiency of all types of pumps although the actual causes may be very different.
Slip factor is defined as the ratio of real tangential velocity (outlet whirl velocity) out of the rotor to its ideal valve (Blade velocity).
In turbo machinery degree of reaction or reaction ratio is defines as the ratio of the static pressure drop in the rotor to the static pressure drop in the stage or as the ratio of static enthalpy drop in the rotor to the static enthalpy drop in the stage. Pressure coefficient is a dimensionless number. It is defined as the ratio of static pressure to the dynamic pressure. It describes the relative pressures throughout a flow field in fluid dynamics.
\(Work factor=\frac{Net work}{Turbin work}\)		 Comments Active
21 Work done by prime mover to run the compressor is minimum if the compression is isothermal adiabatic isentropic polytropic a For compression area under isothermal is minimum, so work done will be minimum for this case maximum for isentropic Comments Active
22 William’s law gives a straight line graph between the rate of steam consumption and pressure of steam temperature of steam volume of steam indicated horse power d Willian’s line method:
* It is method estimating the friction power of compression ignition (CI) engines.
* Willian’s lines for the steam engine is a straight line relationship between the fuel consumption per hour (on vertical axis) and brake power or indicated power (On horizontal axis)
* Here, it is asked for steam engine, therefore replace fuel by steam because steam fulfills the same purpose as fuel in CI engines.		 Comments Active
23 In an internal combustion engine, firing order depends upon crank shaft design arrangement of cylinder number of cylinders none of the above d Should be all of the above.
The order in which the ignition take place in various cylinder of a multi – cylinder engine is called firing order. Every engine cylinder must fire once in every cycle, Three factors must be considered before deciding the firing order in IC engine.
These are
- Arrangement of the cylinder
- Design of crankshaft
- Number of cylinder		 Comments Active
24 The specific fuel consumption of a diesel engine as compared to that for petrol engine is lower higher same for same output none of the above a Diesel engines typically use 20% - 25% less fuel than their petrol equivalents.
A litre of diesel fuels contains roughly 15% more energy than a litre of petrol.		 Comments Active
25 Iso-octane content in a fuel for S.I. Engines retards auto-ignition accelerates auto-ignition does not affect auto ignition none of the above a Octane Rating: The SI engine fuel has been given an anti – knocking rating which is known as octane rating. It basically shows the resistance of the fuel to create knocking by auto – ignition. According to standard practice, the anti knock value of an SI engine fuel is determined by comparing its anti knock property with a mixture of two reference fuels, normal heptane and iso octane . \((C_{7}H_{18})\) \((C_{8}H_{18})\)
Iso octane chemically being very good anti-knock fuel, is arbitrarily assigned a rating of“100 cetane number” normal heptane has very poor antiknock qualities and is given a rating of “0” octane number. \((C_{8}H_{18})\) \( \) \((C_{7}H_{16})\)
Thus, if the content of ISO Octane is high in the fuel then it will ensure high octane number resulting in better fuel anti –knock quality. Because of the high octane number, the fuel will not be auto – ignite as the possibility of knocking has been diminished by the fuel quality.
Hence iso – octane content in fuel for SI engine retards auto – ignition.		 Comments Active
26 The specific speed of a turbine is expressed as N \(NP/H^{\frac{5}{4}}\) \(NH^{\frac{5}{4}}/P\) \(NP/ρH^{\frac{5}{4}}\) \(NH/(gH)^{\frac{5}{4}}\) a The specific speed valve for a turbine is the speed of a geometrically similar. Turbine which would produce unit power (One kilowatt) under unit head (one meter)
\(N_{s}=\frac{NP}{H^{5/4}}\)		 Comments Active
27 Open type impeller centrifugal pump is used to handle water mixture of water, sand, pebbles and clay sewage liquids lighter than water b Open impeller consists of nothing but vanes. Vanes are attached to the central hu(b)
Without any form or side wall or shroud, open impeller are usually used in small, inexpensive pumps or pumps handling abrasive liquids or suspended solids.(In which the impeller rotates between two side plates, between the casing walls of the volute or between the stuffing box head and the suction heads)		 Comments Active
28 Mean diameter of runner of a pelton turbine is 200mm and least diameter of jet is 1 cm. Calculate the jet ratio and number of buckets. 20, 25 200, 115 20, 40 20, 45 a \( Jet ratio=\frac{Runner diameter}{Jet diameter}=\frac{200 mm}{1cm}\)
\(=\frac{200}{10}=20\)
\(No of bucket=\frac{Jet ratio}{2}+15\)
\(-\frac{20}{2}+15=25 (No option is correct)\)		 Comments Active
29 For a reaction turbine, specific flow is given by following expression : Q/D12H Q/D1 \(H\) Q/D1H3/2 Q/D1 \(H\) d Special flow \(=\frac{Q}{D12H}\) Comments Active
30 Which of the following is not a high pressure boiler ? Lancashire boiler La-mont boiler Benson boiler Loeffler boiler a Lancashire boiler is a low pressure boiler. Comments Active
31 The process of supercharging is meant for raising exhaust pressure increasing density of intake air increasing quantity of fuel going into cylinder providing air for cooling b Supercharging is the process of supplying the intake of an engine with air at a density greater than the density of surrounding atmosphere so that each intake cycle of engine gets more oxygen, which turns more fuel thus generating more power. Comments Active
32 A centrifugal pump lifts water through a height h and delivers it at a velocity V(d) The loss of heat through piping is hf. The gross lift is h + hf hf + \(\frac{vd2}{2g}\) h + hf + \(\frac{vd2}{2g}\) h + \(\frac{vd2}{2g}\) c Manometric head/gross lift is defined as the head against which a centrifugal pump has to work.
It can be expressed as:
\(H_{m}=Suction heat (H_{s})+delivery heat (hd)+\)
\(friction head loss in teh suction pipe (h_{fs})+\)
\(friction head loss in the delivery pipe (h_{fd})+\)
\(velocity head of water in the delivery pipe (v^{2}/2g)\)
Gross left = \(h+h_{f}+vd^{2}/2g\)		 Comments Active
33 The ratio of brake power to indicated power of an I.C. engine is called mechanical efficiency thermal efficiency volumetric efficiency relative efficiency a \( η_{mech}=\frac{BP}{IP}=\frac{BP}{BP+IP}\) Comments Active
34 In a 4 – cylinder petrol engine, the standard firing order is 1 – 2 – 3 – 4 1 – 4 – 3 – 2 1 – 3 – 2 – 4 1 – 3 – 4 – 2 d 1 – 3 – 4 – 2 Comments Active
35 The ignition quality of fuels for S.I. engines is determined by Cetane number Octane number Calorific value Volatility of the fuel b In SI engine fuel quality is checked by octane no – while in CI engine it is cetane no. Comments Active
36 The knocking tendency in I.C engines increases with decrease of compression ratio increase of compression ratio increasing the temperature of inlet air increasing cooling water temperature a In CI engine, the knocking tendency increases with a decrease in compression ratio. If the compression ratio is low in CI engine, there will not be enough temperature generation, For ignition of the fuel air mixture, so fuel will get accumulated in the chamber (ignition delay). Comments Active
37 In a variable speed S.I. engine, the maximum torque occurs at the maximum speed brake power indicated power volumetric efficiency b \(\) \(BP=IP-FP\)
The torque developed by an engine is directly proportional to the indicated power or (B)P. Thus maximum torque will occur corresponding to maximum power.
\(IP=P_{m}LAN_{n}\)		 Comments Active
38 The cetane number of automotive diesel fuel used in India lies in which of the following ranges ? 30 – 40 41 – 50 51 – 60 61 – 70 b Cetane number (cetane rating) is an indicator of the combustion speed of diesel fuel and compression needed for ignition. Comments Active
39 A water turbine is usually designed for the given values of N, T and Q P, T and Q P, H and Q P, H and N d Special speed of turbine is used to design it. (also other parameter are used like position)
\(N_{s})_{turbine}=\frac{NP}{H^{5/4}}\)
\(P-Power (h_{P})\)
\(P-Head (m)\)		 Comments Active
40 Performance of an air compressor at high altitudes as compared to that at sea level is better inferior same depends on type of compressor b With the increase in altitude there will be reduction in air pressure, which results in increase in compression ratio, leading to higher discharge temperature and reduced efficiency. As, altitude increases, the atmospheric pressure decreases. Atmospheric pressure is caused by the weight of all the air molecules above you pressing down and compressing the air around you. At higher altitudes, lower atmospheric pressure means that the air molecules are less tightly packed together and have lower density. When an air compressor draws in air as part of its intake process. It draws a fixed volume of air. If the air density is lower, fewer air molecules are drawn in with this air into the compressor. This results in a smaller volume of compressed air and less air is delivered to the receiver tank and tools during each compression cycle. Comments Active
41 In a one ton capacity water cooler, water enters at 30 oC at the rate of 200 lit/hour. Taking specific heat of water as 4.16 kJ/kg K, the outlet temperature of water will be 3.5 oC 6.3 oC 23.7 oC 15 oC d Cooling capacity = 1 ton = 3.5 kJ/s
\(=3.5×3600 KJ/hr\)
Let t is the outlet temperature of water
\(Mass of water flow=Density ×Volume\)
\(=1000kg/m^{3}×200×10^{-3}m^{3}/hr\)
\(200 kg/hr\)
Applying energy balance
\(3.5×3600=4.18×200(30-t)\)
\(∴t=15℃\)		 Comments Active
42 Dew point is the temperature at which the condensation begins when the air is cooled at constant volume entropy pressure enthalpy c Dew point temperature (DPT). Water vapour in air exists in the superheated state (1) and the air is highly unsaturated When it gets coot at constant pressure it meets the saturation vapour curve where the moisture of condensation beings. The temperature at this point is known as the dew point temperature.
States of water vapour in mixture		 Comments Active
43 Which one of the following is not a desirable property of a good refrigerant ? low specific heat high specific volume of vapour large latent heat at evaporator pressure high critical temperature b Good refrigerant posses
* Low specific heat
* Low specific volume
* High latent heat
* High critical temperature
* Low viscosity etc		 Comments Active
44 Finned evaporators are used on air conditioning application to equalize air flow over the cooling coil surface prevent moisture carry over extend the effective area of the cooling surface increase the dehumidifying capacity c The finned evaporator are the bare tube type of evaporator covered with the fins. When the fluid (air or water) to be chilled flows over the bare tube evaporator lots of cooling effect from the refrigerant goes wasted since there is less surface for the transfer of heat from the fluid to the refrigerant. The fluids tends to move between the open spaces of the tubing and does not come in contact with the surface of the coil thus the bare tube evaporators are less effective. The fins on the external surface of the bare tube evaporator increases the contact surface of the metallic tubing with the fluid and increase the heat transfer rate, thus the finned evaporators are more effective than the bare tube evaporators. They help removing the heat from the fluid that otherwise would not have come in contact with the coil. Comments Active
45 Which refrigerant would you choose for 800 TR air conditioning plant using centrifugal compressor ? NH3 CO2 CFC 11 CFC 114 a Big industrial plant uses ammonia as its refrigerant. Comments Active
46 An ideal refrigerator is operating between a condenser temperature of 37 oC and an evaporator temperature of – 3 oC If the machine is functioning as a heat pump, its COP will be 6 6.75 7 7.75 d \( COP)_{ref ideal}=\frac{T_{L}}{T_{H}-T_{L}}=\frac{270}{310-270}\)
\(=\frac{270}{40}=6.75\)
\(COP)_{HP}=1+COP)_{xy}=7.75\)
Or
Directly \(COP)_{HP}=\frac{T_{H}}{T_{H}-T_{L}}\)
\(=\frac{310}{310-270}=7.75\)		 Comments Active
47 Two reversible refrigerators are arranged in series and their COP are 4 and 5 respectively.
The COP of the composite refrigeration system would be		 1.5 2 3 4.5 b Composite COP (in series)
\(=\frac{COP_{1}×COP_{2}}{1+COP_{1}+COP_{2}}\)
\(=\frac{4×5}{1+4×5}=\frac{20}{10}=2\)		 Comments Active
48 The wet bulb depression is zero when relative humidity equals zero 50% 75% 100% d WBD = DBT – WBT
At saturated condition
WBT = DBT = DPT, RH = 100%
WBD = 0 \(∴\)		 Comments Active
49 The brine is an aqueous solution of in water. Calcium chloride Sodium chloride Calcium carbonate Sodium carbonate a Water + Nacl = Brine(aq.) Comments Active
50 The refrigerant commonly used for commercial ice plants is Freon – 12 NH3 CO2 Air b F – 12(R – 12) – Refrigerator domestic
Ammonia is used as refrigerant prominently in the refrigeration system of the food industry like dairies, ice – creams plants, frozen food production plants, cold storage ware houses, processors of fish, poultry and meat and a number of other applications
It have
1. Highest refrigerant capacity per kg
2. Excellent thermal property		 Comments Active
51 When air is at saturated state, which pair, out of the given alternatives is not correctly matched ?
List – I List – II		 Relative humidity-100% DBT-WBT Degree of saturation-1 Specific humidity- 0.01 kg w.v/kgda d Degree of saturation is the ratio of the humidity ratio of moist air to the humidity ratio of saturated moist air at the same temperature and pressure.
So, at saturated condition, humidity ratio of moist air = to the humidity ratio of saturated moist air.
So degree of saturation = 1
At saturated condition
WBT = DBT and RH = 100%		 Comments Active
52 In case of air conditioning of auditoriums, the cooling load which is predominant is lighting load occupancy load load due to fans load due to electronic equipment’s b Heat load in auditorium
- Occupancy (Major load)
- Electronic apparatus
- Electrical components
- Lightening load		 Comments Active
53 In cooling and dehumidifying apparatus, the effect of the by pass factor is to lower the ADP of the cooling coil decrease the COP of the system both (a) & (b) above increase the ADP of the cooling coil and to improve the COP of the system c By pass factor tells about the air which does not come into contact with the surface of cooling coil. It is a kind of loss in COP and ADP of coil will also is shown lower than actual. Comments Active
54 The function of solenoid valve in a refrigeration system is to control the flow of refrigerant in suction line control the flow of refrigerant through expansion valve stop the flow of refrigerant when there is no load on the evaporator stop the flow of refrigerant in liquid line when compressor stops d Compressor
Refrigerant
On
Flow on
Off
Flow off
A solenoid valve is an electromechanical valve frequently used to control the flow of liquid or gas. Solenoid valves are found in many applications and are commonly used in refrigeration and air conditioning systems. Their function is simply to turn refrigerant flow on and off.		 Comments Active
55 Go through the following statements and choose the correct alternative :
1. Wet compression increases COP of ammonia vapour compression system.
2. Use of liquid-vapour heat exchanger in vapour compression system decreases COP in case of ammonia refrigerant.
3. For good performance, a refrigerant must have high critical pressure and low critical temperature.
4. Refrigerants that are not miscible with oils, presents many problems.
5. In flooded evaporators, the liquid refrigerant covers the entire heat transfer surface.		 Statements 1, 2 and 5 are true Statements 1, 2 and 3 are true Statements 3, 4 and 5 are true Statements 2, 4 and 5 are true a Comments Active
56 Due to suction vapour superheating in vapour compression cycle the COP increases in case of the following refrigerant : R 22 NH3 R – 12 None of the above c In case of
\(NH_{3}-COP will decrease\)
\(R-2^{2}-COP will decrease\)
\(R-12-COP will increase\)
As RE and Wc (comp. work) both will increase.		 Comments Active
57 In case of ejector–compression system the power input is in the form of electric power heat mechanical work steam power d Comments Active
58 In a vapour absorption refrigeration system, the refrigeration temperature is – 15 oC and the generator temperature is 110 o(C) If sink temperature is 55 oC, the maximum COP of the system will be 1.00 3.69 0.34 0.90 * Maximum possible COP of VARs.
Is given by
\(COP_{ideal}=\frac{Q_{e}}{Q_{g}}=\frac{Heat absorbed by evaporator}{Heat given to generator}\)
\(=(\frac{T_{e}}{T_{o}-T_{e}}) (\frac{T_{g}-T_{o}}{T_{g}})\)
\(=COP_{carnot}.η_{carnot}\)
\(T_{e}=-15℃ evaporator temperature \)
\(T_{g}=110℃ Generator temperature \)
\(T_{o}=55℃ Sink temperature \)
\(T_{e}=273+(-15)=258K\)
\(T_{o}=273+55=328K\)
\(T_{g}=273+110=383K\)
\(=(\frac{258}{328-258})(\frac{383-328}{383})\)
\(=3.68×0.143\)
\(=0.53\)		 Comments Active
59 Pressure drop in capillary tube (used in refrigeration systems) takes place due to friction change in momentum both (a) & (b) above none of the above c The pressure reduction in a capillary tube occurs due to the following two factors
1. The refrigerant has to overcome the frictional resistance offered by tube walls. This lead to same pressure drop.
2. The liquid refrigerant fleashes (evaporates) into mixture of liquid and vapour as its pressure reduces. The density of vapor is less than that of the liquid Hence, the average density of refrigerant decreases as it flows in the tube. The mass flow rate and tube diameter (hence area) being constant, the velocity of refrigerant increases since = ρV(A) The increase in velocity or acceleration of the refrigerant also requires pressure drop.
Change in velocity means change in momentum.		 Comments Active
60 Which pair, out of following alternatives, is correctly matched. Normal boiling points of different refrigerants (List-I) are given in List-II.
List – IList – II		 R – 12- – 29.8 \(℃\) NH3- – 33.35 \(℃\) R 134 a- – 24.15 \(℃\) R 22- – 40.8 \(℃\) c Normal boiling point of
R – 134a is – 26.3ᵒC
Tetra fluoroethane - \(CH_{2}FeF_{3}\)
Other are wrong.		 Comments Active
61 The design condition of temperature for winter air conditioning is 25 ± 1 oC 27 oC 21 oC none of the above c A standard air conditioner’s temperature should be set to 20 to 23 degrees for cooling in summer, and around 18 to 22 degrees for heating in winter.
This is to ensure maximum energy efficiency from your unit.		 Comments Active
62 What is the storage temperature for milk ? 4 oC 7 oC 2 oC 0.5 oC c Milk storage
Temperature = 2áµ’C (approx.)		 Comments Active
63 In the absorption refrigeration cycle, the compressor of vapour compression refrigeration cycle is replaced by Liquid pump Generator Absorber and generator Absorber, liquid pump and generator d Vapour absorption cycles Comments Active
64 The curved lines on a psychometric chart indicate relative humidity specific humidity dry bulb temperature wet bulb temperature a Lines of constant relative humidity are represented by the curved lines. Comments Active
65 If moist air is passed over chemicals like silica gel, the process which takes place is humidification dehumidification cooling and dehumidification heating and dehumidification b When moist air passes through a bed of silica gel chemical dehumidification process occurs. It increases dry bulb temperature of air due to latent heat of vapour. Comments Active
66 In mechanical refrigeration system, the refrigerant has the maximum temperature before expansion valve between compressor and condenser between condenser and evaporator between compressor and evaporator b The exit condition after evaporation is saturated vapour. Which is compressed in the compressor and become superheated (having the maximum temperature) and then passed to the condenser. Comments Active
67 Vapour absorption system gives noisy operation gives quiet operation requires little power consumption cools below 0 oC b Due to absence of compressor it is silent. Comments Active
68 A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ / kg ; enthalpy after throttling = 125 kJ/kg ; enthalpy before compression = 375 kJ / kg. The COP of refrigeration is 5 3.5 6 not possible to find with this data a \(COP=\frac{RE}{W_{in}}=\frac{h_{1}-h_{3}}{h_{2}-h_{1}}=\frac{375-125}{425-375}=5\) Comments Active
69 Lithium bromide in vapour absorption refrigeration system is used as refrigerant cooling substance auxiliary refrigerant absorbent d In the vapour absorption system, the water is used as the refrigerant while lithium bromide (Li Br) is used as the absorbent. Comments Active
70 Dry ice is solidified carbon dioxide ice free from dissolved air and gases ice free from impurities ice made from transparent distilled water. a Dry ice is the solid form of carbon dioxide. It is used primarily as a cooling agent, but it also used in tag machines at theatres for dramatic effects. Comments Active
71 The chemical formula of Freon – 12 is CClF2 CCl2F3 CCl2F2 CClF c CCl₂F₂ Comments Active
72 The throttling operation in a refrigeration cycle is carried out in Evaporator Discharge valve Capillary tube Expansion valve d Throttling device is another vital part of all the refrigeration systems and air conditioning systems apart from the compressor, condenser and the evaporator.
The throttling device are also called as the expansion valves because when the refrigerant passes through them the pressure of the refrigerant drops down or it expands.
The refrigeration leaving the condenser is at high pressure. The pressure of the refrigerant has to be reduced so that it can vaporize at the required temperature in the evaporator.		 Comments Active
73 The Bell-Coleman refrigeration cycle uses as the working fluid. Hydrogen Carbon dioxide Air Any inert gas c The working fluid of the Bell Coleman refrigeration cycle is Air.  Comments Active
74 A refrigerator working on a reversed Carnot cycle has a COP of 4. If it works as heat pump and consumes 1kW, the heating effect will be 1kW 4kW 5kW 6kW c Reversed carnot cycle = Refrigerator
\(∴COP_{carnot}=COP_{xy}\)
\(∴COP_{HP}=1+COP_{xy}\)
\(∴COP_{HP}=1+4=5\)
\(=\frac{Desired effect}{Work supply}=\frac{Q_{2}}{W}\)
\(5=\frac{Q_{2}}{1}\)
\(Q_{2}=5 kW\)
\(T_{H}>T_{L}\)		 Comments Active
75 On a psychrometric chart what does a vertical downward line represent? Adiabatic saturation Sensible cooling De humidification Humidification c Comments Active
76 Effects of heat pump and refrigeration respectively are obtain at Compressor and condenser evaporator and condenser condenser and evaporator compressor and evaporator c For heat pump
- Desired effect in condenser \(Q_{2}\)
\(- Q_{2}=heat rejected in a space\)
For refrigerator
\(-Desired effect Q_{1} in evaporator\)
\(-Q_{1}=heat absorbed from a body\)
\(In condensser-heat rejection occur\)
\(In evaporator-heat absorbed occur.\)		 Comments Active
77 Temperature recorded by a thermometer which is not affected by moisture is dry bulb temperature wet bulb temperature dew point temperature adiabatic saturation temperature a DBT – measured by thermometer
WBT – measured by wrapping a wet cotton around bulb
Adiabatic saturation temperature
- temperature at which water, by evaporating into air, bring the air to saturation at the same temperature adiabatically (WBT)
- DPT – condensation starts here		 Comments Active
78 In summer air conditioning system fresh air is introduced into the recirculated air to reduce load on equipment exercise easy control over equipment improve air quality by diluting odour and contaminants reduce quantity of supply air c Fresh air improve air quality reduce odour reduce pollutant etc., (which are present in room air) by mixing with room air. Comments Active
79 In a refrigeration system the refrigerant gains heat at Compressor Condenser Expansion valve Evaporator d Comments Active
80 For practical purposes one Ton of refrigeration means 3.48 kW 34.8 kW 348 kW None of these a 1 Tonne = 210 KJ/min
= 3.48 kW		 Comments Active
81 A 0.5 m thick plane wall has its two surfaces kept at 300 oC and 200 oC. Thermal conductivity of the wall varies linearly with temperature and its value at 300 oC and 200 oC are 25 W/mK and 15 W/mK respectively. Then steady heat flux through the wall is 8 kW/m2 5 kW/m2 4 kW/m2 3 kW/m2 c Thermal conductivities are varying linearly with temperature hence
\(K_{avg}=\frac{25+15}{2}=20 W/mk.\)
Now heat flux
\(q=-K_{avg}\frac{∆T}{L}=\frac{Q}{A}\)
\(=-\frac{20(200-300)}{0.5}\)
\(=40×100=4000 W/m^{2}\)
\(=4 kW/m^{2}\)		 Comments Active
82 In a certain heat exchanger, both the fluids have identical mass flow rate and specific heat product. The hot fluid enters at 76 oC and leaves at 47 oC and cold fluid enters at 26 oC and leaves at 55 oC. The effectiveness of the heat exchanger is 0.16 0.58 0.72 1.0 b \( Effectiveness ϵ=\frac{Actual heat transfer}{Max possible heat transfer}\)
\(ϵ=\frac{(mC)_{n}(T_{n1}-T_{n2})}{(m)C_{min} (T_{ni}-T_{ci})}\)
\(Here m_{n}C_{n}=C_{c}:M_{c}\)
\(=(76-47)/(76-26)\)
\(=\frac{29}{50}=0.58\)		 Comments Active
83 The transition Reynolds number for flow over a flat plate is 5 x 105. What is the distance from the leading edge at which transition will occur for flow of water with uniform velocity of 1 m/s ? (For water = 0.858 × 106 m2/s) 1 m 0.43 m 43 m 103 m b Reynold’s number
\(R_{e}=\frac{ρVL}{μ}=\frac{VL}{ν}\)
\(5×10^{5}=\frac{U_{∞}L}{ν}=\frac{1×L}{0.86×10^{-6}}\)
\(L=0.43 m\)		 Comments Active
84 Heat transfer in liquids and gases is essentially due to Conduction Convection Radiation Conduction and Radiation b Heat transfer in fluid and gases is due to mainly bulk motion of molecule i.e. convection. Comments Active
85 In a long cylindrical rod of radius R and for a surface heat flux of 90, the uniform internal heat generation rate is 290/R 290 90/2 R 90/R2 * Heat generation
\(q_{g}=\frac{2q_{0}}{R}=\frac{2×90}{R}=\frac{180}{R}\)		 Comments Active
86 The time constant of the thermocouple is the time required by a thermocouple to reach the following value of initial temperature differences : 63.2% 65% 68% 70.2% a Thermocouple constant
\(\frac{T-T_{1}}{T_{2}-T_{1}}=1-e^{-(\frac{t}{Ï‚})}\)
\(t=elapsed time\)
\(T=Thermistor temperature\)
\(Ï‚=Heat dissipation coefficient\)
and \(T_{1}\) \(T_{2}=temp of thermistor \)
If t = Ï‚
Then
\(\frac{T-T_{1}}{T_{2}-T_{1}}=1-e^{-1}=1-0.368\)
\(=0.632 or 63.2%\)		 Comments Active
87 A thermal transparent body is characterised by absorptivity = 1 reflectivity =1 absorptivity = reflectivity =0 none of the above c For transparent body
Transmissivity \(Ï‚=1\)
Reflectivity \(ρ=0\)
\(Absorptivity α=0\)		 Comments Active
88 Heat transfer rate will be higher in turbulent flow will be lower in turbulent flow will depend only on the fluid will depend only on viscosity a Forced convection produces turbulent flow and gives higher heat transfer. Comments Active
89 The shape factor for radiation heat transfer of a long cylinder of radius r1 enclosed by another concentric long cylinder of radius r2 is 0.25 0.50 0.75 1.0 d \(F_{1-2}=1\)
Whole radiation of cyl – 1 will reach to cyl 2.
\(F_{2-1}=0.5\)
Only half radiation cyl – 2 will reach to cyl - 1		 Comments Active
90 Stefan-Boltzmann law is expressed as Q = σ AT4 Q = σ A2T4 Q = σ AT2 Q = AT4 a Stephan Boltzmann low
\(Q∝T^{4}\)
Heat transfer per unit area (W/m2) absolute temperature \(∝\)
\(\frac{Q}{A}=σT^{4}\)
\(σ=Stefan Boltzman constant \)
\(=5.67×10^{-8} W/m^{2}K^{4}\)
\(Q=σAT^{4}\)		 Comments Active
91 In forced convection, Nusselt number is a function of Reynolds number and dynamic viscosity dynamic viscosity and Prandtl numb Prandtl number and Reynolds number Reynolds number and thickness of boundary layer c \(Nu=\frac{(Heat transfer)convection }{(Heat transfer)conduction}\)
For forced convection
\(Nu=f(Re, Pr)\)
For free convection
\(Nu=f(Gr,Pr )\)		 Comments Active
92 Metals are good thermal conductors since they have free electrons. their atoms are relatively closer. their surfaces reflect. their atoms are of larger size. a Metals are good conductor because of
- Pressure of free \((e^{-})\)		 Comments Active
93 The radial heat transfer rate through hollow cylinder increases as the ratio of outer radius to inner radius : decreases increases constant none of the above a \(R^{th}=\frac{ln(\frac{r_{2}}{r_{1}})}{2Ï€KL}\)
\(Q=\frac{T_{1}-T_{2}}{ln(\frac{r_{2}}{R_{1}})}.2Ï€KL\)
for cylinder \(∴Q∝\frac{1}{ln(\frac{r_{2}}{r_{1}})}\)
for shpere \(Q=\frac{T_{1}-T_{2}}{\frac{r_{1}-R_{2}}{4Ï€Kr_{1}r_{2}}} \)
\(Q=\frac{∆T}{ln(\frac{r_{2}}{r_{1}})/2πKL}\)
So, as decreases, Q increases \(\frac{r_{2}}{r_{1}}\)		 Comments Active
94 Two walls of thickness d and 2d called A and B are made of materials such that their thermal conductivities are KA = 2 K(B) If the difference of temperature on two sides is proportional to thickness, the ratio of heat transfer through A to that through B is 6 4 2 1 c \( K_{A}=2K_{B}, ∆T_{A}=2∆T_{B}\)
\(∆T∝thickness\)
\(Q_{A}=-K_{A}.(A)\frac{dT}{dx}\)
\(=-K_{A}.(A)\frac{∆T_{A}}{d}\)
Same as
\(Q_{B}=-K_{B}.(A)\frac{∆T_{B}}{2d}\)
\(\frac{Q_{A}}{Q_{B}}=\frac{-2K_{B}(A)∆T_{A}/d}{-K_{B}.(A)∆T_{B}/2d}\)
\(=\frac{2.∆T_{A}/d}{∆T_{B}/2d}=4.\frac{∆T_{A}}{∆T_{B}}\)
\(=\frac{4(T_{B})/2}{∆T_{B}} 2:1\)
\([∵\frac{∆T_{A}}{∆T_{B}}=\frac{1}{2}=\frac{x_{A}}{x_{B}}\)		 Comments Active
95 Reynolds analogy states that (St is the Stanton number and Cfx is the skin friction coefficient) St = Cfx/4 St = Cfx/2 St = \(Cfx\) St = 2Cfx b Reynold’s analogy shows inter relationship fluid friction and Newton’s law of viscosity.
It also predicts local convective heat transfer coefficient (Nux) by knowing skin friction coefficient (Cfx)
Now
\((Stx =\frac{Nux}{Rex.Pr}=Stx=\frac{Cfx}{2}\)		 Comments Active
96 For a given heat flow and for the same thickness, the temperature drop across the material will be maximum for Copper Steel Glass wool Refractory brick c Material
Thermal conductivity (K)
Cu
385
Steel
50.2
Glass wool
0.04
Brick
0.6
As we know \(Q=-KA\frac{dT}{dx}\)
So \(∆T∝\frac{1}{K}\)
Lowest K, highest (Temperature drop) \(∆T\)		 Comments Active
97 Choose correct order of metals for increasing conductivity : Cu, Al, Fe, Ag Fe, Al, Cu, Ag Al, Fe, Cu, Ag Cu, Ag, Al, Fe b Material
Thermal conductivity (K)
Fe
79.5
Al
205
Cu
385
Ag
406		 Comments Active
98 Addition of fin to the surface increases the heat transfer if hA/Kp is equal to 1 greater than 1 less than 1 greater than 1 but less than 2 (Notations used have usual meaning) c \(\frac{Q_{with fin}}{Q_{without fin}}=effectiveness of fin\)
\(=\frac{hPKA (T_{0}-T_{∞})tan hmL}{hA(T_{0}-T_{∞})}\)
\(ϵ_{fin}=\frac{tanhmL}{\frac{hA}{KP}}\)
\(if \frac{hA}{KP}<1\)
Then \(ϵ_{fin}>1\)		 Comments Active
99 ________will radiate heat to a large extent. Black polished surface White rough surface White polished surface Black rough surface d A black body absorbs maximum heat and hence radius maximum heat. If the surface is shiny then the heat gets reflected back and the absorption of heat is very less. Hence a black non – shiny or rough body would radiate maximum heat. Comments Active
100 The ratio of thickness of thermal boundary layer to thickness of hydrodynamic boundary layer is equal to (Pr)n where n is – 1/3 2/3 1 –1 a 0 \(\frac{δ_{n}}{δ_{T}}=Pr^{1/3}\)
\(δ_{T}/δ_{n}=\frac{Thickness of thermal boundary layer}{Thickness of hydrodynamic layers}\)
\(=(Prandtl number)^{-1/3}\)		 Comments Active
101 The radiant heat transfer per unit area (W/m2) between two plane parallel gray surfaces (emissivity = 0.9) maintained at 400 K and 300 K is 992 893 464 567 * All options are wrong
\(Q=\frac{σA(T14-T24)}{\frac{1}{ϵ_{1}}+\frac{1}{ϵ_{2}}-1}\)
\(ϵ_{1}=ϵ_{2}=0.9\)
\(σ=5.67×10^{-8}\)
\(T_{1}=400 K\)
\(T_{2}=300 K\)
\(Q=\frac{5.67×10^{-8}(400^{4}-300^{4})}{\frac{1}{0.9}+\frac{1}{0.9}-1}\)
\(Q=813.31 W/m^{2}\)		 Comments Active
102 The value of Prandtl Number of air is about 0.1 0.4 0.7 1.1 c Pr)air = 0.7 Comments Active
103 The Nusselt number in natural convection heat transfer is a function of fluid Prandtl number and Stanton Number Biot Number Grashoff Number Reynolds Number c In natural convection
\(Nu=f(Pr, Gr)\)		 Comments Active
104 What is the value of shape factor for two infinite parallel surfaces separated by a distance x ? 0 ï‚¥ 1 x c Summation Rule.
\(F_{11}+F_{12}=1 (Shape factor)\)
Emitted radiation form a flat plate never return to itself
So,
\(F_{11}=0 ∴F_{12}=1\)
Shape factor = Fraction of radiation leaving a surface that reaches another surface.		 Comments Active
105 In the lumped parameter model, the temperature variation with time is linear cubic exponential sinusoidal c lumper parameter analysis gives.
\(\frac{T-T_{∞}}{T_{s}-T_{∞}}=e^{\frac{-hAC}{ρVC}}\)
\(∴T is varying exponentially\)		 Comments Active
106 What is the equivalent emissivity for radiant heat exchange between a small body (emissivity = 0.4) in a very large enclosure (emissivity = 0.5) ? 0.5 0.4 0.2 0.1 b Equivalent emissivity can be find out by
\(Q_{1-2}=\frac{σA_{1}(T14-T24) }{(\frac{1}{ϵ_{1}}-1)+\frac{1}{F_{12}}+A_{1}/A_{2}(\frac{1}{ϵ_{2}}-1)}\)
body 1 is completely enclosed in body 2 so, \(∵\)
\(F_{12}=1 (Shape factor)\)
\(ϵ_{1}=0.4, ϵ_{2}=0.5\)
\(A-1≪A_{2} ∴A_{1}A_{2}≈0\)
\(∴Q_{12}=\frac{σA_{1}(T14-T24)}{(\frac{1}{ϵ_{1}}-1)+1+0}\)
\(=σA_1.ϵ_1.(T14-T24)\)
\(∴equivalent emissivity\)
\(=ϵ_{1}=0.4\)		 Comments Active
107 The critical thickness of insulation for spheres is given by k/h k/4h h/2k 2k/h d \(t_{cr}=2k/h Sphere\)
\(t_{cr}=K/h Cylinder \)		 Comments Active
108 Two walls of same thickness and cross sectional area have thermal conductivities in the ratio of 1 : 2 . If the same temperature difference is maintained across the wall faces, the ratio of heat flow Q1/Q2 will be 1/2 1 2 4 d \( \frac{K_{1}}{K_{2}}=\frac{1}{2}\)
Fourier’s law \(∴Q=-KA\frac{dT}{dx}\)
\(∴Q∝K\)
\(∴\frac{Q_{1}}{Q_{2}}=\frac{K_{1}}{K_{2}}=\frac{1}{2}\)		 Comments Active
109 The Prandtl number will be the lowest for water liquid metal Aqueous solution lube oil b Liquid Metal
Prandlt number is given as
\(Pr=\frac{Momentum diffusivity}{Thermal diffusivity}\)
\(Pr=\frac{(μ/ρ)}{(K/μC_{p})}\)
Where
S.I. units
Momentum diffusivity
\(m^{2}/s\)
Thermal diffusivity
\(m^{2}/s\)
Dynamic viscosity (𝛍)
Pa – s = NS/m2
Thermal conductivity (K)
W/m.K
Specific heat ( \(C_{P})\)
J/Kg.K
Density (𝛒)
Kg/m3
Pr = around 7.56 for water (at 18áµ’C)
Pr = 13.4 at 0áµ’C for salt solution
Pr = 1000 for sugar
Pr = 0.06 for molten lithium		 Comments Active
110 For an opaque plane surface, the irradiation, radiosity and emissive power are 20,12 and 10 W/m2 respectively. The emissivity of the surface is 0.2 0.4 0.8 1.0 c Irradiation (G) = Incident radiation per unit area.
Radiosity (J) = Radiation leaving surface per unit time per unit are(a)
\(J=E+ρG \)
\(E-Emissive power\)
\(ρ-Reflectivity\)
As we know that
\(α+ρ+ς=1\)
And thermal equilibrium
\(∝=ϵ\) \(ϵ=\frac{E}{E_{b}}\)
\(and ρ+ϵ=1 when ς=0\)
\(∴ρ=1-ϵ\)
\(∴J=ϵ+(1-ϵ)G⇒\)
\(E=ϵE_{b}+(1-ϵ)G \)
\(12=ϵ.10+(1-ϵ)20\)
\(ϵ=0.8\)		 Comments Active
111 A cross-flow type air heater has an area of 50 cm2. The overall heat transfer coefficient is 100 W/m2K and heat capacity of both hot and cold streams are 1000 W/K. The value of NTU is 1000 500 5 0.2 * \( NTU=\frac{AU}{C_{min}}\)
\(A=Surface area\)
\(U=Overall heat transfer coefficient \)
\(C_{min}=minimum heat capacity\)
\(=NTU=\frac{50×10^{-4}×100}{1000}=50×10^{-5}\)
So all options are wrong.
However if we take A = 50m2
Then
\(NTU=\frac{50×100}{1000}=5\)
[Then only option (c) is correct]		 Comments Active
112 During the process of boiling and condensation only a phase change takes place, and one fluid remains at constant temperature throughout the heat exchanger. In terms of number of transfer units (NTU), the effectiveness of such heat exchanger would be NTU/(1 + NTU) 1 – exp (–NTU) 1 – exp (–2NTU)/2 cannot be worked out as heat capacities are unknown b Effectiveness
(for parallel) \(ϵ=\frac{1-e^{-NTU(1+c)}}{1+c}\)
And (for counter flow) \(ϵ=\frac{1-e^{-NTU(1-c)}}{1-ce^{-NTU(1-c)}}\)
c = heat capacity ratio
for boiling and condensation c = 0
\(∴ ϵ_{Parallel}=1-e^{-NTU}\)
\(=ϵ_{counter}\)		 Comments Active
113 If V = volume, A = surface area, h = surface film conductance, ρ = density and C = specific heat, then the time constant of a thermocouple is equal to VρC/Ah Vρ/CAh Ah/VρC VC/ρAh a Lumped system analysis
\(\frac{T-T_{∞}}{T_{1}-T_{∞}}=(\frac{-hA}{ρVC}).t\)
Here = time constant \(ρ\frac{V}{A}.\frac{C}{h}\)		 Comments Active
114 If Q = actual rate of heat transfer and Qmax = maximum possible rate of heat transfer then, heat exchanger effectiveness is equal to Qmax– Q Q/Qmax Qmax/Q (Qmax+ Q )/2 b Effectiveness of a heat exchanger is defined as the ratio of the actual heat transfer to the maximum possible heat transfer
\(ϵ=\frac{Actual heat transfer}{Maximum possible heat transfer}=\frac{Q}{Q_{max}}\)		 Comments Active
115 If ∆Ti and ∆Te are the temperature differences at inlet and at exit of the heat exchanger, then LMTD is equal to (∆Ti-∆Te) / (ln∆Ti/∆Te) \(ln((∆Ti-∆Te) / (l ∆Ti/∆Te))\) (ln∆Ti-∆Te) / (∆Ti/∆Te) (∆Ti-∆Te) / ((ln∆Ti+∆Te)/2) a For parallel flow heat exchanger
LMTD
\(∆T_{lm}=\frac{∆T_{i}-∆T_{0}}{ln(\frac{∆T_{i}}{∆T_{o}})}\)
\(∆T_{i}=T_{ni}-T_{ci}\)
\(∆T_{o}=T_{no}-T_{co}\)
For counter flow heat exchanger
LMTD
\(∆T_{m}=\frac{∆T_{1}-∆T_{2}}{ln(\frac{∆T_{1}}{∆T_{2}})}\)
\(∆T_{1}=T_{ni}-T_{co}\)
\(∆T_{2}=T_{no}-T_{ci}\)		 Comments Active
116 If k = thermal conductivity and h = heat transfer coefficient then the critical thickness of insulation for a cylinder, which will maximize the heat transfer is equal to k/h h/k 2k/h h/2k a For cylinder \(R_{cr}=K/h\)
\(for shpere R_{cr}=2K/h\)		 Comments Active
117 Which pair, out of the following alternatives, is not correctly matched ?
List – IList – II		 Fourier’s law-Conduction Newton’s law of cooling-Convection Stephan-Boltzman law-Radiation Kirchoff’s law-Radiation + Convection d Kirchoff’s law is related only with radiation kirchoff’s low of thermal radiation states that emissive power to the coefficient of absorption in constant for all the substances at a given temperature. We can say that at a given temperature the emissivity of a substance is equal to the coefficient of absorption Comments Active
118 If one radiation shield is placed between two infinite parallel radiating plane surfaces, then the amount of heat radiated becomes one third one fourth half none of the above c Plate 1 emissivity = \(ϵ_{1}\)
emissivity = \(Plate 2 \) \(ϵ_{2}\)
\(Shield plate emissivity= ϵ_{3}\)
\((Q_{12})_{net}=\frac{Aσ_{b}(T14-T24)}{(\frac{1}{ϵ_{3}}+\frac{1}{ϵ_{2}}-1)+(\frac{1}{ϵ_{1}}+\frac{1}{ϵ_{3}}-1)}\)
(Memories this formula)
Radiant energy transfer
\(\frac{with shield}{without shield}=ϵ_{3}=0\)
\(=\frac{(\frac{1}{ϵ_{1}}+\frac{1}{ϵ_{2}}-1) (without)}{((\frac{1}{ϵ_{1}}+\frac{1}{ϵ_{3}}-1)+(\frac{1}{ϵ_{3}}+\frac{1}{ϵ_{2}}-1)) (with)}\)
When the above fraction taken the value . Thus by inserting one shield the heat transfer rate is reduced to one – half of the original value. Only half rad. Will be transmitterd \(ϵ_{1}=ϵ_{2}=ϵ_{3}, \) \(\frac{1}{2}\)		 Comments Active
119 The rate of heat transfer from a solid surface to a fluid is obtained from Newton’s law of cooling Fourier’s law Kirchhoff’s law Stefan’s law a Convection heat transfer is energy transport due to bulk fluid motion. Convection heat transfer occurs through gases and liquids from a solid boundary results from the fluid motion along the surface.
Newton determined that the heat transfer/are(a) Q/A, is proportional to the fluid solid temperature difference \(T_{s}:T_{f}\)
The temperature difference usually occurs across a thin layer of fluid adjacent to the solid surface. This thin fluid layer is called a boundary layer. The constant of proportionality is called the heat transfer coefficient n
Newtonion’s eqn
\(\frac{Q}{A}=n(T_{s}-T_{f})\)
The heat transfer coefficient depends on the type of fluid and the fluid velocity. The heat flux, depending on the area of interest, is the local or area average(d)		 Comments Active
120 The rate of heat transfer through a hollow cylinder of inner and outer radii r1 and r2, respectively, depends on difference of radii, (r2 – r1) sum of radii, (r2 + r1) product of radii, (r1 r2) ratio of radii, r2/r1 d \( R^{th}=\frac{ln(\frac{r_{2}}{r_{1}})}{2πKL}\)
\(Q=\frac{T_{1}-T_{2}}{ln(\frac{r_{2}}{R_{1}})}.2Ï€KL\)
for cylinder \(∴Q∝\frac{1}{ln(\frac{r_{2}}{r_{1}})}\)
for shpere \(Q=\frac{T_{1}-T_{2}}{\frac{r_{1}-R_{2}}{4Ï€Kr_{1}r_{2}}} \)		 Comments Active
121 A container carrying water is moved in a horizontal direction with an acceleration of2.45 m/s2. The angle of inclination of free water surface to the horizontal is 14.03o 67.8o 45o 0o a \( tan θ=\frac{-a}{g}\)
P = Pressure force parallel to new surface
m = Mass of fluid element.
\(→Pcos (90-θ)+ma=0\)
\(Psin θ+ma=0…..(1)\)
\(→Psin(90-θ)=mg\)
\(Pcosθ=mg……..(2)\)
Dividing eq (1)÷ (2)
\(tan θ=\frac{-a}{g}\)
Numerically
\(tanθ=\frac{a}{g}=\frac{2.45}{9.8}\)
\(θ=tan^{-1}(0.25)=14.03°\)
Another method
\(∵tanθ<1 astanθ=0.25\)
So, \(θ must be>0°\)
\(∴θ=14.03° is only quantity in options\)		 Comments Active
122 Surface tension is a phenomenon due to Cohesion only Viscous forces only Adhesion between liquid and solid molecules Difference in magnitude between the forces due to adhesion and cohesion d The phenomenon of surface tension arises due to the two kinds of intermolecular forces.
(1) Cohesion: The force of attraction between the molecules of a liquid by virtue of which they are bound to each other to remain as one assemblage of particles is known as the force of cohesion. This property enables the liquid to resist tensile stress.
(2) Adhesion: The force of attraction between the unlike molecules, i.e. between the molecules of different liquids or between the molecules of a liquid and those of a solid body when they are in contact with the each other, is known as the force of adhesion. This force enables two different liquids to adhere to each other or a liquid to adhere to a solid body or surface.
* A and B experience equal force of cohesion in all directions, C experiences a net force inferior of the liquid. The net force is maximum for D since it is at surface.
The magnitude of surface tension is defined as the tensile force acting across imaginary short and straight elemental line divided by the length of the line.		 Comments Active
123 Boundary layer on a flat plate is called laminar
boundary layer if Reynold’s number is less than		 2000 4000 5 × 105 None of the above c If Re < 2000 Laminar flow
Re > 4000 Turbulent flow
Re < Laminar flow over a flat plate. \(5×10^{5}\)		 Comments Active
124 In order to have a continuous flow through a syphon, no portion of the pipe be higher than __________ measured above the hydraulic grade line : 10m 10.33m 5.5m 7.75m d Syphon – Long bend pipe
HGL – Hydraulic grade line
10.3 – 7.75 = 2.58		 Comments Active
125 Head loss in sudden expansion is given by V12-V22/2g (V1-V2)3/2g (V1-V2)2/2g 2(V12-V2)2/g
(Notations used have usual meanings)		 a Due to expansion
\(h_{L}=\frac{(v_{1}-v_{2})^{2}}{2g}\)
Loss due to contraction
\(h_{L}=\frac{(v_{c}-v_{2})^{2}}{2g}\) \((V_{c}=Velocity at contraction)\)
Loss due to bend contraction
\(h_{L}=\frac{V^{2}}{2g}\)
K = Constt depends radius bend and angle of ben(d)		 Comments Active
126 The parameters which determine the friction factor for turbulent flow in a rough pipe are Froude number and relative roughness Froude number and Mach number Reynolds number and relative roughness Mach number and relative roughness c In case of a laminar fully developed flow through pipes, the friction factor f is found form the exact solution of the Navier – Stokes equation is given by
\(f=64/Re\)
* In the case of turbulent flow, friction factor depends on both the Reynolds number and the roughness of pipe surface.
* The friction factor f at a given Reynolds number in the turbulent region, depends on the relative roughness, defined as the ratio of average roughness to the diameter of the pipe rather than the absolute roughness.		 Comments Active
127 The velocity distribution in turbulent flow is a function of the distance y measured from the boundary surface and the friction velocity V*, and follows a parabolic law logarithmic law hyperbolic law linear law b Velocity distribution for turbulent flow in pipe is given by prandtl’s universal equation
\(U=U_{max}+2.5 u^{*}Log_{e}(y/P)\)
Where \(U^{*}=shear or friction velocity\)
\(=\frac{ς_{0}}{ρ}\)		 Comments Active
128 The Darcy-Weisbach friction factor which is a direct measure of resistance to flow in pipes is dependent on Roughness height, diameter and velocity Relative roughness, diameter and viscosity Relative roughness, velocity and viscosity Roughness height, diameter, velocity and kinematic viscosity d The Darcy Weisbach equation is used to drop across a pipe for a fluid mathematically.
Where P is pressure, f is friction factor, L is pipe lenth, V is flow velocity, D is pipe diameter, and ρ is fluid density. \(∆P=(fLV^{2}ρ)/2D\)
and Roughness \(f∝\frac{1}{Re}\)		 Comments Active
129 Kinematic similarity between model and prototype means the similarity of forces shape motions discharge c Kinematic similarity means the similarity of motion between model and prototype.
The ratio of velocity and acceleration at the corresponding point in the model and at corresponding linear dimension in the model and prototype are equal.
Dynamic similarity means the similarity of forces between model and prototype.		 Comments Active
130 The velocity distribution in laminar flow through a circular pipe follows the Linear law Parabolic law Logarithmic law None of the above b Velocity distribution for laminar flow is given by
\(U=-\frac{1}{4μ}\frac{∂P}{∂x} (R^{2}-r^{2}) \)
i.e. parabola of profile		 Comments Active
131 An ideal fluid is defined as the fluid which is compressible is incompressible is incompressible and inviscid has negligible surface tension d An ideal fluid is a fluid that is incompressible and no internal resistance to flow (zero viscosity). In addition ideal fluid particles undergo no rotation about their centre of mass (irrotational).
An ideal fluid can flow in a circular pattern, but the individual fluid particles are irrotational.		 Comments Active
132 Reynold’s number is defined as the ratio of inertia force to gravity force pressure force elastic force viscous force d Reynold’s number is a dimensionless number that determines the nature of the flow of liquid through a pipe. It is defined as the ratio of the inertial force to the viscous force. Comments Active
133 The continuity equation in fluid mechanics is a mathematical statement embodying the principle of conservation of energy conservation of mass conservation of momentum none of above b Velocity in mass flow per unit will be same at inlet and outlet.
\(m_{1}=m_{2}\)
\(ρA_{1}V_{1}=ρA_{2}V_{2}\)
\(A_{1}V_{1}=A_{2}V_{2}=0(discharge)\)		 Comments Active
134 Match the following and choose the correct alternatives :
List – IList – II		 3635 3641 2452 2615 b \( Fraude no= \frac{Inertial force}{Gravity force}\)
\(\frac{V}{gD} v=flow velocity\)
\(D=hydraulic depth\)
\(Weber no= \frac{Pressure force}{Surface tension}\)
\(=\frac{v^{2}ρL}{σ}=\frac{v}{\frac{σ}{ρL}}\)
\(σ=surface tension\)
\(L=length\)
\(Eulers no= \frac{Inertia force}{Pressure force}\)
\(=\frac{v}{P/ρ}=\frac{ρv^{2}}{P}\)
\(P=Pressure\)
\(ρ=Density\)
\(Mach no= \frac{Inertia force}{Elastic force}\)
(Velocity of sound) \(=\frac{v}{K/ρ }=\frac{V}{C}\)
Note: Some texts write Euler’s no. as
\(\frac{Pressure force}{Inertia force}=P/ρV^{2}\)
Here option 1 is already wrong.		 Comments Active
135 A box of rectangular base 2m × 3m contains gasoline (Sp. Gravity 0.7) up-to a height of 5m. The force on the base and two vertical surfaces, 2m × 5m and 3m × 5m respectively. 206 kN, 258 kN and 172 kN 21 kN, 17.5 kN and 26.3 kN 258 kN, 172 kN and 206 kN 206 kN, 172 kN and 258 kN d \( ρ_{oil}=Sp. gravity ×ρ_{water}\)
\( =0.7×1000\)
Force on base
\((2×3)m^{2}=ρgAh\)
\(=(0.7×1000)×9.8×6×5\)
\(=700×500\)
\(=206000 N=206 kN\)
Force on side surface = \(ρgAh\)
\((2×5)m^{2}=(0.7×1000)×9.8×10×2.5\)
\(=171500N=171.5 kN\)
Same as we can calculate force on face as 258 kN \(3×5m^{2}\)		 Comments Active
136 Which of the following is a Mechanical Gauge? Diaphragm gauge Dead weight pressure gauge Bourdon tube pressure gauge All the above d Mechanical gauges are used to measure pressure, dimension, level etc. Comments Active
137 The resultant force on a floating body will act vertically upwards through centre of buoyancy vertically downwards through centre of buoyancy vertically upwards through meta centre vertically downwards through meta centre a When a body is either wholly or partially immersed in a fluid, the hydrostatic left due to the net vertical component of the hydrostatic pressure forces experienced by the body is called the “Buoyant force” and the phenomenon is called “Buoyancy”. The buoyancy is an upward force exerted by the fluid on the body when the body is immersed in a fluid or floating on a liquid. This upward force is fluid displaced by the body. It acts along center of buoyancy. Comments Active
138 Which of the following head loss is significant in a pipe flow? Loss of head due to gradual contraction Loss of head due to friction Loss of head due to sudden enlargement Loss of head due to sudden contraction b Minor losses include head loss or pressure drop due to pipe fittings, entrance and exit of pipe, sudden contraction or expansion et(c) These are considered as minor because drop is small compared to the major of frictional head loss. Minor head losses are.
1. Head loss at the entrance of pipe
2. Head loss at the exit of pipe
3. Head loss due to obstruction in pipe
4. Head loss due to sudden expansion in pipe
5. Head loss due to pipe fitting		 Comments Active
139 A metal piece having density exactly equal to the density of a fluid is placed in the liquid. The metal piece will sink to the bottom float on the surface will be partly immersed will be wholly immersed d Buoyant force: An upward force exerted by a fluid that opposes the weight of an immersed object.
Archimedes Principle: The buoyant force exerted on the body immersed in a fluid is equal to the weight of the fluid, the body displaces. If an object is exactly the same density as the liquid it will not move up or down. It will not move up or down. It will just stay right where it is (unless it is pushed around by water currents). The object when placed in the liquid it will submerged wholly and its upper surface will consider with free surface of fluid.		 Comments Active
140 Hammer blow in pipes occurs due to sudden sharp bends sudden contraction sudden stoppage of flow sudden release of fluid from pipe c Water hammer in pipe lines takes place when flowing fluid is suddenly brought to rest by closing a valve. When the water flowing in a long pipe is suddenly brought to rest by closing the value or by any similar cause, there will be sudden rise in pressure due to the momentum or the moving water being destroyed. This causes a wave of high pressure to be transmitted along the pipe which creates noise known as knocking. This phenomenon of sudden rise in pressure in the pipe is known as water hammer or hammer blow. Comments Active
141 Froude number is defined as the ratio of inertia force to viscous force inertia force to gravity force inertia force to elastic force inertia force to pressure force b The Froude number is a measurement of bulk flow characteristics such as waves, sand bed forms, flow/depth interactions at a cross – section or between boarders.
The fraude number Fr, is a dimensionless value that describes different flow regimes of open channel flow.
The Froude number is a ratio of inertial and gravitational forces.
Gravity (numerator) – moves water downhill.
Inertial (denominator) – reflects its willingness to do so
\(F_{r}=\frac{v}{9D}\)
Where:
V = Water velocity
D = Hydraulic depth (cross – sectional area of flow (top width)
g = Gravity
When
\(F_{r}=1, Critical flow\)
Supercritical flow \(F_{r}>1, \)
Subcritical flow (slow) \(F_{r}<1,\)		 Comments Active
142 The separation of boundary layer takes place in case of negative pressure gradient positive pressure gradient zero pressure gradient none of the above b At the verge of separation
\(\frac{∂P}{∂x}>0 Pressure gradient\)
\(\frac{∂u}{∂y}=0 velocity gradient\)
shear stress \(ζ=μ (du/dy)\)		 Comments Active
143 For the laminar flow through a circular pipe the ratio of maximum velocity and the average velocity is 1.5 2.0 None of the above b Discharge in a laminar.
\(Q=\frac{πU_{max }R^{2}}{2} ……..(1)\)
\(Q=A×V\)
\(Q=πR^{2}×V_{avg} ………(2)\)
(1) = (2)
\(Ï€R^{2}V=\frac{Ï€U_{max}R^{2}}{2}\)
\((V_{avg}=\frac{U_{max}}{2})\)
Average or local velocity occur at r = 0.707 R.		 Comments Active
144 For irrotational and incompressible flow, the velocity potential and stream function are given by and ψ respectively. Which one of the following sets is correct? 2=0, 2ψ=0 20, 2ψ=0 20,2ψ0 20, 2ψ=0 a Velocity function – Scalar function of space and time such that \((∅)\)
\(\frac{-∂∅}{∂x}=u, \frac{-∂∅}{∂y}=v, \frac{-∂∅}{∂z}=ω\)
- 3D function
- if ∅ exists then flow is irrotational
- If ∅ satisfies Laplace equation then flow is incompressible and irritation flow.
or \(\frac{∂^{2}∅}{∂x^{2}}+\frac{∂^{2}∅}{∂y^{2}}+\frac{∂^{2}∅}{∂z^{2}}=0\) \(∇^{2}∅=0\)
Stream function \((Ψ)\)
- Scalar function of space and time such that
\(\frac{∂Ψ}{∂x}=v, \frac{∂Ψ}{∂y}=u\)
- 2D function
- If Ψ exists flow may be irrotational or rotational
- If Ψ satisfies the Laplace equation it is possible case of an irrotational flow.
\(\frac{∂^{2}Ψ}{∂x^{2}}+\frac{∂^{2}Ψ}{∂y^{2}}=θ, ∇^{2}Ψ=0\)		 Comments Active
145 Using Blasius Equation, the friction factor for turbulent flow through pipes varies as 1/Re 1/Re0.5 1/Re 0.33 1/Re 0.25 d Friction factor for laminar flow
\(f=\frac{64}{Re}=4×coefficient of friction\)
\(=4×\frac{16}{Re}\)
Friction factor for turbulent flow
\(f=\frac{0.316}{Re^{1/4}}\)
\(∴f∝\frac{1}{Re^{0.25}}\)		 Comments Active
146 At a distance x from the leading edge of a plate, the thickness of laminar boundary layer varies as 1/ x x 4/5 x1/2 x2 c Thickness of Laminar boundary layer δ:
Local Reynold’s number
\(Re_{x}=\frac{ρVL}{μ} or\frac{ρVx}{μ} [L=x\)]
\(δ=\frac{5x}{Re_{x}}=\frac{5x}{xev/μ}\)
\(δ∝x^{1-1/2}\)
\(δ∝x^{1/2}\)		 Comments Active
147 An oil of kinematic viscosity 0.5 stokes flows through a pipe of 5 cm diameter. The flow is critical at a velocity of about 0.2 m/s 2 m/s 2.5 m/s 4 m/s b \( μ_{K}=0.5 stokes\)
\(d=5 cm in pipe\)
For critical velocity in pipe.
\(R_{e}=\frac{ρVD}{μ} or\frac{vD}{μ_{K}}=2000\)
Kinematic viscosity
\(μ_{K}=\frac{μ}{ρ}= \frac{Dynamic viscosity}{density}\)
\(2000=\frac{v×5×10^{-2}×10}{0.5×10^{-4}}\)
\(v=2 m/s\)		 Comments Active
148 Local acceleration in fluid is due to unsteady nature of flow non uniformity of flow turbulence in flow Irrotational nature of flow a Unsteady – change w.r.t time
Local acceleration is basically defined as the rate of increase of velocity with respect to time at a given point in the flow field Local acceleration is basically due to the change in local velocity of the fluid particle as function of time.
It the velocity vector of any fluid flow is a function of space and time, then it can change with space as well as with time. Thus the acceleration, or change in velocity experienced by the fluid particles can be due to the change of velocity with the acceleration of fluid particles due to change in velocity in space is called convective acceleration and acceleration due to change in velocity in time is called local or temporal acceleration.		 Comments Active
149 The mathematical expression for lift force is given by FL = CL. ρUA FL = (CL. ρU2/2)A FL = CL. ρU2A None of the above
(Notations used have usual meaning.)		 b A fluid flowing around the surface of an object exerts force on it, lift is the component of this force that is perpendicular to the oncoming flow direction. It contrasts with the drag force, which is the component of the force parallel to the flow direction.
Lift depends on the density of the air, the square of the velocity, the air’s viscosity and compressibility, the surface area over which the air flows the shape of the body and the body’s inclination to the flow. In general, the dependence on body shape, inclination air viscosity and compressibility is very complex. The lifting force acting on the body in a fluid flow can be calculated
\(F_{L}=C_{L}.\frac{1}{2} ev^{2} A\)
Where
\(F_{L}=Lifting force (N)\)
\(C_{L}=Lifting coefficient\)
\(ρ=Density of fluid (kg/m^{3})\)
\(v=Flow velocity (m/s)\)
\(A=body area (m^{2})\)
Drag force
The drag force acting on a body in fluid flow can be calculated
\(F_{D}=C_{D}.\frac{1}{2} ρv^{2}A\)
Where
\(F_{D}=Drag force (N)\)
\(C_{d}=Drag coefficient\)
\(ρ=Density of fluid (Kg/m^{3})\)
\(v=Flow velocity (m/s)\)
\(A=Body area (m^{2})\)		 Comments Active
150 For the viscous flow the coefficient of friction is given by f = 8/Re f = 16/Re f = 32/Re f = 60/Re b Coefficient of friction
\(f^{1}=\frac{16}{Re}\)
Friction factor
\(f=4f^{1}=\frac{64}{Re} \)		 Comments Active
151 The velocity profile is approximated by a parabola u/U= 3/2 (y/δ)² - 1/2 (y/δ)3 , where the displacement thickness for the profile is 3/8 δ 5/8 δ 11/8 δ None of the above a \(\) \(\frac{u}{U^{'}}=\frac{3}{2}(\frac{Y}{δ})^{2}-\frac{1}{2}(\frac{Y}{δ})^{3}\)
Displacement thickness
\(δ^{*}=0δ(1-\frac{u}{U})_{dy}\)
\(=0δ(1-(\frac{3}{2}(\frac{Y}{δ})^{2}-\frac{1}{2}(\frac{Y}{δ})^{3})) dy\)
\(=(Y-\frac{3}{2}:\frac{Y^{3}}{3δ^{2}}-\frac{1}{2}\frac{Y^{4}}{4.δ^{3}})0δ\)
\(δ^{*}=δ-\frac{δ^{3}}{2δ^{2}}-\frac{δ^{4}}{8δ^{3}}\)
\(δ^{*}=δ-\frac{δ}{2}-\frac{δ}{8}\)
\(δ^{*}=\frac{δ}{2}-\frac{δ}{8}=\frac{4δ-δ}{8}\)
\(=\frac{3δ}{8}\)		 Comments Active
152 Rayleigh’s method of dimensional analysis is used for determining the expression for a variable which depends on maximum following number of variables. 4 2 6 a Rayleigh’s method: Used to find expression for maximum 3 – 4 variable.
As time period
\(T=2K\frac{L}{g}\)
Centrifugal force
\(F_{c}=mv^{2}/r\)		 Comments Active
153 Which is the flow measuring device through which fluid does not flow? Orifice plate Venturi meter Pitot tube Elbow meter c Pitot tube Comments Active
154 A streamlined body is defined as a body about which the drag is zero the flow is laminar the flow is along streamlines the flow separation is suppressed d A streamlined body is a shape of that lowers the friction drag between fluid, like air and water, and an object moving through that fluid Drag is a force that slows down motion, friction drag is special kind of drag. A stream lined body is defined as the body whose surface coincides with the stream lines, when the body is placed in a flow. In that case the separation of flow will take place only at trailing edge (or rearmost part of the body). Behind a stream lined body, wake formation zone will be very small and consequently the pressure drag will be very small. Thus, the total drag on the stream lined body will be due to friction (shear) only. Comments Active
155 Mach number is defined as the square root of the ratio of the inertia force to the pressure force inertia force to the surface tension force inertia force to the elastic force none of the above c Mach Number
\(M=\frac{Inertial force}{Elastic force}=\frac{eAV^{2}}{KA}\)
\(=\frac{v}{K/ρ }=\frac{v}{velocity of sound}\)		 Comments Active
156 A pitot tube is used for measuring velocity of flow pressure of flow flow rate total energy a A pitot tube is used in wind tunnel experiments fluid flow experiments and an air planes to measure flow speed It’s a slender tube that has two holes in it. The front hole is placed in the air stream to measure what’s called, the stagnation pressure. The side hole measures static pressure. By measuring the difference between these pressures you get the dynamic pressure. Which can be used to calculate air speed Comments Active
157 The concept of stream function which is based on the principle of continuity is applicable to irrotational flow only two-dimensional flow only three-dimensional flow only uniform flow only b Velocity potential function is for 3D and stream function is for 2D flow. Comments Active
158 The type of flow in which the velocity at any given time does not change with respect to space is called Steady flow Compressible flow Uniform flow Rotational flow c This uniform fluid flow is defined as the type of flow in which the velocity at any given time does not change with respect to space (i.e. length of direction of the flow) Comments Active
159 A fully developed pipe flow implies that the flow should be laminar flow should be turbulent velocity profile should be uniform velocity profile should not change in the direction of flow. d No change in velocity along direction of flow i.e. x – direction. Comments Active
160 Application of Bernoulli’s equation requires that the duct is two dimensional the flow is laminar the duct is frictionless the fluid is inviscid and incompressible d For Bernouli’s equation to be applied, the following assumption must be met:
* The flow must be steady. (Velocity – pressure and density cannot change at any point)
* The flow must be incompressible – even when the pressure varies, the density must remain constant along the streamline friction by viscous forces must be minimal.		 Comments Active
161 In an Otto cycle, the heat addition and heat rejection take place at constant volume and at constant pressure respectively constant volume and at constant volume respectively constant pressure and at constant volume respectively constant pressure and at constant pressure respectively b 1 -2: Isentropic compression
2 – 3: Constant volume combustion
3 – 4: Isentropic expansion
4 – 1: Constant volume heat rejection		 Comments Active
162 Which pair of the following alternatives is correctly matched?
List – IList – II		 HeatPoint function EnergyPath function EntropySecond law of thermodynamics Gibbs functionPath function c Heat – Path function
Energy (ΔE) – Point function
Gibb’s function – Point function		 Comments Active
163 The most practical vapour power cycle is Carnot Joule Rankine Binary c We know that the carnot cycle is most efficient cycle b/w two specified temperature limits. However, the carnot cycle is not a suitable model for steam power cycle since.
The Rankine cycle is the ideal cycle for vapour power plants, it includes the following four reversible processes.
1 -2: Isentropic compression water enters the pump as state 1 as saturated liquid and is compressed isentropically to the operating pressure of the boiler.
2 – 3: Constant preheat addition saturated water enters the boiler and leaves it as superheated vapor at state – 3.
3 – 4: Isentropic expansion superheated vapor expands isentropically in turbine and produces work.
4 – 1: Constant pressure heat rejection high quality steam is condensed in the condenser.		 Comments Active
164 In an irreversible process there is a loss of heat no loss of heat gain of heat no gain of heat a Irreversibility is mainly due to friction which produces heat in action. So supplied energy is dissipated or wasted in form of heat. Comments Active
165 The second law of thermodynamics defines Heat Enthalpy Internal Energy Entropy d The second law of thermodynamics states that the total entropy of an isolated system always increases over time, or remains constant in ideal cases where the system is in a steady state or undergoing a reversible process. The increase in entropy accounts for the irreversibility of natural processes.
Zeroth law defines: Temperature
First law defines: Internal energy
Second law defines: Entropy		 Comments Active
166 The property of a working substance which increases or decreases as the heat is supplied or removed in a reversible manner is known as Enthalpy Entropy Internal energy External Energy b Entropy: Entropy transfer is associated with heat transfer if the heat is added to the system, then its entropy increases and if heat is lost from the system its entropy decreases Comments Active
167 Carnot cycle is not considered as a practical cycle because its p – V diagram is narrow its thermal efficiency is low heat addition takes place at high pressure heat rejection takes place at high temperature a The most efficient heat engine cycle is the Carnot cycle, consisting of two isothermal processes and two adiabatic processes. The carnot cycle can be thought of as the most efficient heat engine cycle allowed by physical laws.
In order to approach the carnot efficiency, the processes involved in the heat engine cycle must be reversible and involve no change in entropy. This means that the carnot cycle is an idealization, since no real engine processes are reversible and all real physical processes involve some increase in entropy. It is not a practical engine cycle because the heat transfer into the engine in the isothermal process is too slow to be of practical value.		 Comments Active
168 Fuel injector is used in steam engines gas engines spark ignition engines compression ignition engines d Compressed Ignition engine Comments Active
169 The change in entropy is zero during Hyperbolic process Constant pressure process Reversible adiabatic process Polytropic process c Change in entropy = Change in heat of the system / Temp in Kalvin = \(∆S)\) \(\frac{θ}{T}\)
\(∆S=dq_{rev}/T\)
For adiabatic process \(dq_{rev}=0\)
or S = Constant \(So, ∆S=0\)		 Comments Active
170 Parson’s reaction turbine has fixed blades only moving blades only identical fixed and moving blades fixed and moving blades of different shapes. c The parson’s reaction turbine has identical fixed & moving blades. Since in the parson’s reaction turbine, the blade of turbine acts, as nozzle and hence the relative velocity as exit will be more than that at inlet.
Degree of reaction = \(\frac{1}{2}=0.5\)		 Comments Active
171 Which of the following is not a property of thermodynamic system? Pressure Energy Heat Volume c Heat is a path function not a property which are known as point function. Comments Active
172 The main cause of the irreversibility is Mechanical and Fluid Friction Unrestricted expansion Heat transfer with a finite temperature difference All of the above d Four of the most common causes of irreversibility are friction, unrestrained expansion of a fluid, heat transfer through a finite temperature difference, and mixing of two different substances. These factors are present in real, irreversible processes and prevent these processes and prevent these processes from being reversible. Comments Active
173 The gas constant R is equal to the sum of two specific heats. difference of two specific heats. product of two specific heats. ratio of two specific heats. b Mayer’s formula
\(C_{P}-C_{V}=R\)		 Comments Active
174 Kelvin Planck’s law deals with conservation of energy conservation of heat conservation of mass conversion of heat into work d Statement: It is impossible to construct an engine that an operating in a cycle, will produce no effect other than the extraction of heat from a reservoir & the performance of an equal amount of work. Comments Active
175 The efficiency of an ideal Carnot engine depends on working substance the temperature of the source only the temperature of the sink only the temperature of both source and sink. d \( η_{carnot}=1-\frac{T_{L}}{T_{H}} \)
\(η_{Carnot}∝sinte an source temperature both.\)		 Comments Active
176 For complete burning of 1 kg of carbon, the air required will be about 2.67 kg 11.6 kg 12.7 kg 14.5 kg a 8/3 kg of O2 = 2.66 kg Comments Active
177 The unit of work is kW kWh kW/h kJ/s b \( Work=Power×time\)
\(=KW×hr\)
\(=KWhr\)
\(Power→*Watt\)
* Kilo watt
* \(\frac{N-m}{S}\)
* KJ/s or J/S		 Comments Active
178 Choose the correct alternative :
1. First law for a closed system undergoing a cycle
Q – W = ∆E
2. Two reversible adiabatic paths cannot intersect each other.
3. If two fluids are mixed, the entropy of universe remains unchanged.
4. Clausius statement – Heat can flow from low to high temperature body without the aid of external work.
5. The efficiency of a reversible heat engine is independent of nature of working substance undergoing a cycle.		 All statements are true. Statements 2 and 5 are true. Statements 1, 3, 4 and 5 are true. Statements 1, 3 and 5 are true. b 1st law of thermodynamics for closed system.
\(Q-W=∆E\) \([For cycle dW=dQ\)
If it undergoing a reversible process
\(∆E→∆U\)
\(Q-W=∆U\)
- By mixing 2 fluids entropy of system will increase
- Heat cannot flow from lower to high temperature without external air. (Clausius statement)		 Comments Active
179 For the same compression ratio and same heat rejection, the efficiency of Otto cycle is same as that of Diesel Cycle not comparable to that of Diesel Cycle less than that of Diesel Cycle more than that of Diesel Cycle d \( η=1-\frac{Q_{rej}}{Q_{supp}}\)
Compression ratio (is same for all) \(=\frac{V_{2}}{V_{1}}\)
Least heat rejection will give highest efficiency because is same heat rejection is least for 4” – 1. \(Q_{supp}\)
\(∴η_{otto}>η_{dual}>η_{diesel}\)		 Comments Active
180 Steam flows through an adiabatic steady flow turbine from state 1 to state 2. with respect to a base temperature T0, the unavailable energy is T0(I1 – I2) T0 (S1 – S2) (I1 – I2) – T0 (S1 – S2) I2 + I0 (S1 – S2) b \( To (S_{1}-S_{2})\)
Carnot cycle
Friction of supplied heat which can be converted into useful work – Available energy area with in cycle.
Remaining portion of energy – Unavailable energy area under process 1 – 2, correspond to base temperature.		 Comments Active
181 The slope of an isobar on the h – s coordinates is equal to the Gibb’s function Helmholtz function Pressure Absolute saturation temperature at that pressure d Mollier diagram
We know that Tds = dh – Vdp
At P = Constant
Tds = dh
\((\frac{dh}{ds})_{p=c}=T \)		 Comments Active
182 The value of compressibility factor for a Vander Waals gas is equal to 1.0 0.375 0.2 to 0.3 0.35 b Compressibility factor
\(Z=\frac{PV}{RT}\)
At critical point
\(P=\frac{a}{27b^{2}}\)
\(V=3b\)
\(T=\frac{8a}{27R_{b}}\)
\(Z=\frac{a}{\frac{27b^{2}}{R.\frac{8a}{2TR.b}}}.3b\)
\(=\frac{3}{8}=0.375\)		 Comments Active
183 A Carnot heat pump absorbs heat from atmosphere at 10 oC and supplies it to a room maintained at 25 oC A temperature difference of 5 oC exists between working fluid and atmosphere on one hand, and the required room temperature on the other hand If the heat pump consumes 1 kW power, the heat delivered to the room will be 12.1 kW 14.9 kW 1.67 kW 19.9 kW b \( COP=\frac{T_{H}}{T_{H}-T_{L}}\)
\(\frac{Q_{2}}{W}=\frac{298}{298-278}\)
\(Q_{2}=14.9 kW\)		 Comments Active
184 A system is taken from state A to state B along two different paths 1 and 2. The heat absorbed and work done by the system along these paths are Q1 and Q2 and W1 and W2 respectively, then Q1 = Q2 W1 + Q1 = Q2 + W2 W1 = W2 Q1 – W1 = Q2 – W2 d A system is taken from state A to state B along 2 different paths 1 and 2. The heat absorbed and work done by the system along these 2 paths are and \(Q_{1}, Q_{2}\) \(W_{1}, W_{2}\)
By first law of thermodynamics and since the end condition are same so change in internal energy will be equal because it is a property of system.
\(U_{1}=U_{2}\)
So is correct. \(Q_{1}-W_{1}=Q_{2}=W_{2}\)		 Comments Active
185 The latent heat of steam with increase in pressure does not change increases decreases remains unpredictable c With the increase in pressure the boiling point of the liquid increases and a lesser amount of energy needed to overcome the intermolecular force thus the latent heat of steam required is decreased. Comments Active
186 Steam coming out of the whistle of a pressure cooker is dry saturated vapour wet vapour Superheated vapour Ideal gas c Practically vapor will be of wet type but in theory:
* Inside a pressure cooker, the pressure can be increased to more than 1 atm.
* As the pressure increases, the boiling point of water also increases and its value increases.
* The temperature will also become more than saturation temperature and will go in the superheated region.		 Comments Active
187 The air standard efficiency of an Otto cycle for a compression ratio of 5 and index γ = 1.4 is 60% 50% 47.47% 40% c Here
\(r=5, γ=1.4\)
\(η=1-\frac{1}{r^{γ-1}}=1-\frac{1}{5^{(1.4-1)}}\)
\(η=1-\frac{1}{5^{0.4}}=47.47%\)
All other options complete no but c option is appropriate.		 Comments Active
188 A mixture of gases expands from 0.03 m3 to 0.06 m3 at constant pressure of 1MPa and absorbs 84 KJ of heat during the process. The change in internal energy of the mixture is 30 kJ 54 kJ 84 kJ 114 kJ b \( V_{1}=0.03 m^{3},\) \(V_{2}=0.06 m^{3}\)
\(P=1 MPa, Q=84 KJ\)
\(=10^{3} kPa\)
1st law
\(Q-W=∆U\)
\(-((0.06-0.03)×10^{3})=W\)
\(∆U=84-0.03×10^{3}\)
\( =84-30=54 KJ\)		 Comments Active
189 Air is compressed isothermally by performing work equal to 16 kJ upon it. The change in internal energy is –16 kJ Zero 16 kJ 32 kJ b As we know that internal energy depends of temperature
\(U=f(T)\)
As this process is isothermal then no change occur in internal energy.		 Comments Active
190 An engine receives 15152 J/s of heat and produces 5 kW of power. The efficiency of the engine is 25% 27.5% 30% 33% d \( η=\frac{W}{Q_{in}}=\frac{5000}{15152}=0.3299\)
\(=33%\)		 Comments Active
191 Vander Waal’s equation may be written as (p + ) (v – b) = RT \(\frac{a}{v}\) (p + ) (v – b) = RT \(\frac{a}{v2}\) (p + ) (v2 – b) = RT \(\frac{a}{v}\) (p + ) (v – b) = RT2 \(\frac{a}{v}\)
(Notations used have usual meaning.)		 b Real gas Eqn: \((p+\frac{a}{v^{2}})(v-b)=RT\)
Ideal gas equation is PV = mRT		 Comments Active
192 At the critical point, any substance will exist in all the three phases simultaneously will change directly from solid to vapour will lose phase distinction between liquid and vapour will behave as an ideal gas b Critical point – the set of conditions under which a liquid and its vapor become identical. The liquid expands and becomes less dense until, at the critical point, the densities of liquid and vapor becomes equal and eliminating the boundary between the two phases. At the critical point, the particles in a closed container are thought to be vaporizing at such a rapid rate that density of liquid and vapor are equal, and thus form a supercritical fluid. As a result of the high rates of change, the surface tension of the liquid eventually disappears. You will have noticed that this liquid vapor equilibrium curve has a top limit, which is temperature and pressure corresponding to this are known as the critical temperature and critical pressure. It you increase the pressure on a gas (vapor) at a temperature lower than the critical temperature you will eventually cross the liquid vapor equilibrium line and the vapor will condensate to given a liquid. Comments Active
193 The Clausius equation for a reversible cycle is < 0 \(\frac{δQ}{T}\) = 0 \(\frac{δQ}{T}\) < 0 \(\frac{δQ}{T}\) ≤ 0 \(\frac{δQ}{T}\)
(Notations used have usual meaning)		 b Clausius Inequality
\(\frac{δQ}{T}≤0\)
If
\(\frac{δQ}{T}=0-Reversible process\)
\(\frac{δQ}{T}>0-Impossible process\)
\(\frac{δQ}{T}<0-Irreversible process\)		 Comments Active
194 The work in a closed system undergoing an isentropic process is given by / ()mR (T1 – T2 )) \(γ\) \(γ-1\) ()/ mR (T1 – T2 ) \(γ-1\) \(γ\) / ()mR (T1 – T2 ) \(1\) \(γ-1\) / () m (T1 – T2 ) \(γ\) \(γ-1\)
(Notations used have usual meaning)		 c WD for isentropic process
\(=\frac{P_{1}V_{1}-P_{2}V_{2}}{Ï…-1}=\frac{mRT-mRT_{2}}{Ï…-1}\)
\(=\frac{mR(T_{1}-T_{2})}{υ-1} (∵PV=mRT)\)		 Comments Active
195 Which of the following is an irreversible process? An isothermal process An isentropic process An isobaric process An isenthalpic process d A throttling process is a thermodynamics process in which the enthalpy of the gas or medium remains constant (h = const). In fact, the throttling process is one of isenthalpic processes. During the throttling process no work is done by or on the system (dW = 0), and usually there is no heat transfer (adiabatic) form or into the system (dQ = 0) on the other the throttling process cannot be isentropic, it is a fundamentally irreversible process.
Characteristics of throttling process.
1. No heat transfer
2. No work transfer
3. Irreversible process
4. Isentropic process
5. Throttling of the flow causes significant reduction in pressure because a throttling device causes a local pressure loss. A throttling can be achieved by introducing a restriction into a line through which a gas or liquid flows. This restriction is commonly done by means of a partially open valve or a porous plug. Once the pressure is released is cannot be re – attained without doing compression, so first expansion then compression is required.		 Comments Active
196 Which of the following constituents of a fuel does not contribute to its calorific value on combustion? Carbon Hydrogen Sulphur Nitrogen d \( N_{2}\)
are responsible for combustion N2 does not help in combustion. \(C, H_{2}, S\)		 Comments Active
197 The thermal efficiency of a theoretical Otto cycle increases with increase in compression ratio increases with increase in isentropic index does not depend on the pressure ratio follows all the above d Thermal efficiency of otto cycle
\(η_{th}=1-\frac{1}{r^{γ-1}}\)
\(r=compression ratio\)
decrease \(-as r increases,\frac{1}{r^{γ-1}}\)
Hence increases \(η_{th}\)
increases, \(-as γ\) \(\frac{1}{r^{γ-1}}\)
Decreases hence increases \(η_{th}\)		 Comments Active
198 A frictionless heat engine can be 100% efficient only if the exhaust temperature is equal to its input temperature less than its input temperature 0°C 0°K d Friction less heat engine means Carnot engine
Now
\(η_{carnot}=1-\frac{T_{Lower}}{T_{higher}}=1-\frac{T_{L}}{T_{h}}\)
then or 100% \(if T_{L}=0°K\) \(η=1\)		 Comments Active
199 In a reversible adiabatic process the ratio (T1/T2) is equal to (P1/P2) \(\frac{γ-1}{γ}\) (V1/V2) \(\frac{γ-1}{γ}\) (V1/V2) \(\frac{γ-1}{2γ}\) (V1/V2)γ
where γ is ratio of specific heats?		 a For reversible adiabatic process.
\(PV^{γ}=C\)
\(\frac{P_{1}}{P_{2}}=(\frac{v_{2}}{v_{1}})^{γ}=(\frac{T_{1}}{T_{2}})^{\frac{γ}{γ-1}}\)
\(\frac{T_{1}}{T_{2}}=(\frac{P_{1}}{P_{2}})^{\frac{γ-1}{γ}}\)
[By using equation PV = mRT above relation will come]		 Comments Active
200 An inventor claims to have developed an engine that takes in 105 MJ at a temperature of 400 K, rejects 42 MJ at a temperature of 200 K and delivers 15 kWh of mechanical work. Is this engine feasible? Yes No The information given is not sufficient to find answer. May be or may not be depends upon several factors. b For inventor’s engine, efficiency
\(η_{e}=1-\frac{Q_{rej}}{Q_{add}}\)
\(η_{e}=1-\frac{42}{105}=0.6\)
Now if this engine is an carnot engine then
Efficiency \(η_{c}=1-\frac{T_{low}}{T_{high}}\)
\(η_{c}=1-\frac{200}{400}=0.5\)
Here which is not possible. Between same temperature limits carnot engine will have maximum efficiency. \(η_{e}>η_{c}\)		 Comments Active