1. The first law of thermodynamics takes the from W= – \Delta H when applied to:
(a) A closed system undergoing a reversible adiabatic process.
(b) An open system undergoing an adiabatic process with negligible changes in kinetic and potential energies.
(c) A closed system undergoing a reversible constant volume process.
(d) A closed system undergoing a reversible constant pressure process.
(1 Mark, 1993)

Ans: b
Explanation:
Steady flow energy equation:
\dot { m } ({ h }_{ 1 }+\frac { { v }_{ 1 }^{ 2 } }{ 2 } +g{ z }_{ 1 })+\dot { Q } =\dot { m } ({ h }_{ 2 }+\frac { { v }_{ 2 }^{ 2 } }{ 2 } +g{ z }_{ 2 })+\dot { W }For an open system undergoing an adiabatic process with negligible changes in kinetic and potential energies
\dot { m } ({ h }_{ 1 })=\dot { m } ({ h }_{ 2 })+\dot { W }=> \dot { W } ={ H }_{ 1 }{ H }_{ 2 }=\Delta H
2. A steel ball of mass 1 kg and specific heat 0.4 kJ/kgK is at a temperature of 60^{0}C. It is dropped into 1 kg water at 20^{0}C. The final steady state temperature of water is
(a) 23.5^{0}C
(b) 30^{0}C
(c) 35^{0}C
(d) 40^{0}C
(1 Mark, 1993)

Ans: a
Explanation:
Specific heat of water = 4.2 kJ/kgK
{ \left( mC \Delta T \right) }_{ sb }={ \left( mC\Delta T \right) }_{ w }=> 1\times 0.4\times (60T)=1\times 4.2\times (T20)
=> T = 23.5^{0}C
3. A vertical cylinder with a freely floating piston contains 0.1 kg air at 1.2 bar and a small electrical resistor. The resistor is wired to an external 12 Volt battery. When a current of 1.5 amps is passed through the resistor for 90 secs, the piston sweeps a volume of 0.01 m^{3}. Assume (i) piston and the cylinder are insulated and (ii) air behaves as a ideal gas with C_{V} = 700 J/kg K. Find the rise in temperature of air in ^{0}C.
(1 Mark, 1993)

Ans: 6
Explanation:
Electrical work (W_{1}) = VIt = –12\times 1.5\times 90 = 1620J [Work on the system is ve]
Boundary work (W_{2}) = Pdv = 1.2\times { 10 }^{ 5 }\times 0.01 = 1200 J
Now, \delta W= = 1620 + 1200 = – 420 J
Again, \delta Q=dU+\delta W
=> dU=\delta W[\delta Q=0 because piston and cylinder are insulated.]
=> m{ C }_{ v }dT = 420 J
=> 0.1\times 700\times dT = 420 J
=> dT = 6^{0}C
4. For reversible adiabatic compression in a steady flow process, the work transfer per unit mass is
(a) \int { pdv }
(b) \int { vdP }
(c) \int { Tds}
(d) \int { sdT }
(1 Mark, 1996)

Ans: b
5. A steam turbine receives steam steadily at 10 bars with an enthalpy of 3000 kJ/kg and discharges at 1 bar with an enthalpy of 2700 kJ/kg. The work output is 250 kJ/kg. The changes in kinetic and potential energies are negligible. The heat transfer from the turbine casing to the surroundings is equal to
(a) 0 kJ
(b) 50 kJ
(c) 150 kJ
(d) 250 kJ
(1 Mark, 2000)

Ans: b
Explanation:
Applying SFEE in steam turbine:
h_{1} + q = h_{2 }+ w
=> q = 2700 + 250 – 3000 = 50 kJ/kg
( ve sign indicates heat is transferred from the steam turbine)
6. A small steam whistle (perfectly insulated and doing no shaft work) causes a drop of 0.8 kJ/kg in the enthalpy of steam from entry to exit. If the kinetic energy of the steam at entry is negligible, the velocity of the steam at exit is
(a) 4 m/s
(b) 40 m/s
(c) 80 m/s
(d) 120 m/s
(2 Mark, 2001)

Ans: b
Explanation:
Steady flow energy equation:
\dot { m } ({ h }_{ 1 }+\frac { { v }_{ 1 }^{ 2 } }{ 2 } +g{ z }_{ 1 })+\dot { Q } =\dot { m } ({ h }_{ 2 }+\frac { { v }_{ 2 }^{ 2 } }{ 2 } +g{ z }_{ 2 })+\dot { W }According to the question:
=> { h }_{ 1 }{ h }_{ 2 }=\frac { { V }_{ 2 }^{ 2 } }{ 2 }
=> { V }_{ 2 }=\sqrt { 800\times 2 } =40m/s
7. What is the speed of sound in Neon gas at a temperature of 500K (Gas constant of Neon is 0.4210 kJ/kgK)?
(a) 492 m/s
(b) 460 m/s
(c) 592 m/s
(d) 543 m/s
(2 Mark, 2002)

Ans: c
Explanation:
We know, { v=\sqrt { \gamma RT } =\sqrt { 1.67\times 421\times 500 } }
=> v = 592 m/s
8. A 2 kW, 40 litre water heater is switched on for 20 minutes. The heat capacity C_{P} for water is 4.2 kJ/kg K. Assuming all the electrical energy has gone into heating the water, increase of the water temperature in degree centigrade is
(a) 2.7
(b) 4.0
(c) 14.3
(d) 25.25
(1 Mark, 2003)

Ans: c
Explanation:
According to the question:
{ \left( m{ C }_{ p }\Delta T \right) }_{ w }=Power\times time=> { 40\times 4.2\times }{ 10 }^{ 3 }\times \Delta T=2\times { 10 }^{ 3 }\times (20\times 60)
=> \Delta T={ 14.28 }^{ 0 }C
9. A gas contained in a cylinder is compressed, the work required for compression being 5000 kJ. During the process, heat interaction of 2000 kJ causes the surroundings to the heated. The change in internal energy of the gas during the process is
(a) 7000 kJ
(b) –3000 kJ
(c) +3000 kJ
(d) + 7000 kJ
(1 Mark, 2004)

Ans: c
Explanation:
Work required to compression (\delta W)=5000kJ
Heat interaction causes the surrounding heated. So, \delta Q=2000kJ
We know: \delta Q=dE+\delta W
=> dE=2000+5000=3000kJ
Statement for Linked Answer Questions 10 & 11:
A football was inflated to a gauge pressure of 1 bar when the ambient temperature was 15°C. When the game started next day, the air temperature at the stadium was 5°C. Assume that the volume of the football remains constant at 2500 cm^{3}.
10. The amount of heat lost by the air in the football and the gauge pressure of air in the football at the stadium respectively equal
(a) 30.6 J, 1.94 bar
(b) 21.8 J, 0.93 bar
(c) 61.1 J, 1.94 bar
(d) 43.7 J, 0.93 bar
(2 Mark, 2006)

Ans: d
Explanation:
Given: P_{1 }= 1 + 1.01325 = 2.01325 bar
T_{1 }= 273 + 15 = 288 K
T_{2 }= 273 + 5 = 278 K
From ideal gas equation: \frac { { P }_{ 1 } }{ { P }_{ 2 } } =\frac { { T }_{ 1 } }{ { T }_{ 2 } }[Volume is constant]
=> { P }_{ 2 }=\frac { 278\times 2.01235 }{ 288 } =1.94bar
Mass of air inside the ball(m) = \frac { PV }{ RT } =\frac { 2.01325\times { 10 }^{ 5 }\times 2500\times { 10 }^{ 6 } }{ 287\times 288 } = 0.006 kg
Amount of heat lost = (\delta Q)=m{ C }_{ v }dT
=> \delta Q=0.006\times 718\times 10 = 43.7 J
11. Gauge pressure of air to which the ball must have been originally inflated so that it would equal 1 bar gauge at the stadium is:
(a) 2.23 bar
(b) 1.94 bar
(c) 1.07 bar
(d) 1.00 bar
(2 Mark, 2006)

Ans: c
Explanation:
Given:
P_{2}= 1 + 1.01325 = 2.01325 bar
From Ideal gas equation: \frac { { P }_{ 1 } }{ { P }_{ 2 } } =\frac { { T }_{ 1 } }{ { T }_{ 2 } }
=> P_{1 }= \frac { 2.01325\times 288 }{ 278 } = 2.08567 bar
P_{1 }= 2.08567 – 1.01325 = 1.07 bar
12. A 100 W electric bulb was switched on in a 2.5 m × 3 m × 3 m size thermally insulated room having a temperature of 20°C. The room temperature at the end of 24 hours will be
(a) 321°C
(b) 341°C
(c) 450°C
(d) 470°C
(2 Mark, 2006)

Ans: d
Explanation:
Q = P*t = 100*24*3600 = 8.64 MJ
m = pv/RT = (101235*22.5)/(287*293) = 27.1 kg
The Process is constant volume,
Q= m. Cv. dT
8.64* 10^{6 }= 27.1*718*(T – 20)
T \simeq 470^{0} C
13. A balloon containing an ideal gas is initially kept in an evacuated and insulated room. The balloon ruptures and the gas fills up the entire room. Which one of the following statements is TRUE at the end of above process?
(a) The internal energy of the gas decreases from its initial value, but the enthalpy remains constant.
(b) The internal energy of the gas increases from its initial value, but the enthalpy remains constant.
(c) Both internal energy and enthalpy of the gas remain constant.
(d) Both internal energy and enthalpy of the gas increase.
(2 Mark, 2008)

Ans: c
Explanation:
The question is a case of free expansion.
In free expansion the final and initial temperature are same.
As both internal energy and enthalpy are functions of temperature.
So, they remain constant.
14. In a steady state steady flow process taking place in a device with a single inlet and a single outlet, the work done per unit mass flow rate is given by W = – \int _{ inlet }^{ outlet }{ vdp } , where v is the specific volume and P is the pressure. The expression for ‘W’ given above
(a) is valid only if the process is both reversible and adiabatic
(b) is valid only if the process is both reversible and isothermal
(c) is valid for any reversible process
(d) is incorrect, it must be W = \int _{ inlet }^{ outlet }{ pdv }
(2 Mark, 2008)

Ans: c
15. A rigid, insulated tank is initially evacuated. The tank is connected with a supply line through which air (assumed to be ideal gas with constant specific heats) passes at 1 MPa, 350°C. A valve connected with the supply line is opened and the tank is charged with air until the final pressure inside the tank reaches 1 MPa. The final temperature inside the tank
(a) is greater than 350°C
(b) is less than 350°C
(c) is equal to 350°C
(d) may be greater than, less than, or equal to350°C, depending on the volume of the tank
(2 Mark, 2008)

Ans: a
Explanation:
The final temperature inside the tank (T_{tank}) = \gamma \times { T }_{ pipe }
=> T_{tank }= 1.4\times 623K={ 599.2 }^{ 0 }C
For more understanding: See Unsteady Energy flow equation.
16. A compressor undergoes a reversible, steady flow process. The gas at inlet and outlet of the compressor is designated as state 1 and state 2 respectively. Potential and kinetic energy changes are to be ignored. The following notations are used:
v = specific volume and P = pressure of the gas.
The specific work required to be supplied to the compressor for this gas compression process is
(a) \int _{ 1 }^{ 2 }{ Pdv }
(b) \int _{ 1 }^{ 2 }{ vdp }
(c) v_{1}(P_{2 }– P_{1})
(d) –P_{2}(v_{1 }– v_{2})
(2 Mark, 2009)

Ans: b
Statement for Linked Answer Questions: 17 & 18.
The temperature and pressure of air in a large reservoir are 400K and 3 bar respectively. A converging–diverging nozzle of exit area 0.005 m^{2} is fitted to the wall of the reservoir as shown in the figure. The static pressure of air at the exit section for isentropic flow through the nozzle is 50kPa. The characteristic gas constant and the ratio of specific heats of air are 0.287kJ/kgK and 1.4 respectively.
17. The density of air in kg/m^{3} at the nozzle exit is
(a) 0.560
(b) 0.600
(c) 0.727
(d) 0.800
(2 Mark, 2011)

Ans: c
Explanation:
Density of air at inlet \left( { \rho }_{ 1 } \right) =\frac { { p }_{ 1 } }{ R{ T }_{ 1 } }
=> { \rho }_{ 1 }=\frac { 3\times { 10 }^{ 5 } }{ 287\times 400 } =2.61kg/{ m }^{ 3 }
In an isentropic flow process:
\frac { { p }_{ 1 } }{ { p }_{ 2 } } ={ \left( \frac { { \rho }_{ 1 } }{ { \rho }_{ 2 } } \right) }^{ \gamma }=> \frac { 300 }{ 50 } ={ \left( \frac { 261 }{ { \rho }_{ 2 } } \right) }^{ \gamma }
=> { \rho }_{ 2 }=0.727kg/{ m }^{ 3 }
18. The mass flow rate of air through the nozzle in kg/s is
(a) 1.30
(b) 1.77
(c) 1.85
(d) 2.06
(2 Mark, 2011)

Ans: d
Explanation:
For isentropic process:
\frac { { T }_{ 1 } }{ { T }_{ 2 } } ={ \left( \frac { { P }_{ 1 } }{ { P }_{ 2 } } \right) }^{ \frac { \gamma 1 }{ \gamma } }=> \frac { 400 }{ { T }_{ 2 } } ={ 6 }^{ \frac { 0.4 }{ 1.4 } }
=> { T }_{ 2 }=239.73K
Using steady flow energy equation:
{ h }_{ 1 }={ h }_{ 2 }+\frac { { V }^{ 2 } }{ 2000 }=> { h }_{ 1 }={ h }_{ 2 }+\frac { { V }^{ 2 } }{ 2000 }
=> \frac { { V }^{ 2 } }{ 2000 } ={ c }_{ p }\left( { T }_{ 1 }{ T }_{ 2 } \right)
=> \frac { { V }^{ 2 } }{ 2000 } =1\left( 400239.73 \right)
=> V = 567.57 m/s
Mass flow rate at the exit (m) = Volume rate \times density
=> m = 0.005\times 567.57\times 0.727
=> m = 2.06 kg/s
19. The contents of a wellinsulated tank are heated by a resistor of 23 Ω in which 10 A current is flowing. Consider the tank along with its contents as a thermodynamic system. The work done by the system and the heat transfer to the system are positive. The rates of heat (Q), work (W) and change in internal energy ( U) during the process in kW are
(a) Q = 0, W = −2.3, U = +2.3
(b) Q = +2.3, W = 0, U = +2.3
(c) Q = −2.3, W = 0, U = −2.3
(d) Q = 0, W = +2.3, U = −2.3
(1 Mark, 2011)

Ans: a
Explanation:
As the tank is wellinsulated: So, Q = 0
Work is in the form of electrical work.
As work is provided into the system, So it is negative.
W= I^{2}R = – { 10 }^{ 2 }\times 23 = 2.3 kW
Using First law of thermodynamics:
Q=\Delta U+W=> \Delta U=W = 2.3 kW
20. A pump handling a liquid raises its pressure from 1 bar to 30 bar. Take the density of the liquid as 990 kg/m^{3}. The isentropic specific work done by the pump in kJ/kg is
(a) 0.10
(b) 0.30
(c) 2.50
(d) 2.93
(2 Mark, 2011)

Ans: d
Explanation:
The isentropic specific work done by the pump = \upsilon \left( { p }_{ 2 }{ p }_{ 1 } \right)
=> W = \frac { 1 }{ \rho } \left( { p }_{ 2 }{ p }_{ 1 } \right)
=> W = \frac { 1 }{ 990 } \left( 301 \right) \times { 10 }^{ 2 }
=> W = 2.93 kJ/kg
21. Steam enters an adiabatic turbine operating at steady state with an enthalpy of 3251.0 kJ/kg and leaves as a saturated mixture at 15 kPa with quality (dryness fraction) 0.9. The enthalpies of the saturated liquid and vapour at 15 kPa are h_{f} = 225.94 kJ/kg and h_{g} = 2598.3 kJ/kg respectively. The mass flow rate of steam is 10 kg/s. Kinetic and potential energy changes are negligible. The power output of the turbine in MW is
(a) 6.5
(b) 8.9
(c) 9.1
(d) 27.0
(2 Mark, 2012)

Ans: b
Explanation:
Enthalpy at the exit:
{ h }_{ 2 }={ h }_{ f }+x({ h }_{ g }{ h }_{ f })=> { h }_{ 2 }=225.94+0.9(2598.3225.9)
=> { h }_{ 2 }=2361.1kJ/kg
Turbine power (P)= \dot { m } ({ h }_{ 1 }{ h }_{ 2 })
=> P = 10\times (32512361.1)
=> P = 8900 kW = 8.9 MW
Statement for Linked Answer Questions 22 and 23:
Air enters an adiabatic nozzle at 300 kPa, 500 K with a velocity of 10 m/s. It leaves the nozzle at 100 kPa with a velocity of 180 m/s. The inlet area is 80 cm^{2}. The specific heat of air C_{p} is 1008 J/kg.K.
22. The exit temperature of the air is
(a) 516 K
(b) 532 K
(c) 484 K
(d) 468 K
(2 Mark, 2012)

Ans: c
Explanation:
Using steady flow energy equation for a nozzle
{ h }_{ i }+\frac { { v }_{ i }^{ 2 } }{ 2000 } ={ h }_{ o }+\frac { { v }_{ o }^{ 2 } }{ 2000 } .
{ { c }_{ p }T }_{ i }+\frac { { v }_{ i }^{ 2 } }{ 2000 } ={ { c }_{ p }T }_{ o }+\frac { { v }_{ o }^{ 2 } }{ 2000 } .
1.008\times 500+\frac { { 10 }^{ 2 } }{ 2000 } ={ 1.008\times T }_{ o }+\frac { { 180 }^{ 2 } }{ 2000 } .
So, { T }_{ o }=484K
23. The exit area of the nozzle in cm^{2} is
(a) 90.1
(b) 56.3
(c) 4.4
(d) 12.9
(2 Mark, 2012)

Ans: d
Explanation:
Density can be calculated from the ideal gas equation.
\rho =\frac { p }{ RT }Using continuity equation in the nozzle:
{ \rho }_{ i }{ A }_{ i }{ V }_{ i }={ \rho }_{ 0 }{ A }_{ 0 }{ V }_{ 0 }=> \frac { { A }_{ i }{ V }_{ i }{ p }_{ i } }{ { T }_{ i } } { =\frac { { A }_{ 0 }{ V }_{ 0 }{ p }_{ 0 } }{ { T }_{ 0 } } }
=> \frac { 80\times 10\times 300 }{ 500 } =\frac { { A }_{ 0 }\times 180\times 100 }{ 484 }
=> { A }_{ 0 }=12.9{ cm }^{ 2 }
24. Specific enthalpy and velocity of steam at inlet and exit of a steam turbine, running under steady state, are as given below:
The rate of heat loss from the turbine per kg of steam flow rate is 5 kW. Neglecting changes in potential energy of steam, the power developed in kW by the steam turbine per kg of steam flow rate, is
(a) 901.2
(b) 911.2
(c) 17072.5
(d) 17082.5
(2 Mark, 2013)

Ans: a
Explanation:
Using steady flow energy equation:
{ h }_{ 1 }+\frac { { v }_{ 1 }^{ 2 } }{ 2000 } +Q={ h }_{ 2 }+\frac { { v }_{ 2 }^{ 2 } }{ 2 } +W=> 3250+\frac { { 180 }^{ 2 } }{ 2000 } 5=2360+\frac { 5^{ 2 } }{ 2000 } +W
=> W = 901.2 kW/kg
25. 1.5 kg of water is in saturated liquid state at 2 bar (v_{f} = 0.001061 m^{3}/kg, u_{f} = 504.0 kJ/kg, h_{f} = 505 kJ/kg). Heat is added in a constant pressure process till the temperature of water reaches 400°C (v = 1.5493 m^{3}/kg, u = 2967.0 kJ/kg, h = 3277.0 kJ/kg). The heat added (in kJ) in the process is ____.
(2 Mark, 2014[1])

Ans: 4158
Explanation:
At constant pressure the heat addition:
Q = m(h_{2} – h_{1}) = 1.5(3277 – 505) = 4158 kJ
26. A pure substance at 8 MPa and 400°C is having a specific internal energy of 2864 kJ/kg and a specific volume of 0.03432 m^{3}/kg. Its specific enthalpy (in kJ/kg) is ___.
(1 Mark, 2014[2])

Ans: 3138.5
Explanation:
h = u + pv
=> h = 2864 + 8000 * 0.03432 = 3138.56 kJ/kg
26. A well insulated rigid container of volume 1m^{3} contains 1.0 kg of an ideal gas [C_{p} = 1000 J/(kg.K) and C_{v} = 800 J/(kg.K)] at a pressure of 10^{5} Pa. A stirrer is rotated at constant rpm in the container for 1000 rotations and the applied torque is 100 Nm. The final temperature of the gas (in K) is
(a) 500.0
(b) 773.0
(c) 785.4
(d) 1285.4
(2 Mark, 2015[1])

Ans: d
Explanation:
Gas constant (R) = C_{p }– C_{v }= 0.2 kJ/kgK
{ T }_{ 1 }=\frac { { p }_{ 1 }{ V }_{ 1 } }{ mR } =\frac { 100\times 1 }{ 0.2\times 1 } =500KWork done (W) = 2\pi \times 100\times 1000=628.318kJ
[Work done is negative as it is one on the system]
According to first law of thermodynamics:
\delta Q\delta W=dE=> 0(628.318)=m{ c }_{ v }({ T }_{ 2 }{ T }_{ 1 })
=> 628.318=1\times 0.8({ T }_{ 2 }500)
=> { T }_{ 2 }=1285.4K
27. Work is done on an adiabatic system due to which its velocity changes from 10 m/s to 20 m/s, elevation increases by 20 m and temperature increases by 1 K. The mass of the system is 10 kg, C_{v} = 100 J/(kg.K) and gravitational acceleration is 10 m/s^{2}. If there is no change in any other component of the energy of the system, the magnitude of total work done (in kJ) on the system is ____.
(2 Mark, 2015[2])

Ans: 4.5
Explanation:
Steady flow energy equation:
\dot { m } ({ h }_{ 1 }+\frac { { v }_{ 1 }^{ 2 } }{ 2 } +g{ z }_{ 1 })+\dot { Q } =\dot { m } ({ h }_{ 2 }+\frac { { v }_{ 2 }^{ 2 } }{ 2} +g{ z }_{ 2 })+\dot { W }=> \frac { { 10 }^{ 2 } }{ 2 } =100\times 1+\frac { { 20 }^{ 2 } }{ 2 } +10\times 20+W
=> W = – 4.5 kJ
So, Work done on the system is 4.5 kJ.
28. Steam enters a turbine at 30 bar, 300°C (u = 2750 kJ/kg, h = 2993 kJ/kg) and exits the turbine as saturated liquid at 15 kPa (u = 225 kJ/kg, h = 226 kJ/kg). Heat loss to the surrounding is 50 kJ/kg of steam flowing through the turbine. Neglecting changes in kinetic energy and potential energy, the work output of the turbine (in kJ/kg of steam) is ___.
(2 Mark, 2015[3])

Ans: 2717
Explanation:
Steady flow energy equation:
\dot { m } ({ h }_{ 1 }+\frac { { v }_{ 1 }^{ 2 } }{ 2 } +g{ z }_{ 1 })+\dot { Q } =\dot { m } ({ h }_{ 2 }+\frac { { v }_{ 2 }^{ 2 } }{ 2} +g{ z }_{ 2 })+\dot { W }For the steam turbine: h_{1 }+ Q = h_{2 }+ W
=> W = 2993 – 50 – 226 = 2717 kJ/kg
29. A pistoncylinder device initially contains 0.4 m^{3} of air (to be treated as an ideal gas) at 100 kPa and 80^{0}C. The air is now isothermally compressed to 0.1 m^{3}. The work done during this process is ____ kJ.
(Take the sign convention such that work done on the system is negative)
(2 Mark, 2016[2])

Ans: 55.45
Explanation:
Work done in isothermal process = { P }_{ 1 }{ V }_{ 1 }ln\frac { { V }_{ 2 } }{ { V }_{ 1 } }
=> W = 100\times 0.4\times ln\frac { 0.1 }{ 0.4 } = 55.45 kJ
30. The internal energy of an ideal gas is a function of
(a) Temperature and pressure
(b) Volume and pressure
(c) Entropy and pressure
(d) Temperature only
(1 Mark, 2016[2])

Ans: d
31. Steam at an initial enthalpy of 100 kJ/kg and inlet velocity of 100 m/s, enters an insulated horizontal nozzle. It leaves the nozzle at 200 m/s. The exit enthalpy (in kJ/kg) is ______.
(2 Mark, 2016[3])

Ans: 85
Explanation:
Steady flow energy equation for nozzle:
{ h }_{ 1 }+\frac { { v }_{ 1 }^{ 2 } }{ 2000 } ={ h }_{ 2 }+\frac { { v }_{ 2 }^{ 2 } }{ 2000 }=> 100+\frac { { 100 }^{ 2 } }{ 2000 } ={ h }_{ 2 }+\frac { { 200 }^{ 2 } }{ 2000 }
=> h_{2 }= 85 kJ/kg
32. Water (density = 1000 kg/m^{3}) at ambient temperature flows through a horizontal pipe of uniform cross section at the rate of 1 kg/s. If the pressure drop across the pipe is 100 kPa, the minimum power required to pump the water across the pipe, in watts, is ____.
(1 Mark, 2017[1])

Ans: 100
Explanation:
Given: v=\frac { m }{ \rho } =\frac { 1 }{ 1000 } { m }^{ 3 }/s
Minimum power required to pump = v(P_{2} – P_{1}) = \frac { 1 }{ 1000 } \times 100\times { 10 }^{ 3 } = 100 watt
33. The molar specific heat at constant volume of an ideal gas is equal to 2.5 times the universal gas constant (8.314 J/mol.K). When the temperature increases by 100 K, the change in molar specific enthalpy is ____ J/mol.
(1 Mark, 2017[1])

Ans: 2909.9
Explanation:
Given: C_{v }= 2.5R
We know: C_{p }– C_{v }= R
=> C_{p }= 3.5R
The temperature increases (dT) = 100 K
Change in molar specific enthalpy (dh) = C_{p}(dT) = 3.5\times 8.314\times 100 = 2909.9 J/mol
35.A calorically perfect gas (specific heat at constant pressure 1000 j/kgK) enters and leaves a gas turbine with the same velocity. The temperatures of the gas at turbine entry and exit are 1100 K and 400 K, respectively. The power produced is 4.6 MW and heat escapes at the rate of 300 kJ/s through the turbine casing. The mass flow rate of the gas (in kg/s) through the turbine is
(a) 6.14
(b) 7.00
(c) 7.50
(d) 8.00
(2 Mark, 2017[2])

Ans: b
Explanation:
Steady flow energy equation:
\dot { m } ({ h }_{ 1 }+\frac { { v }_{ 1 }^{ 2 } }{ 2 } +g{ z }_{ 1 })+\dot { Q } =\dot { m } ({ h }_{ 2 }+\frac { { v }_{ 2 }^{ 2 } }{ 2} +g{ z }_{ 2 })+\dot { W }=> \dot { m } { h }_{ 1 }+\dot { Q } =\dot { m } { h }_{ 2 }+\dot { W }
=> \dot { m } { c }_{ p }({ T }_{ 1 }{ T }_{ 2 })=\dot { W } \dot { Q }
=> \dot { m } { \times }1\times (1100400)=4600+300
=> \dot { m } = 7 kg/s
36. Steam flows through a nozzle at a mass flow rate of \dot { m } = 0.1 kg/s with a heat loss of 5 kW. The enthalpies at inlet and exit are 2500kJ/kg and 2350 kJ/kg, respectively. Assuming negligible velocity at inlet (C_{1} \approx 0), the velocity (C_{2}) of steam (in m/s) at the nozzle exit is ______ (correct to two decimal places).
(2 Mark, 2018[1])

Ans: 447.213
Explanation:
Steady flow energy equation:
\dot { m } ({ h }_{ 1 }+\frac { { v }_{ 1 }^{ 2 } }{ 2 } +g{ z }_{ 1 })+\dot { Q } =\dot { m } ({ h }_{ 2 }+\frac { { v }_{ 2 }^{ 2 } }{ 2} +g{ z }_{ 2 })+\dot { W }=> \dot { m } { h }_{ 1 }+\dot { Q } =\dot { m } \left( { h }_{ 2 }+\frac { { v }_{ 2 }^{ 2 } }{ 2000 } \right)
=> 0.1\times 25005=0.1\times 2350+0.1\times \frac { { c }_{ 2 }^{ 2 } }{ 2000 }
=> { c }_{ 2 }=447.213m/s
37. During a nonflow thermodynamic process (12) executed by a perfect gas, the heat interaction is equal to the work interaction \left( { Q }_{ 12 }={ W }_{ 12 } \right) when the process is
(A) Isentropic
(B) Isothermal
(C) Polytropic
(D) Adiabatic
(1 Mark, 2019[1])

Ans: b
Explanation:
According to first law of thermodynamics for a process:
\delta Q=dU+\delta WFor isothermal process:
Change in internal energy (dE) = 0
So, Q_{12} = W_{12}
38. A gas is heated in a duct as it flows over a resistance heater. Consider a 101 kW electric heating system. The gas enters the heating section of the duct at 100 kPa and 27°C with a volume flow rate of 15 m^{3}/s. If heat is lost from the gas in the duct to the surroundings at a rate of 51 kW, the exit temperature of the gas is (Assume constant pressure, ideal gas, negligible change in kinetic and potential energies and constant specific heat; C_{p} = 1 kJ/kg.K; R = 0.5 kJ/kg.K).
(a) 53°C
(b) 32°C
(c) 37°C
(d) 76°C
(2 Mark, 2019[1])

Ans: b
Explanation:
From Ideal gas equation:
{ P }_{ 1 }{ \dot { V } }_{ 1 }=\dot { m } R{ T }_{ 1 }=> \dot { m } =\frac { { P }_{ 1 }{ \dot { V } }_{ 1 } }{ R{ T }_{ 1 } } =\frac { 100\times 15 }{ 0.5\times 300 } =10kg/s
Applying steady flow energy equation:
\dot { m } { h }_{ 1 }+Q=\dot { m } { h }_{ 2 }+W=> \dot { m } { C }_{ P }T_{ 1 }+Q=\dot { m } { { C }_{ P }T }_{ 2 }+W……….(1)
Given:
Heat is lost to the surrounding. So, Q = 51 kW
Work input by electric heating system. So, W = 101 kW
Putting the abive values in equation (1), We get:
T_{2 }= 305 K or 32^{0}C