Category: Thermodynamics Objective Questions

Thermodynamics Multiple Choice Questions on “Clausius-Clapeyron Equation”.

1. During phase transitions like vaporization, melting and sublimation
a) pressure and temperature remains constant
b) volume and entropy changes
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: This is what happens during a phase transition.

2. Which of the following requirement is satisfied by a phase change of the first order?
a) there are changes of volume and entropy
b) the first-order derivative of the Gibbs function changes discontinuously
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: These requirements must be satisfied for a phase change to be of first order.

3. The Clausius-Clapeyron equation is given by
a) dp/dT = l / T(vf+vi)
b) dp/dT = l / T(vf-vi)
c) dT/dp = l / T(vf+vi)
d) dT/dp = l / T(vf-vi)
Answer: b
Clarification: Here vf is the final specific volume and vi is the initial specific volume and l is the latent heat.

4. Water ____ on melting and has the fusion curve with a ____ slope.
a) contracts, negative
b) contracts, positive
c) expands, negative
d) expands, positive
Answer: a
Clarification: Unlike other substances which expands on melting, water contracts on melting and hence the slope of the fusion curve is negative.

5. The vapour pressure curve is of the form ln(p) = A + B/T + C*lnT + DT.
a) true
b) false
Answer: a
Clarification: This is the form of vapour pressure curve where A,B,C, and D are constants.

6. According to Trouton’s rule, the ratio of latent heat of vaporization to the boiling point at 1.013 bar is
a) 77 kJ/kgmol K
b) 88 kJ/kgmol K
c) 99 kJ/kgmol K
d) 100 kJ/kgmol K
Answer: b
Clarification: This is the statement of Trouton’s rule.

7. The vapour pressure p in kPa at temperature T can be given by the relation
a) p = 101.325 exp (88/R)(1+T/Tb)
b) p = 101.325 exp (88/R)(1+Tb/T)
c) p = 101.325 exp (88/R)(1-T/Tb)
d) p = 101.325 exp (88/R)(1-Tb/T)
Answer: d
Clarification: Here Tb is the boiling point at 1.013 bar and this relation comes from the latent heat of vaporization and Trouton’s rule.

8. At the triple point, l(sublimation) = l(vaporization) – l(fusion).
a) true
b) false
Answer: b
Clarification: At the triple point, l(sublimation) = l(vaporization) + l(fusion), where l is the latent heat.

9. The slope of sublimation curve is ____ the slope of the vaporization curve at triple point.
a) equal to
b) less than
c) greater than
d) none of the mentioned
Answer: c
Clarification: This is because at triple point, l(sublimation) > l(vaporization).

10. Latent heat of sublimation is given by
a) l(sublimation) = -2.303*(R)*(d(1/T)/d(log p ))
b) l(sublimation) = +2.303*(R)*(d(log p )/d(1/T))
c) l(sublimation) = +2.303*(R)*(d(1/T)/d(log p ))
d) l(sublimation) = -2.303*(R)*(d(log p )/d(1/T))
Answer: d
Clarification: This is the expression for finding the latent heat of sublimation.

11. An application requires R-12 at −140°C. The triple-point temperature is −157°C. Find the pressure of the saturated vapour at the required condition.
a) 0.0058 kPa
b) 0.0098 kPa
c) 0.0068 kPa
d) 0.0088 kPa
Answer: b
Clarification: The lowest temperature for R-12 is -90°C, so it must be extended to -140°C using the Clapeyron equation.
at T1= -90°C = 183.2 K, P1 = 2.8 kPa
R = 8.3145/120.914 = 0.068 76 kJ/kg K
ln P/P1 = (hfg/R)(T-T1)/(T*T1)
= (189.748/0.068 76)[(133.2 – 183.2)/(133.2 × 183.2)] = -5.6543
P = 2.8 exp(-5.6543) = 0.0098 kPa.

12. Ice (solid water) at −3°C and 100 kPa, is compressed isothermally until it becomes liquid. Find the required pressure.
a) 20461 kPa
b) 30461 kPa
c) 40461 kPa
d) 50461 kPa
Answer: c
Clarification: Water, triple point T = 0.01°C, P = 0.6113 kPa, vf = 0.001 m^3/kg,
hf = 0.01 kJ/kg, vi= 0.001 0908 m^3/kg, hi = -333.4 kJ/kg
dPif/dT = (hf – hi)/[(vf – vi)T] = 333.4/(-0.0000908 × 273.16) = -13442 kPa/K
∆P = (dPif/dT)*∆T = -13442(-3 – 0.01) = 40460 kPa
P = P(tp) + ∆P = 40461 kPa.

13. Estimate the freezing temperature of liquid water at a pressure of 30 MPa.
a) -2.2°C
b) 0°C
c) -0.2°C
d) -1.2°C
Answer: a
Clarification: At the triple point,
vif = vf – vi = 0.001000 – 0.0010908 = -0.0000908 m^3/kg
hif = hf – hi = 0.01 – (-333.40) = 333.41 kJ/kg
dPif/dT = 333.41/[(273.16)(-0.0000908)] = -13 442 kPa/K
at P = 30 MPa, T = 0.01 + (30 000-0.6)/(-13 442) = = -2.2°C.

Thermodynamics Multiple Choice Questions on “Diesel Cycle and Dual Cycle”.

1. In SI engines,
a) air-fuel mixture is compressed
b) compression ratio is limited
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: The onset of engine knock or auto-ignition limits the compression ratio in SI engines.

2. In CI engines,
a) during compression stroke, only air is compressed
b) compression ratios can be much higher
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: This is an advantage of CI engine over SI engine.

3. The correct sequence of processes in CI engine is
a) intake->fuel injection and combustion->compression->expansion->exhaust
b) intake->compression->fuel injection and combustion->expansion->exhaust
c) intake->compression->expansion->fuel injection and combustion->exhaust
d) intake->compression->exhaust->fuel injection and combustion->expansion
Answer: b
Clarification: The correct sequence of processes in CI engine is intake->compression->fuel injection and combustion->expansion->exhaust.

4. The processes in CI engine cycle is completed in ____ strokes of piston and ____ revolutions of crankshaft.
a) four, four
b) two, two
c) two, four
d) four, two
Answer: d
Clarification: There are four strokes and number of revolutions of crankshaft required are two.

5. The Diesel cycle consists of
a) two reversible isotherms and two reversible isobars
b) one reversible isochore and two reversible adiabatics and one reversible isobar
c) one reversible isotherm and two reversible isochores and one reversible isobar
d) two reversible isobars and two reversible adiabatics
Answer: b
Clarification: These four processes comprises Diesel cycle.

6. Which of the following is the relation between compression ratio, expansion ratio and cut-off ratio?
a) rc=(rk)*(re)
b) re=(rk)*(rc)
c) rk=(re)*(rc)
d) none of the mentioned
Answer: c
Clarification: Here rk=compression ratio, re=expansion ratio and rc=cut-off ratio.

7. The efficiency of Diesel cycle is ____ the efficiency of Otto cycle.
a) less than
b) greater than
c) equal to
d) none of the mentioned
Answer: a
Clarification: This comes from the formula for the efficiency of Diesel cycle.

8. In Dual cycle,
a) all the heat is added at constant volume
b) all the heat is added at constant pressure
c) some heat is added at constant volume and remaining at constant pressure
d) none of the mentioned
Answer: c
Clarification: This is the reason why Dual cycle is also called Mixed cycle.

9. The constant volume pressure ratio is given by the ratio of pressures of constant volume heat addition.
a) true
b) false
Answer: a
Clarification: This ratio is used in Dual cycle.

10. Detonation in SI engine is
a) noisy and destructive combustion phenomenon
b) limits the compression ratio
c) it depends on engine design and fuel
d) all of the mentioned
Answer: d
Clarification: This is what detonation means and its causes.

11. The premature ignition of fuel is called ____
a) engine knock
b) auto-ignition
c) detonation
d) all of the mentioned
Answer: b
Clarification: Engine knock or detonation is the audible noise produces by auto-ignition.

12. The auto-ignition
a) reduces performance of engine
b) can cause damage to engine
c) sets upper limit to compression ratios used in SI engines
d) all of the mentioned
Answer: d
Clarification: Thus auto-ignition should always be avoided.

13. Many a times, tetraethyl ether is added to gasoline.
a) true
b) false
Answer: a
Clarification: This is done to raise the octane rating of fuel.

14. For a given compression raise, the highest efficiency can be obtained by using
a) triatomic gases
b) diatomic gases
c) monoatomic gases
d) all of the mentioned
Answer: c
Clarification: Monoatomic gases like helium and argon has highest value of ɣ.

15. ɣ ____ with temperature.
a) increases
b) decreases
c) remains constant
d) none of the mentioned
Answer: b
Clarification: This is a property of ɣ .

Thermodynamics Multiple Choice Questions on “Volumetric Efficiency”.

1. The temperature and pressure conditions at free air delivery are
a) 27 degree Celsius, 100 bar
b) 15 degree Celsius, 101.325 bar
c) 27 degree Celsius, 101.325 bar
d) 15 degree Celsius, 100 bar
Answer: b
Clarification: This is known as FAD(free air delivery).

2. The volumetric efficiency is defined as the ratio of
a) total volume / piston displacement volume
b) total volume / gas volume taken during suction
c) gas volume taken during suction / swept volume
d) swept volume / gas volume taken during suction
Answer: c
Clarification: The swept volume is also called piston displacement volume.

3. Clearance is given by
a) total volume / swept volume
b) total volume / clearance volume
c) swept volume / clearance volume
d) clearance volume / swept volume
Answer: d
Clarification: Here C=Vc/Vs.

4. The clearance volumetric efficiency is equal to
a) 1 + C + C(p2/p1)^(1/n)
b) 1 – C – C(p2/p1)^(1/n)
c) 1 – C + C(p2/p1)^(1/n)
d) 1 + C – C(p2/p1)^(1/n)
Answer: d
Clarification: Here C is the clearance and p1,p2 are pressures.

5. As clearance and pressure ratio increases, volumetric efficiency ____
a) decreases
b) increases
c) remains constant
d) none of the mentioned
Answer: a
Clarification: This comes from the equation of volumetric efficiency.

6. To get maximum flow capacity, compressors are built with maximum practical clearance.
a) true
b) false
Answer: b
Clarification: To get maximum flow capacity, compressors are built with minimum practical clearance.

7. When the clearance volume is at minimum level,
a) volumetric efficiency is maximum
b) flow through machine is maximum
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: We can get this from volumetric efficiency equation.

8. For a given pressure ratio, volumetric efficiency is zero when maximum clearance is
a) 1 / ((p2/p1)^(1/n) +1)
b) 1 / ((p2/p1)^(1/n) -1)
c) 1 / ((p1/p2)^(1/n) -1)
d) 1 / ((p1/p2)^(1/n) +1)
Answer: b
Clarification: This comes from the equation of volumetric efficiency when we put volumetric efficiency equal to zero.

9. Increasing the pressure ratio, increases the volumetric efficiency.
a) true
b) false
Answer: b
Clarification: For a fixed clearance, increasing the pressure ratio decreases the volumetric efficiency.

10. Which of the following is correct?
a) p2max/p1 = (1 + 1/C)ⁿ
b) p2max/p1 = (1 – 1/C)ⁿ
c) p2max/p1 = (1 – C)ⁿ
d) p2max/p1 = (1 + C)ⁿ
Answer: a
Clarification: the maximum pressure ratio which can be attained by a reciprocating compressor cylinder is limited by the clearance.

Thermodynamics Multiple Choice Questions on “Conduction, Convection and Radiation”.

1. The sun shines on a 150 m² road surface so it is at 45°C. Below the 5cm thick asphalt(average conductivity of 0.06 W/m K), is a layer of rubbles at 15°C. Find the rate of heat transfer to the rubbles.
a) 5300 W
b) 5400 W
c) 5500 W
d) 5600 W
Answer: b
Clarification: There is conduction through the asphalt layer.
heat transfer rate = k A ∆T/∆x = 0.06 × 150 ×(45-15)/0.05
= 5400 W.

2. A pot of steel(conductivity 50 W/m K), with a 5 mm thick bottom is filled with liquid water at 15°C. The pot has a radius of 10 cm and is now placed on a stove that delivers 250 W as heat transfer. Find the temperature on the outer pot bottom surface assuming the inner surface to be at 15°C.
a) 15.8°C
b) 16.8°C
c) 18.8°C
d) 19.8°C
Answer: a
Clarification: Steady conduction, Q = k A ∆T/∆x ⇒ ∆Τ = Q ∆x / kΑ
∆T = 250 × 0.005/(50 × π/4 × 0.2²) = 0.796
T = 15 + 0.796 = 15.8°C.

3. A water-heater is covered with insulation boards over a total surface area of 3 m². The inside board surface is at 75°C and the outside being at 20°C and the conductivity of material being 0.08 W/m K. Find the thickness of board to limit the heat transfer loss to 200 W ?
a) 0.036 m
b) 0.046 m
c) 0.056 m
d) 0.066 m
Answer: d
Clarification: Steady state conduction through board.
Q = k A ∆T/∆x ⇒ ∆Τ = Q ∆x / kΑ
∆x = 0.08 × 3 ×(75 − 20)/200 = 0.066 m.

4. On a winter day with atmospheric air at −15°C, the outside front wind-shield of a car has surface temperature of +2°C, maintained by blowing hot air on the inside surface. If the wind-shield is 0.5 m² and the outside convection coefficient is 250 W/Km², find the rate of energy loss through front wind-shield.
a) 125 W
b) 1125 W
c) 2125 W
d) 3125 W
Answer: c
Clarification: Q (conv) = h A ∆Τ = 250 × 0.5 × [2 − ( −15)]
= 250 × 0.5 × 17 = 2125 W.

5. A large heat exchanger transfers a total of 100 MW. Assume the wall separating steam and seawater is 4 mm of steel, conductivity 15 W/m K and that a maximum of 5°C difference between the two fluids is allowed. Find the required minimum area for the heat transfer.
a) 180 m²
b) 280 m²
c) 380 m²
d) 480 m²
Answer: d
Clarification: Steady conduction
Q = k A ∆T/∆x ⇒ Α = Q ∆x / k∆Τ
A = 100 × 10^6 × 0.004 / (15 × 5) = 480 m².

6. The black grille on the back of a refrigerator has a surface temperature of 35°C with a surface area of 1 m². Heat transfer to the room air at 20°C takes place with convective heat transfer coefficient of 15 W/Km^2. How much energy is removed during 15 minutes of operation?
a) 202.5 kJ
b) 212.5 kJ
c) 222.5 kJ
d) 232.5 kJ
Answer: a
Clarification: Q = hA ∆T ∆t, Q = 15 × 1 × (35-20)×15×60 = 202500 J = 202.5 kJ.

7. A small light bulb (25 W) inside a refrigerator is kept on and 50 W of energy from the outside seeps into the refrigerated space. How much of temperature difference to the ambient(at 20°C) must the refrigerator have in its heat exchanger having an area of 1 m² and heat transfer coefficient of 15 W/Km² to reject the leak of energy.
a) 0°C
b) 5°C
c) 10°C
d) 15°C
Answer: b
Clarification: Total energy that goes out = 50+25 = 75 W
75 = hA∆T = 15 × 1 × ∆T hence ∆T = 5°C.

8. As the car slows down, the brake shoe and steel drum continuously absorbs 25 W. Assume a total outside surface area of 0.1 m² with a convective heat transfer coefficient of 10 W/Km² to the air at 20°C. How hot does the outside brake and drum surface become when steady conditions are reached?
a) 25°C
b) 35°C
c) 45°C
d) 55°C
Answer: c
Clarification: ∆Τ = heat / hA hence ∆T = [ Τ(BRAKE) − 20 ] = 25/(10 × 0.1) = 25°C
Τ(BRAKE) = 20 + 25 = 45°C.

9. A burning wood in the fireplace has a surface temperature of 450°C. Assume the emissivity to be 1 and find the radiant emission of energy per unit area.
a) 15.5 kW/m²
b) 16.5 kW/m²
c) 17.5 kW/m²
d) 18.5 kW/m²
Answer: a
Clarification: Q /A = 1 × σ T^4
= 5.67 × 10^–8×( 273.15 + 450)⁴
= 15505 W/m² = 15.5 kW/m².

10. A radiant heat lamp is a rod, 0.5 m long, 0.5 cm in diameter, through which 400 W of electric energy is deposited. Assume the surface emissivity to be 0.9 and neglecting incoming radiation, find the rod surface temperature?
a) 700K
b) 800K
c) 900K
d) 1000K
Answer: d
Clarification: Outgoing power equals electric power
T⁴= electric energy / εσA
= 400 / (0.9 × 5.67 ×10^-8× 0.5 × π × 0.005)
= 9.9803 ×10^11 K⁴ ⇒ T = 1000K.

11. A water-heater is covered up with insulation boards over a total surface area of 3 m². The inside board surface is at 75°C and the outside surface is at 20°C and the board material has a conductivity of 0.08 W/m K. How thick a board should it be to limit the heat transfer loss to 200 W ?
a) 0.066 m
b) 0.166 m
c) 0.266 m
d) 0.366 m
Answer: a
Clarification: Steady state conduction through a single layer board.
Δx = kA(ΔT)/Q
Δx = (0.08*3)*(75-20)/200 = 0.066 m.

12. Find the rate of conduction heat transfer through a 1.5 cm thick hardwood board, k = 0.16 W/m K, with a temperature difference between the two sides of 20°C.
a) 113 W/m²
b) 213 W/m²
c) 230 W/m²
d) 312 W/m²
Answer: b
Clarification: . q = .Q/A = k ΔT/Δx = 0.16 Wm /K × 20K/0.015 m = 213 W/m².

13. A 2 m² window has a surface temperature of 15°C and the outside wind is blowing air at 2°C across it with a convection heat transfer coefficient of h = 125 W/m2K. What is the total heat transfer loss?
a) 2350 W
b) 1250 W
c) 2250 W
d) 3250 W
Answer: d
Clarification: .Q = h A ΔT = 125 W/m2K × 2 m2 × (15 – 2) K = 3250 W.

14. A radiant heating lamp has a surface temperature of 1000 K with ε = 0.8. How large a surface area is needed to provide 250 W of radiation heat transfer?
a) 0.0035 m2
b) 0.0045 m2
c) 0.0055 m2
d) 0.0065 m2
Answer: c
Clarification: .Q = εσAT^4
A = .Q/(εσT4) = 250/(0.8 × 5.67 × 10-8 × 10004)
= 0.0055 m².

Thermodynamics Multiple Choice Questions on “Carnot Theorem, Carnot Cycle and Reversed Heat Engine”.

1. Carnot cycle is a reversible cycle.
a) true
b) false
Answer: a
Clarification: A reversible cycle is an ideal hypothetical cycle in which all processes are reversible.

2. A reversible cycle has following processes.
a) 4 isothermal processes
b) 4 adiabatic processes
c) 2 isothermal and 2 adiabatic processes
d) none of the mentioned
Answer: c
Clarification: Two reversible isotherms and two reversible adiabatics constitute a Carnot cycle.

3. The correct sequence of the processes taking place in a carnot cycle is
a) adiabatic -> adiabatic -> isothermal -> isothermal
b) adiabatic -> isothermal -> adiabatic -> isothermal
c) isothermal -> isothermal -> adiabatic -> adiabatic
d) isothermal -> adiabatic -> isothermal -> adiabatic
Answer: d
Clarification: Carnot cycle consists if these four processes in succession.

4. The reversed heat engine takes heat from a ___ temperature body, then discharges it to a ___ temperature body and ___ an inward flow of network.
a) high, low, receives
b) low, high, receives
c) high, low, gives
d) low, high, gives
Answer: b
Clarification: In reversed heat engine, the magnitude of energy transfers remains same and only directions change.

5. Example of reversed heat engine is
a) heat pump
b) refrigerator
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: Heat pump and refrigerator are the types of reversed heat engine.

6. According to Carnot’s theorem, all heat engines operating between a given constant temperature source and sink, none has a higher efficiency than a reversible engine.
a) true
b) false
Answer: a
Clarification: This is the statement of Carnot’s theorem .

7. The efficiency of all reversible heat engines operating between the same heat reservoirs is
a) same
b) independent of the nature of working substance
c) independent of the amount of working substance
d) all of the mentioned
Answer: d
Clarification: This statement is a corollary of Carnot’s theorem.

8. Efficiency of a reversible heat engine is given by
a) 1-(T1/T2)
b) 1-(T2/T1)
c) (T1/T2)-1
d) (T2/T1)-1
Answer: b
Clarification: Efficiency=1-(Q2/Q1) and T2,T1 are temperatures at which heat is rejected and received.

9. For a reversible refrigerator, Coefficient of Performance is given by
a) T2/(T1-T2)
b) T1/(T1-T2)
c) T2/(T2-T1)
d) T1/(T2-T1)
Answer: a
Clarification: For a reversible refrigerator, (Q1/Q2)=(T1/T2).

10. For a reversible heat pump, COP is given by
a) T2/(T1-T2)
b) T1/(T1-T2)
c) T2/(T2-T1)
d) T1/(T2-T1)
Answer: b
Clarification: For a reversible heat pump we have, (Q1/Q2)=(T1/T2).

Thermodynamics Multiple Choice Questions on “Work in a Reversible Process-1”.

1. Hot air at 1500 K expands in a polytropic process to a volume 6 times as large with n = 1.5. Find the specific boundary work.
a) 309.5 kJ/kg
b) 409.5 kJ/kg
c) 509.5 kJ/kg
d) 609.5 kJ/kg
Answer: c
Clarification: u1 = 444.6 kJ/kg, u2 = 1205.25 kJ/kg
T2 = T1(v1/v2)^(n-1) = 1500(1/6)^0.5 = 612.4 K
1w2 = R(T2-T1)/(1-n) = 0.287(612.4 – 1500)/(1 – 1.5) = 509.5 kJ/kg.

2. In a Carnot-cycle heat pump, heat is rejected from R-22 at 40°C, during which the R-22 changes from saturated vapor to saturated liquid. The heat is transferred to the R-22 at 0°C. Determine the COP for the cycle.
a) 6.83
b) 7.83
c) 8.83
d) 9.83
Answer: b
Clarification: s4 = s3 = 0.3417 kJ/kg K = 0.1751 + x4(0.7518) => x4 = 0.2216
s1 = s2 = 0.8746 kJ/kg K = 0.1751 + x1(0.7518) => x1 = 0.9304
β′ = q/w = Th/(Th – Tl) = 313.2/40 = 7.83.

3. 1kg of ammonia in a piston/cylinder at 50°C, 1000 kPa is expanded in a reversible isothermal process to 100 kPa. Find the work for this process.
a) 333.75 kJ
b) 343.75 kJ
c) 353.75 kJ
d) 363.75 kJ
Answer: d
Clarification: 1W2 = ⌠ PdV
State 1: u1 = 1391.3 kJ/kg; s1 = 5.265 kJ/kg K
State 2: u2 = 1424.7 kJ/kg; s2 = 6.494 kJ/kg K;
v2 = 1.5658 m^3/kg; h2 = 1581.2 kJ/kg
1Q2 = 1 kg (273 + 50) K (6.494 – 5.265) kJ/kg K = 396.967 kJ
1W2 = 1Q2 – m(u2 – u1) = 363.75 kJ.

4. 1kg of ammonia in a piston/cylinder at 50°C, 1000 kPa is expanded in a reversible isobaric process to 140°C. Find the work in the process.
a) 50.5 kJ
b) 60.5 kJ
c) 70.5 kJ
d) 80.5 kJ
Answer: a
Clarification: 1W2 = mP(v2 – v1)
v1 = 0.145 m^3/kg, u1 = 1391.3 kJ/kg
v2 = 0.1955 m^3/kg, u2 = 1566.7 kJ/kg
1W2 = 1 × 1000(0.1955 – 0.145) = 50.5 kJ.

5. 1kg of ammonia in a piston/cylinder at 50°C, 1000 kPa is expanded in a reversible adiabatic process to 100 kPa. Find the work for this process.
a) 222.4 kJ
b) 232.4 kJ
c) 242.4 kJ
d) 252.4 kJ
Answer: b
Clarification: 1Q2 = 0 ⇒ s2 = s1 and u1 = 1391.3 kJ/kg, s1 = 5.2654 kJ/kg K
sg2 = 5.8404 kJ/kg K, sf = 0.1192 kJ/kg K; x2 = (s – sf)/sfg
x2 = (5.2654 − 0.1192)/5.7212 = 0.90;
u2 = uf + x2 ufg = 27.66 + 0.9×1257.0 = 1158.9 kJ/kg
1W2 = 1 × (1391.3 – 1158.9) = 232.4 kJ.

6. A cylinder-piston contains ammonia at 50°C, 20% quality, volume being 1 L. The ammonia expands slowly, and heat is transferred to maintain a constant temperature. The process continues until all liquid is gone. Determine the work for this process.
a) 7.11 kJ
b) 9.11 kJ
c) 5.11 kJ
d) 8.11 kJ
Answer: a
Clarification: T1 = 50°C, x1 = 0.20, V1 = 1 L, v1 = 0.001777 + 0.2 ×0.06159 = 0.014095 m^3/kg
s1 = 1.5121 + 0.2 × 3.2493 = 2.1620 kJ/kg K,
m = V1/v1 = 0.001/0.014095 = 0.071 kg
v2 = vg = 0.06336 m^3/kg, s2 = sg = 4.7613 kJ/kg K
Process: T = constant to x2 = 1.0, P = constant = 2.033 MPa
1W2 = ⌠PdV = Pm(v2 – v1) = 2033 × 0.071 × (0.06336 – 0.014095)
= 7.11 kJ.

7. An insulated cylinder fitted with a piston contains 0.1 kg of water at 100°C and 90% quality. The piston is moved, compressing the water till it reaches a pressure of 1.2 MPa. How much work is required in the process?
a) -27.5 kJ
b) -47.5 kJ
c) -17.5 kJ
d) -37.5 kJ
Answer: d
Clarification: 1Q2 = 0 = m(u2 – u1) + 1W2
State 1: 100°C, x1 = 0.90: s1 = 1.3068 + 0.90×6.048 = 6.7500 kJ/kg K
u1 = 418.91 + 0.9 × 2087.58 = 2297.7 kJ/kg
State 2: s2 = s1 = 6.7500 and P2 = 1.2 MPa which gives
T2 = 232.3°C and u2 = 2672.9 kJ/kg
1W2 = -m(u2 – u1) = -0.1(2672.9 – 2297.7) = -37.5 kJ.

8. Compression and heat transfer brings R-134a from 50°C, 500 kPa to saturated vapour in an isothermal process. Find the specific work.
a) -24.25 kJ/kg
b) -25.25 kJ/kg
c) -26.25 kJ/kg
d) -27.25 kJ/kg
Answer: c
Clarification: Process: T = C and assume reversible ⇒ 1q2 = T (s2 – s1)
u1 = 415.91 kJ/kg, s1 = 1.827 kJ/kg K
u2 = 403.98 kJ/kg, s2 = 1.7088 kJ/kg K
1q2 = (273 + 50) × (1.7088 – 1.827) = -38.18 kJ/kg
w2 = 1q2 + u1 – u2 = -38.18 + 415.91 – 403.98 = -26.25 kJ/kg.

9. 1kg of water at 300°C expands against a piston in a cylinder until it reaches 100 kPa, at which point the water has a quality of 90.2%. The expansion is reversible and adiabatic. How much work is done by the water?
a) 371.2 kJ
b) 471.2 kJ
c) 571.2 kJ
d) 671.2 kJ
Answer: b
Clarification: Process: Adiabatic Q = 0 and reversible => s2 = s1
P2 = 100 kPa, x2 = 0.902, thus s2 = 1.3026 + 0.902 × 6.0568 = 6.7658 kJ/kg K
s2 = 1.3026 + 0.902 × 6.0568 = 6.7658 kJ/kg K
State 1 At T1 = 300°C, s1 = 6.7658 and ⇒ P1 = 2000 kPa, u1 = 2772.6 kJ/kg
1W2 = m(u1 – u2) = 1(2772.6 – 2301.4) = 471.2 kJ.

10. A piston/cylinder has 2kg ammonia at 100 kPa, 50°C which is compressed to 1000 kPa. The temperature is assumed to be constant. Find the work for the process assuming it to be reversible.
a) -727.6 kJ
b) -794.2 kJ
c) -723.6 kJ
d) -743.2 kJ
Answer: a
Clarification: Process: T = constant and assume reversible process
v1 = 1.5658 m^3/kg, u1 = 1424.7 kJ/kg, s1 = 6.4943 kJ/kg K
v2 = 0.1450 m^3/kg, u2 = 1391.3 kJ/kg, s2 = 5.2654 kJ/kg K
1Q2 = mT(s2 − s1) = 2 × 323.15 (5.2654 – 6.4943) = -794.2 kJ
1W2 = 1Q2 – m(u2 – u1) = -794.24 – 2(1391.3 – 1424.62)
= -727.6 kJ.

11. A piston cylinder has R-134a at 100 kPa, –20°C which is compressed to 500 kPa in a reversible adiabatic process. Find the specific work.
a) -41.63 kJ/kg
b) -11.63 kJ/kg
c) -21.63 kJ/kg
d) -31.63 kJ/kg
Answer: d
Clarification: Process: Adiabatic and reversible => s2 = s1
u1 = 367.36 kJ/kg, s1 = 1.7665 kJ/kg K
P2 = 500 kPa, s2 = s1 = 1.7665 kJ/kg K
very close at 30°C, u2 = 398.99 kJ/kg
1w2 = u2 – u1 = 367.36 – 398.99 = -31.63 kJ/kg.

12. A cylinder containing R-134a at 150 kPa, 10°C has an initial volume of 20 L. A piston compresses the R-134a in a isothermal, reversible process until it reaches the saturated vapour state. Calculate the required work in the process.
a) -1.197 kJ
b) -2.197 kJ
c) -3.197 kJ
d) -4.197 kJ
Answer: c
Clarification: Process: T = constant, reversible
u1 = 388.36 kJ/kg, s1 = 1.822 kJ/kg K, m = V/v1 = 0.02/0.148283 = 0.1349 kg
u2 = 383.67 kJ/kg, s2 = 1.7218 kJ/kg K
1Q2 = ⌠Tds = mT(s2 – s1) = 0.1349 × 283.15 × (1.7218 – 1.822) = -3.83 kJ
1W2 = m(u1 – u2) + 1Q2 = 0.1349 × (388.36 – 383.67) – 3.83 = -3.197 kJ.

13. A piston/cylinder has 2kg water at 250°C, 1000 kPa which is now cooled with a constant load on the piston. This isobaric process ends when the water has reached a state of saturated liquid. Find the work.
a) -363.1 kJ
b) -463.1 kJ
c) -563.1 kJ
d) -663.1 kJ
Answer: b
Clarification: Process: P = C => W = ∫ P dV = P(V2 − V1)
State 1: v1 = 0.23268 m^3/kg, s1 = 6.9246 kJ/kg K, u1 = 2709.91 kJ/kg
State 2: v2 = 0.001127 m^3/kg, s2 = 2.1386 kJ/kg K, u2 = 761.67 kJ/kg
1W2 = m P (v2 − v1) = 2 × 1000 (0.001127 – 0.23268) = -463.1 kJ.

14. Water at 250°C, 1000 kPa is brought to saturated vapour in a piston/cylinder with an isothermal process. Find the specific work.
a) -38 kJ/kg
b) -138 kJ/kg
c) -238 kJ/kg
d) -338 kJ/kg
Answer: d
Clarification: Process: T = constant, reversible
State 1: v1 = 0.23268 m^3/kg; u1 = 2709.91 kJ/kg; s1 = 6.9246 kJ/kg K
State 2: v2 = 0.05013 m^3/kg, u2 = 2602.37 kJ/kg, s2 = 6.0729 kJ/kg K
1q2 = ∫ T ds = T(s2 − s1) = (250 + 273) (6.0729 – 6.9246) = -445.6 kJ/kg
1w2 = 1q2 + u1 − u2 = -445.6 + 2709.91 – 2602.37 = -338 kJ/kg.

15. Water at 250°C, 1000 kPa is brought to saturated vapour in a rigid container. Find the specific heat transfer in this isometric process.
a) −132 kJ/kg
b) −232 kJ/kg
c) −332 kJ/kg
d) −432 kJ/kg
Answer: a
Clarification: Process: v = constant => 1w2 = 0
State 1: u1 = 2709.91 kJ/kg, v1 = 0.23268 m^3/kg
State 2: x = 1 and v2 = v1, thus P2=800 kPa
T2 = 170 + 5 × (0.23268 – 0.24283)/(0.2168 – 0.24283)
= 170 + 5 × 0.38993 = 171.95°C
u2 = 2576.46 + 0.38993 × (2580.19 – 2576.46) = 2577.9 kJ/kg
1q2 = u2 − u1 = 2577.9 – 2709.91 = −132 kJ/kg.