Thermodynamics MCQ Set 1
1. Variable flow processes include
a) filling up a gas cylinder
b) evacuating a gas cylinder
c) all of the mentioned
d) none of the mentioned
Answer
Answer: c [Reason:] These are variable flow processes which can be analysed by the control volume technique.
2. The rate at which the mass of fluid within the control volume is accumulated is equal to the net rate of mass flow across the control surface.
a) true
b) false
Answer
Answer: a [Reason:] (dm/dt)=w1-w2, where m is the mass of fluid within the control volume at any instant.
3. Rate of energy increase within the control volume is given by
a) rate of energy inflow + rate of energy outflow
b) rate of energy inflow – rate of energy outflow
c) rate of energy inflow = rate of energy outflow
d) none of the mentioned
Answer
Answer: b [Reason:] The rate of accumulation of energy within the control volume is equal to the net energy flow across the control surface.
4. Which of the following is true for steady flow?
a)(dE/dt)=0
b)(dE/dt)>0
c)(dE/dt)<0
d) none of the mentioned
Answer
Answer: a [Reason:] Rate of change of energy of fluid with respect to time within the control volume is constant.
5. Variable flow processes can be analysed by
a) system technique
b) constant volume technique
c) both of the mentioned
d) none of the mentioned
Answer
Answer: c [Reason:] These two are the main techniques used for analysing variable flow process.
6. Using system technique, energy balance for the process comes out to be
a) m2u2+m1u1+(m2-m1)(((V^2)/2)+(h/2))
b) m2u2-m1u1+(m2-m1)(((V^2)/2)+(h/2))
c) m2u2+m1u1-(m2-m1)(((V^2)/2)+(h/2))
d) m2u2-m1u1-(m2-m1)(((V^2)/2)+(h/2))
Answer
Answer: d [Reason:] This comes from the first law by neglecting PE, KE and E is the energy of the gas.
7. Both the techniques for analysing variable flow processes gives same result.
a) true
b) false
Answer
Answer: a [Reason:] This is because these techniques have same initial assumptions and hence give same result.
8. In ____ filling a bottle with air at 300K, the gas temperature rises to 420K due to flow work being converted to ____ increase.
a) adiabatically, heat
b) adiabatically, internal energy
c) constant pressure, heat
d) none of the mentioned
Answer
Answer: b [Reason:] In the energy equation, m1=0, Q=0, h>>(V^2)/2, we will get we will get that flow work is converted to increase in molecular internal energy.
9. Which of the following is true for a discharging tank?
a) the process is adiabatic
b) the process is quasi-static
c) dQ=0
d) all of the mentioned
Answer
Answer: d [Reason:] Applying first law to the control volume and dW=0, dm=0 and KE and PE of the fluid are assumed to be small.
10. For charging a tank,
a) enthalpy is converted to work done
b) work done is converted to enthalpy
c) enthalpy is converted to internal energy
d) internal energy is converted to work done
Answer
Answer: c [Reason:] Tank is initially taken to be empty and ΔU=(m2u2-m1u1)=(mh) at constant state of the fluid in the pipeline.
11. For a variable flow process
a) P.E. terms are neglected
b) K.E. of the fluid is assumed to be small
c) the process is not steady
d) all of the mentioned
Answer
Answer: d [Reason:] These are some of the basic assumptions for a variable flow process.
Thermodynamics MCQ Set 2
1. An incompressible fluid is one for which density does not change with change in ____
a) pressure
b) temperature
c) velocity
d) all of the mentioned
Answer
Answer: d [Reason:] In an incompressible fluid, either density does not change or changes very little.
2. A compressible fluid is one for which density changes with change in temperature, pressure or velocity.
a) true
b) false
Answer
Answer: a [Reason:] This is opposite to an incompressible fluid.
3. Liquids are ____ and gases are ____
a) both are compressible
b) both are incompressible
c) incompressible, compressible
d) compressible, incompressible
Answer
Answer: c [Reason:] Liquids are incompressible whereas gases are compressible.
4. The pressure wave velocity c is given by
a) 1 / (dp/d(density))
b) sqrt(dp/d(density))
c) sqrt(d(density)/dp)
d) (dp/d(density))
Answer
Answer: b [Reason:] Here dp is the change in pressure.
5. For an ideal gas, velocity of sound is given by
a) 1/£*R*T
b) sqrt(1/£*R*T)
c) sqrt(£/RT)
d) sqrt(£*R*T)
Answer
Answer: d [Reason:] Here R is the characteristic gas constant.
6. The lower the molecular weight of fluid, ____ the value of £ and ____ is the sonic velocity at same temperature.
a) lower, lower
b) higher, higher
c) lower, higher
d) higher. lower
Answer
Answer: b [Reason:] This comes from the expression of velocity of sound in an ideal gas.
7. The Mach number is given by
a) (c/V)2
b) (V/c)2
c) V/c
d) c/V
Answer
Answer: c [Reason:] Here V is the actual velocity and c is the sonic velocity.
8. For the isentropic stagnation state,
a) it is a reference state
b) designated with subscript zero
c) both of the mentioned
d) none of the mentioned
Answer
Answer: c [Reason:] It is a reference state in compressible fluid flow.
9. The reference temperature To and normal temperature T are related by
a) (To/T) = 1 + (V2)/(2*cp*T)
b) (To/T) = 1 – (V2)/(2*cp*T)
c) (To/T) = 1 + (V2)/(cp*T)
d) (To/T) = 1 – (V2)/(cp*T)
Answer
Answer: a [Reason:] Here cp is the specific heat at constant pressure and V is the actual velocity.
10. The reference temperature To and normal temperature T can be related in terms of M by
a) (To/T) = 1 + (£-1)*(M2)
b) (To/T) = 1 + (£-1)*(M2)/2
c) (To/T) = 1 – (£-1)*(M2)/2
d) (To/T) = 1 – (£-1)*(M2)
Answer
Answer: b [Reason:] Here M is the Mach number.
Thermodynamics MCQ Set 3
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
Answer: b [Reason:] 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
Answer: c [Reason:] 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
Answer: d [Reason:] 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
Answer: d [Reason:] 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
Answer: a [Reason:] 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
Answer: b [Reason:] 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
Answer: c [Reason:] 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
Answer: b [Reason:] 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
Answer: b [Reason:] 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
Answer: a [Reason:] the maximum pressure ratio which can be attained by a reciprocating compressor cylinder is limited by the clearance.
Thermodynamics MCQ Set 4
1. A gas compression process is
a) adiabatic
b) involves heat transfer
c) both of the mentioned
d) none of the mentioned
Answer
Answer: c [Reason:] A gas compression process can be either adiabatic or can involve heat transfer.
2. If the gas is cooled during compression, work required will be ____ the adiabatic compression work.
a) more than
b) less than
c) equal to
d) none of the mentioned
Answer
Answer: b [Reason:] Here the work required will be less than that required for adiabatic compression.
3. Which of the following is an advantage of cooling?
a) less pipe friction losses
b) reduction in volume of gas
c) both of the mentioned
d) none of the mentioned
Answer
Answer: c [Reason:] These are the two advantages of cooling.
4. We use after-coolers to cool the gas which leaves the compressor.
a) true
b) false
Answer
Answer: a [Reason:] This is done because compression process is somewhat ineffective.
5. The work of compression is ____ the shaft work.
a) positive of
b) negative of
c) equal to
d) less than
Answer
Answer: b [Reason:] This is true for reversible adiabatic compression.
6. For £>n>1 and for same pressure ratio p2/p1, the maximum work is needed for
a) isothermal compression
b) adiabatic compression
c) polytropic compression
d) all need same work
Answer
Answer: b [Reason:] This comes when these three reversible compression processes are plotted on the p-V diagram.
7. In isothermal compression, all work done on gas is transformed into
a) heat added into system
b) heat going out of system
c) internal energy increase
d) none of the mentioned
Answer
Answer: c [Reason:] This is the case of isothermal compression.
8. When isothermal compression is taken as ideal process, the energy imparted
a) raises the temperature of gas
b) raises the pressure of gas
c) both of the mentioned
d) none of the mentioned
Answer
Answer: b [Reason:] In isothermal compression considered as ideal process, no energy is imparted to the gas.
9. The adiabatic efficiency is given by
a) Ws/Wc
b) Ws/Wt
c) Wt/Wc
d) Wt/Ws
Answer
Answer: a [Reason:] This is the efficiency of compressor working in a steady flow process.
10. The isothermal efficiency is given by
a) Ws/Wc
b) Ws/Wt
c) Wt/Wc
d) Wt/Ws
Answer
Answer: c [Reason:] This is the efficiency of compressor working in a steady flow process and Wt=work in reversible isothermal compression.
11. The adiabatic efficiency of real compressor can be ____
a) less than unity
b) greater than unity
c) equal to unity
d) none of the mentioned
Answer
Answer: b [Reason:] This is due to the effects of cooling.
12. For an adiabatic machine, work of compression is greater than enthalpy rise of gas.
a) true
b) false
Answer
Answer: b [Reason:] For an adiabatic machine, work of compression is equal to the enthalpy rise of gas.
13. Argon is kept in a 5 m3 tank at −30°C and 3 MPa. Determine the mass using compressibility factor.
a) 208.75 kg
b) 308.75 kg
c) 303.75 kg
d) 203.75 kg
Answer
Answer: b [Reason:] Tr = 243.15/150.8 = 1.612 and Pr = 3000/4870 = 0.616 hence Z = 0.96
m = PV/ZRT = (3000 × 5)/(0.96 × 0.2081 × 243.2)
= 308.75 kg.
14. Find the error in specific volume if ideal gas model is used to represent the behaviour of superheated ammonia at 40°C and 500 kPa?
a) 1.5%
b) 3.5%
c) 4.5%
d) 2.5%
Answer
Answer: c [Reason:] NH3, T = 40°C = 313.15 K, Tc = 405.5 K, Pc = 11.35 MPa
v = 0.2923 m3/kg
Ideal gas: v = RT/P = (0.48819 × 313)/(500) = 0.3056 m3/kg
thus error = 4.5%.
15. Find the volume of ethylene having mass of 125 kg at 7.5 MPa and 296.5 K.
a) 0.369 m3
b) 0.669 m3
c) 0.569 m3
d) 0.469 m3
Answer
Answer: d [Reason:] For ethylene, Tc = 282.4 K and Pc = 5.04 MPa
Tr = T/Tc = 296.5 / 282.4 = 1.05 and Pr = P/Pc = 7.5 / 5.04 = 1.49
thus Z = 0.32
hence V = mZRT / P = 125 × 0.32 × 0.2964 × 296.5 / 7500 = 0.469 m3.
Thermodynamics MCQ Set 5
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
Answer: c [Reason:] 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
Answer: b [Reason:] 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
Answer: d [Reason:] 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
Answer: a [Reason:] 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
Answer: b [Reason:] 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
Answer: a [Reason:] 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
Answer: d [Reason:] 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
Answer: c [Reason:] 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
Answer: b [Reason:] 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
Answer: a [Reason:] 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
Answer: d [Reason:] 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
Answer: c [Reason:] 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
Answer: b [Reason:] 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
Answer: d [Reason:] 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
Answer: a [Reason:] 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.
Thermodynamics MCQ Set 6
1. Water at 250°C, 1000 kPa is brought to saturated vapour in a piston/cylinder with an isobaric process. Find the specific work.
a) -18.28 kJ/kg
b) -48.28 kJ/kg
c) -28.28 kJ/kg
d) -38.28 kJ/kg
Answer
Answer: d [Reason:] Process: P = C => w = ∫ P dv = P(v2 − v1)
1: v1 = 0.23268 m3/kg, s1= 6.9246 kJ/kgK, u1 = 2709.91 kJ/kg
2: v2 = 0.19444 m3/kg, s2 = 6.5864 kJ/kg K,
u2 = 2583.64 kJ/kg, T2 = 179.91°C
1w2 = P (v2 − v1) = 1000 (0.1944 – 0.23268) = -38.28 kJ/kg.
2. A heavily insulated cylinder/piston contains ammonia at 60°C, 1200 kPa. The piston is moved, expanding the ammonia in a reversible process until the temperature is −20°C during which 600 kJ of work is given out by ammonia. What was the initial volume of the cylinder?
a) 0.285 m3
b) 0.385 m3
c) 0.485 m3
d) 0.585 m3
Answer
Answer: b [Reason:] State 1: v1 = 0.1238 m3/kg, s1 = 5.2357 kJ/kg K,
u1 = h – Pv = 1553.3 – 1200×0.1238 = 1404.9 kJ/kg
Process: reversible (1S2(gen) = 0) and adiabatic (dQ = 0) => s2 = s1
State 2: T2, s2 ⇒ x2 = (5.2357 – 0.3657)/5.2498 = 0.928
u2 = 88.76 + 0.928×1210.7 = 1211.95 kJ/kg
1Q2 = 0 = m(u2 – u1) + 1W2 = m(1211.95 – 1404.9) + 600 ⇒ m = 3.110 kg
V1 = mv1 = 3.11 × 0.1238 = 0.385 m3.
3. Water at 250°C, 1000 kPa is brought to saturated vapor in a piston/cylinder with an adiabatic process. Find the specific work.
a) 139.35 kJ/kg
b) 149.35 kJ/kg
c) 159.35 kJ/kg
d) 169.35 kJ/kg
Answer
Answer: c [Reason:] State 1: v1 = 0.23268 m3/kg, u1 = 2709.91 kJ/kg, s1 = 6.9246 kJ/kg K
State 2: x = 1 and s2 = s1 = 6.9246 kJ/kg K
T2 = 140.56°C, P2 = 367.34 kPa, v2 = 0.50187 m3/kg, u2= 2550.56 kJ/kg
1w2 = u1 – u2 = 2709.91 – 2550.56 = 159.35 kJ/kg.
4. A piston/cylinder contains 2kg water at 200°C, 10 MPa. The water expands in an isothermal process to a pressure of 200 kPa. Any heat transfer takes place with an ambient at 200°C and whole process is be assumed reversible. Calculate the total work.
a) 1290.3 kJ
b) 1390.3 kJ
c) 1490.3 kJ
d) 1590.3 kJ
Answer
Answer: a [Reason:] State 1: v1 = 0.001148 m3/kg, u1 = 844.49 kJ/kg,
s1 = 2.3178 kJ/kg K, V1 = mv1 = 0.0023 m3
State 2: v2 = 1.08034 m3/kg, u2 = 2654.4 kJ/kg,
s2 = 7.5066 kJ/kg K, V2 = mv2 = 2.1607 m3
1Q2 = mT(s2 − s1) = 2 × 473.15 (7.5066 – 2.3178) = 4910 kJ
1W2 = 1Q2 – m(u2 – u1) = 1290.3 kJ.
5. A piston/cylinder of total 1kg steel contains 0.5 kg ammonia at 1600 kPa both masses at 120°C with minimum volume being 0.02 m3. The whole system is cooled down to 30°C by heat transfer to the ambient at 20°C, and during the process the steel keeps same temperature as the ammonia. Find the work.
a) – 28.14 kJ
b) – 38.14 kJ
c) – 48.14 kJ
d) – 58.14 kJ
Answer
Answer: d [Reason:] 1 : v1 = 0.11265 m3/kg, u1 = 1516.6 kJ/kg,
s1 = 5.5018 kJ/kg K, V1 = mv1 = 0.05634 m3
2 : 30°C < T(stop) so v2 = v(stop) = 0.04 m3/kg
1W2= ∫ P dV = (P1)m(v2-v1) = 1600 × 0.5 (0.04 – 0.11265)
= – 58.14 kJ.
6. A mass of 1 kg of air contained in a cylinder at 1000 K, 1.5 MPa, expands in a reversible isothermal process to a volume 10 times larger. Calculate the heat transfer during the process.
a) 460.84 kJ
b) 560.84 kJ
c) 660.84 kJ
d) 760.84 kJ
Answer
Answer: c [Reason:] Process: T = constant so with ideal gas => u2 = u1
1Q2 = 1W2 = ⌠PdV = P1V1 ln (V2/V1) = mRT1 ln (V2/V1)
= 1 × 0.287 × 1000 ln (10) = 660.84 kJ.
7. A piston/cylinder contains air at 400 K, 100 kPa which is compressed to a final pressure of 1000 kPa. Consider the process to be a reversible adiabatic process. Find the specific work.
a) -166.7 kJ/kg
b) -266.7 kJ/kg
c) -366.7 kJ/kg
d) -466.7 kJ/kg
Answer
Answer: b [Reason:] We have constant s, an isentropic process
T2 = T1( P2 / P1)^[(k-1)/k] = 400(1000/100)^(0.4/1.4)
= 400 × 10^(0.28575) = 772 K
1w2 = u1 – u2 = Cv(T1 – T2) = 0.717(400 – 772) = -266.7 kJ/kg.
8. A piston/cylinder contains air at 400 K, 100 kPa which is compressed to a final pressure of 1000 kPa. Consider the process to be a reversible isothermal process. Find the specific work.
a) −264 kJ/kg
b) −364 kJ/kg
c) −464 kJ/kg
d) −564 kJ/kg
Answer
Answer: a [Reason:] For this process T2 = T1 so since ideal gas we get u2 = u1 and also 1w2 = 1q2
1w2 = 1q2 = T(s2 – s1) = −RT ln(P2/P1)
= − 0.287 × 400 ln 10 = −264 kJ/kg.
9. Consider a small air pistol with a cylinder volume of 1 cm3 at 27°C, 250 kPa. The bullet acts as a piston and is released so the air expands in an adiabatic process. If the pressure should be 100 kPa as the bullet leaves the cylinder find the work done by the air.
a) 0.115 J
b) 0.125 J
c) 0.135 J
d) 0.145 J
Answer
Answer: d [Reason:] Process: Adiabatic 1q2 = 0 Reversible 1s2(gen) = 0
this is an isentropic expansion process giving s2 = s1
T2 = T1( P2 / P1)^[(k-1)/k] = 300(100/250)^(0.4/1.4) = 300 × 0.4^(0.28575) = 230.9 K
V2 = V1 P1 T2/P2 T1 = 1 × 250 × 230.9/100 × 300 = 1.92 cm3
Work = [1/(k-1)](P2V2 – P1V1) = [1/(1-1.4)](100 × 1.92 – 250 × 1) ×10^(-6)
= 0.145 J.
10. A spring loaded piston cylinder contains 1.5 kg air at 160 kPa and 27°C. It is heated in a process where pressure is linear in volume, P = A + BV, to twice the initial volume where it reaches 900 K. Find the work assuming a source at 900 K.
a) 61.4 kJ
b) 161.4 kJ
c) 261.4 kJ
d) 361.4 kJ
Answer
Answer: b [Reason:] State 1: u1 = 214.36 kJ/kg, V1 = mRT1/ P1 = (1.5 × 0.287 ×300) / 160 = 0.8072 m3
State 2: u2 = 674.824 kJ/kg,
P2 = RT2/v2 = RT2/(2v1) = T2 P1/(2T1)= P1(T2/2)T1
= 160 × 900 / 2 × 300 = 240 kPa
1W2 = ∫ PdV = 0.5 × (P1 + P2) (V2 – V1) = 0.5 × (P1 + P2) V1
= 0.5 × (160 + 240) 0.8072 = 161.4 kJ.
11. Helium contained in a cylinder at ambient conditions, 100 kPa, 20°C, is compressed in a reversible isothermal process to 600 kPa, after which the gas is expanded back to 100 kPa in a reversible adiabatic process. Calculate the net work per kilogram of helium.
a) -623.6 kJ/kg
b) +467.4 kJ/kg
c) -1091.0 kJ/kg
d) none of the mentioned
Answer
Answer: a [Reason:] The adiabatic reversible expansion gives constant s
T3 = T2(P3/P2)^[(k-1)/k] = 293.15 (100/600)^0.4 = 143.15 K
The isothermal process: 1w2 = -RT1 ln(P2/P1)
= -2.0771 × 293.15 × ln(600/100) = -1091.0 kJ/kg
The adiabatic process: 2w3 = CVo(T2-T3) = 3.116 (293.15 – 143.15) = +467.4 kJ/kg
The net work is the sum: w(NET) = -1091.0 + 467.4 = -623.6 kJ/kg.
12. A cylinder/piston contains 1kg methane gas at 100 kPa, 20°C. The gas is compressed reversibly to a pressure of 800 kPa. Calculate the work required if the process is adiabatic.
a) -112.0 kJ
b) -212.0 kJ
c) -312.0 kJ
d) -412.0 kJ
Answer
Answer: c [Reason:] Process: 1Q2 = 0 => s2 = s1 thus isentropic process
T2 = T1(P2/P1)^[(k-1)/k] = 293.2(800/100)^0.230 = 473.0 K
1W2 = -mCv(T2 – T1) = -1 × 1.7354 (473.0 – 293.2)
= -312.0 kJ.
13. A piston/cylinder contains air at 100 kPa, 300 K. It is now compressed in a reversible adiabatic process to a volume 7 times as small. Use constant heat capacity and find the specific work.
a) -233.6 kJ/kg
b) -243.6 kJ/kg
c) -253.6 kJ/kg
d) -263.6 kJ/kg
Answer
Answer: c [Reason:] v2/ v1 = 1/7; P2 /P1 = (v2/v1)^(-k) = 7^(1.4) = 15.245
P2 = P1[7^(1.4)] = 100 × 15.245 = 1524.5 kPa
T2 = T1 (v1/v2)^(k-1) = 300 × 7^(0.4) = 653.4 K
1q2 = 0 kJ/kg; work = R(T2-T1)/(1-k) = 0.287*(653.4 – 300)/(-0.4)
= -253.6 kJ/kg.
14. A gas confined in a piston-cylinder is compressed in a quasi-static process from 80 kPa and 0.1 m3 to 400 kPa and 0.03m3. If the pressure and volume are related by PV^n= constant, calculate the work involved in the process.
a) – 12.87 kJ
b) 12.87 kJ
c) – 11.87 kJ
d) 11.87 kJ
Answer
Answer: c [Reason:] n = ln(P2/P1)/ln(V1/V2) = ln(400/80)/ln(0.1/0.03) = 1.337
Work involved in the process (1W2) = (P2V2 – P1V1)(1 – n) = – 11.87 kJ.
15. One kg of steam at 200 kPa with 20% quality is heated at constant pressure to 400°C. Calculate the work done by the system .
a) 274.261 kJ
b) 374.261 kJ
c) 474.261 kJ
d) 574.261 kJ
Answer
Answer: a [Reason:] v1 = 0.001061 + 0.2*0.88467 = 0.177995 m3/kg; v2 = 1.5493 m3/kg
work done by the system during this process (1W2) = mP(v2 – v1)
= 1*200*(1.5493 – 0.177995) = 274.261 kJ.
Thermodynamics MCQ Set 6
1. The magnitude of mechanical work is the
a) product of the force and distance travelled perpendicular to the force
b) product of the force and distance travelled parallel to the force
c) sum of the force and distance travelled perpendicular to the force
d) sum of the force and distance travelled parallel to the force
Answer
Answer: b [Reason:] The work is done by a force as it acts upon a body moving in the direction of the force.
2. Work done by a system is taken to be
a) positive
b) negative
c) zero
d) varies according to situation
Answer
Answer: a [Reason:] In thermodynamics, work done by a system is take to be positive.
3. Work done on a system is taken to be
a) positive
b) negative
c) zero
d) varies according to situation
Answer
Answer: b [Reason:] In thermodynamics, work done on a system is take to be negative.
4. Work is a
a) point function
b) path function
c) depends on the state
d) none of the mentioned
Answer
Answer: b [Reason:] Amount of work done depends on the path the system follows.
5. Thermodynamic properties are
a) point function
b) path function
c) depends on the state
d) none of the mentioned
Answer
Answer: a [Reason:] For a given state there is a definite value for each property.
6. The differentials of point functions are
a) perfect differentials
b) exact differentials
c) all of the mentioned
d) none of the mentioned
Answer
Answer: c [Reason:] Change in thermodynamic property is independent of path and depends only on initial and final states of the system.
7. In the equation dV=(1/p)dW, (1/p) is known as
a) volume factor
b) pressure factor
c) differential factor
d) integration factor
Answer
Answer: d [Reason:] Used to convert inexact differential dW into exact differential dV.
8. Cyclic integral of a property is always
a) zero
b) one
c) infinite value
d) none of the mentioned
Answer
Answer: a [Reason:] The initial and final states of the system for a cyclic process are the same.
9. Constant pressure process is also known as
a) isopiestic process
b) isobaric process
c) all of the mentioned
d) none of the mentioned
Answer
Answer: c [Reason:] Isobaric and isopiestic means pressure being constant.
10. Work done in a quasi-static process
a) depends on the path followed
b) independent of the path followed
c) depends only on the initial and final states
d) none of the mentioned
Answer
Answer: a [Reason:] This is because work done is a path function.
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