1. 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
Explanation: Tank is initially taken to be empty and ΔU=(m2u2-m1u1)=(mh) at constant state of the fluid in the pipeline.
2. 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
Explanation: These are some of the basic assumptions for a variable flow process.
3. A polytropic process(n = − 1) starts with P = 0, V = 0 and ends with P= 600 kPa, V = 0.01 m3. Find the boundary work done.
a) 1 kJ
b) 2 kJ
c) 3 kJ
d) 4 kJ
Explanation: W = ⌠ PdV
= (1/2)(P1 + P2)(V2 – V1)
= (1/2)(P2 + 0)( V2 – 0)
= (1/2)(600*0.1)
= 3 kJ.
4. The piston/cylinder contains carbon dioxide at 300 kPa, with volume of 0.2 m3 and at 100°C. Mass is added at such that the gas compresses with PV^(1.2) = constant to a final temperature of 200°C. Determine the work done during the process
a) -80.4 kJ
b) -40.4 kJ
c) -60.4 kJ
d) -50.4 kJ
Explanation: Work done = (P2V2 – P1V1)/(1-n) and mR = (P1V1)/T1 = 0.1608 kJ/K
Work done = 0.1608(473.2 – 373.2)/(1 – 1.2) = -80.4 kJ.
5. Neon at 400 kPa, 20°C is brought to 100°C in a polytropic process with n = 1.4. Find the work done
a) -52.39 kJ/kg
b) -62.39 kJ/kg
c) -72.39 kJ/kg
d) -82.39 kJ/kg
Explanation: For Neon, k = γ = 1.667 so n < k, Cv = 0.618, R = 0.412
1w2 = [R/(1-n)](T2 – T1) = -82.39 kJ/kg.
6. A mass of 1kg of air contained in a cylinder at 1000 K, 1.5 MPa, expands in a reversible adiabatic process to 100 kPa. Calculate the work done during the process using Constant specific heat.
a) 286.5 kJ
b) 386.5 kJ
c) 486.5 kJ
d) 586.5 kJ
Explanation: Process: 1Q2 = 0, 1S2 gen = 0 => s2 = s1
T2 = T1(P2/P1)^[(k-1)/k] = 1000(0.1/1.5)0.286 = 460.9 K
1W2 = -(U2 – U1) = mCv(T1 – T2)
= 1 × 0.717(1000 – 460.9) = 386.5 kJ.
7. 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 isothermal.
a) -216.0 kJ
b) -316.0 kJ
c) -416.0 kJ
d) -516.0 kJ
Explanation: Process: T = constant. For ideal gas then u2 = u1 1W2 = 1Q2 and ∫ dQ/T = 1Q2/T
1W2 = 1Q2 = mT(s2 – s1) = -mRT ln(P2/P1)
= -0.51835× 293.2 ln(800/100) = -316.0 kJ.
8. 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 polytropic, with exponent n = 1.15.
a) -314.5 kJ
b) -414.5 kJ
c) -514.5 kJ
d) -614.5 kJ
Explanation: Process: Pv^(n) = constant with n = 1.15 ;
T2 = T1(P2/P1)^[(n-1)/n] = 293.2(800/100)^0.130 = 384.2 K
1W2 = ∫ mP dv = m(P2v2 – P1v1)/(1 – n) = mR (T2 – T1)/(1 – n)
= 1*0.51835(384.2 – 293.2)/(1 – 1.15) = -314.5 kJ.
9. Helium in a piston/cylinder at 20°C, 100 kPa is brought to 400 K in a reversible polytropic process with exponent n = 1.25. Helium can be assumed to be an ideal gas with constant specific heat. Find the specific work.
a) -587.7 kJ/kg
b) -687.7 kJ/kg
c) -787.7 kJ/kg
d) -887.7 kJ/kg
Explanation: Process: Pv^(n) = C & Pv = RT => Tv^(n-1) = C
Cv = 3.116 kJ/kg K, R = 2.0771 kJ/kg K
v2 / v1 = (T1 / T2 )^[1/(n-1)] = 0.2885
P2 / P1 = (v1 / v2)^(n) = 4.73 => P2 = 473 kPa
W = (P2 v2 – P1 v1)/(1-n) = R(T2-T1)/(1-n) = -887.7 kJ/kg.
10. Consider air in a cylinder volume of 0.2 L at 7 MPa, 1800K. It now expands in a reversible polytropic process with exponent, n = 1.5, through a volume ratio of 8:1. Calculate the work for the process.
a) 1.61 kJ
b) 1.71 kJ
c) 1.81 kJ
d) 1.91 kJ
Explanation: Process: PV^(1.50) = constant, V2/V1 = 8
State 1: P1 = 7 MPa, T1 = 1800 K, V1 = 0.2 L, m1=P1V1/RT1 = 2.71×10-3 kg
State 2: T2 = T1 (V1/V2)^(n-1) = 1800(1/8)^(0.5) = 636.4 K
1W2 = ⌠ PdV = mR(T2 – T1)/(1 – n)
= 2.71×10^(-3) × 0.287(636.4 – 1800)/(1-1.5) = 1.81 kJ.c