§ME 231 Homework 10

Sep 25, 2025

§6.81

Air in a piston-cylinder assembly is compressed isentropically from state 1, where T1 = 35°C, to state 2, where the specific volume is one-tenth of the specific volume at state 1. Applying the ideal gas model and assuming variations in specific heat, determine (a) T2, in °C, and (b) the work, in kJ/kg.

T_1=35\degree C

v_2=\frac{1}{10}v_1

From table A-22 with

T_1=35\degree C=308.15K\approx310K

v_{r1}=572.3

h_1=310.24kJ/kg

\frac{v_{r2}}{v_{r1}}=\frac{v_2}{v_1}

v_{r2}=\frac{v_2 v_{r1}}{v_1}

v_{r2}=\frac{\frac{1}{10}v_1 v_{r1}}{v_1}

v_{r2}=\frac{v_{r1}}{10}

v_{r2}=\frac{v_{r1}}{10}=\frac{572.3}{10}=57.23

From table A-22 with

v_r=57.23\approx57.63

T_2=750K=\boxed{476.85\degree C}

h_2=767.29kJ/kg

w=h_2-h_1=767.29kJ/kg-310.24kJ/kg=\boxed{457.15kJ/kg}

§6.82

Steam undergoes an isentropic compression in an insulated piston–cylinder assembly from an initial state where T1 = 120°C, p1 = 1 bar to a final state where the pressure p2 = 100 bar. Determine the final temperature, in °C, and the work, in kJ per kg of steam.

Q=0

T_1=120\degree C

P_1=1bar

P_2=100bar

From table A-4 at

p=1bar

and

T=120\degree C

s_1=s_2=7.4668\frac{kJ}{kg*K}

u_1=2537.3kJ/kg

Excerpt from table A-4 at

p=100bar

$T$	$s$	$u$
$700\degree C$	$7.1687\frac{kJ}{kg*K}$	$3434.7kJ/kg$
$740\degree C$	$7.2670\frac{kJ}{kg*K}$	$3512.1kJ/kg$

Extrapolation (not interpolation):

T_2=700+(740-700)*\frac{7.4668-7.1687}{7.2670-7.1687}=\boxed{821.3\degree C}

u_2=3434.7+(3512.1-3434.7)*\frac{7.4668-7.1687}{7.2670-7.1687}=3669.4kJ/kg

w=u_2-u_1=3669.4kJ/kg-2537.3kJ/kg=\boxed{1132kj/kg}

§6.83

NOTE

Propane undergoes an isentropic expansion from an initial state where T1 = 40°C, p1 = 1 MPa to a final state where the temperature and pressure are T2, p2, respectively. Determinea. p2 in kPa, when T2 = −50°C.b. T2, in °C, when p2 = 0.7 MPa.

T_1=40\degree C

P_1=1MPa

From table A-18 with

P=1MPa

and

T=40\degree C

s_1=s_2=1.810\frac{kJ}{kg*K}

§6.83 (a)

T_2=-50\degree C

From table A-18 with

T=-50\degree C

and

s=1.810\frac{kJ}{kg*K}\approx=1.871\frac{kJ}{kg*K}

P_2=0.5bar=\boxed{50kPa}

§6.83 (b)

P_2=0.7MPa

From table A-18 with

P=0.7MPa

and

s=1.810\frac{kJ}{kg*K} \approx 1.840\frac{kJ}{kg*K}

T_2=\boxed{30\degree C}

§6.92

Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adiabatically to an exit state of 1 bar, 160°C. Kinetic and potential energy effects are negligible. Determine for the turbine (a) the power developed, in kW, (b) the rate of entropy production, in kW/K, and (c) the isentropic turbine efficiency.

P_1=5bar

T_1=320\degree C

\dot{m}_1=\dot{m}_2=\dot{m}

\dot{V}_1=0.65m^3/s

\dot{Q}=0

P_2=1bar

T_2=160\degree C

KE=PE=0

From table A-4 with

P=5bar

and

T=320\degree C

v_1=0.5416m^3/kg

h_1=3105.6kJ/kg

s_1=7.5308\frac{kJ}{kg*K}

From table A-4 with

P=1bar

and

T=160\degree C

h_2=2796.2kJ/kg

s_2=7.6597\frac{kJ}{kg*K}

v=\frac{V}{m}

m=\frac{V}{v}

\dot{m}=\frac{\dot{V}_1}{v_1}=\frac{0.65m^3/s}{0.5416m^3/kg}=1.2kg/s

0=\cancel{\dot{Q}}-\dot{W}+\dot{m}\left[h_1-h_2+\cancel{\frac{\vec{v}_1^2-\vec{v}_2^2}{2}}+\cancel{g(z_1-z_2)}\right]

0=-\dot{W}+\dot{m}\left[h_1-h_2\right]

\dot{W}=\dot{m}[h_1-h_2]=1.2kg/s * [3105.6kJ/kg - 2796.2kJ/kg]=\boxed{371kW}

0=\cancel{\sum\frac{\dot{Q}_j}{T_j}}+\dot{m}(s_1-s_2)+\dot{\sigma}_{cv}

-\dot{\sigma}=\dot{m}(s_1-s_2)

\dot{\sigma}=\dot{m}(s_2-s_1)

\dot{\sigma}=1.2kg/s * (7.6597\frac{kJ}{kg*K}-7.5308\frac{kJ}{kg*K})=\boxed{0.1547\frac{kW}{K}}

s_{2s}=s_1=7.5308\frac{kJ}{kg*K}

Excerpt from table A-4 with

P=1bar

and around

s=s_{2s}=7.5308\frac{kJ}{kg*K}

$h$	$s$
$2716.6\frac{kJ}{kg}$	$7.4668\frac{kJ}{kg*K}$
$2796.2\frac{kJ}{kg}$	$7.6597\frac{kJ}{kg*K}$

h_{2s}=2716.6+(2796.2-2716.6)\frac{7.5308-7.4668}{7.6597-7.4668}=2743.0\frac{kJ}{kg}

\dot{W}_s=\dot{m}[h_1-h_{2s}]=1.2kg/s * [3105.6kJ/kg - 2743.0kJ/kg]=435kW

\eta=\frac{\dot{W}}{\dot{W}_s}=\frac{371kW}{435kW}=0.8529=\boxed{85.29\%}

§6.98

Air enters the compressor of a gas turbine power plant operating at steady state at 290 K, 100 kPa and exits at 330 kPa. Stray heat transfer and kinetic and potential energy effects are negligible. The isentropic compressor efficiency is 90.3%. Using the ideal gas model for air, determine the work input, in kJ per kg of air flowing.

T_1=290K

P_1=100kPa

P_2=330kPa

\dot{m}=\text{const}

\dot{Q}=0

KE=PE=0

\eta=90.3\%=0.903

From table A-22 with

T=290K

h_1=290.16\frac{kJ}{kg}

From table A-20 with

T=290K\approx300K

k=1.400

\frac{T_{2s}}{T_1}=\left(\frac{P_2}{P_1}\right)^\frac{k-1}{k}

T_{2s}=T_1\left(\frac{P_2}{P_1}\right)^\frac{k-1}{k}=290K * \left(\frac{330kPa}{100kPa}\right)^\frac{1.400-1}{1.400}=407.89K

Excerpt from table A-22 around

T=407.89K

$T$	$h$
$400$	$400.98kJ/kg$
$410$	$411.12kJ/kg$

h_{2s}=400.98+(411.12-400.98)\frac{407.89-400}{410-400}=408.98kJ/kg

0=\cancel{\dot{Q}}-\dot{W}_s+\dot{m}\left[h_1-h_2+\cancel{\frac{\vec{v}_1^2-\vec{v}_2^2}{2}}+\cancel{g(z_1-z_2)}\right]

0=-\dot{W}_s+\dot{m}\left[h_1-h_2\right]

\dot{W}_s=\dot{m}[h_1-h_2]

\frac{\dot{W}_s}{\dot{m}}=\frac{W_s}{m}=w_s=h_1-h_{2s}=290.16kJ/kg-408.98kJ/kg=-118.8kJ/kg

\eta=\frac{W}{W_s}=\frac{w}{w_s}

w=\eta w_s=0.903 * -118.8kJ/kg=\boxed{-107.28kJ/kg}

The negative sign indicates that this is from the perspective of the compressor which is receiving the work.