§ME 231 Homework 3

Sep 25, 2025

Hello, grader of my homework! In order to preserve my sanity, I am moving away from writing most of my homeworks for most classes on paper/Google Docs. I am now trying out Markdown and

\LaTeX

with the GitHub flavor so my homeworks should be prettier for you and easier for me to write.

§3.33

A closed, rigid tank contains Refrigerant 134a, initially at 100°C. The refrigerant is cooled until it becomes saturated vapor at 20°C. For the refrigerant, determine the initial and final pressures, each in bar, and the heat transfer, in kJ/kg. Kinetic and potential energy effects can be ignored.

From table A-10 with

T_2=20\degree C

\boxed{P_2=5.7160bar}

u_2=u_{2g}=237.91kJ/kg

The total specific volume is constant.

v_1=v_2=v_{2g}=0.0358m^3/kg

From table A-10 with

T_1=100\degree C

v_{1f}=1.5443*10^{-3}m^3/kg=0.0015443m^3/kg

v_{1g}=0.0027m^3/kg

v_1>v_{1g}

The liquid at

100\degree C

is superheated.Two entries from table A-12 with

T=100\degree C

surrounding my target

v=0.0358m^3/kg

$P$	$v$	$u$
$7.0bar$	$0.04064m^3/kg$	$309.74kJ/kg$
$8.0bar$	$0.03519m^3/kg$	$308.93kJ/kg$

Interpolation time 😨

P_1=7.00bar+(8.0bar-7.00bar)\frac{0.0358m^3/kg-0.04064m^3/kg}{0.03519m^3/kg-0.04064m^3/kg}=\boxed{7.89bar}

u_1=309.74kJ/kg+(308.93kJ/kg-309.74kJ/kg)\frac{0.0358m^3/kg-0.04064m^3/kg}{0.03519m^3/kg-0.04064m^3/kg}=309kJ/kg

Now specific

Q

can be solved for.

\Delta U=Q-\cancel{W}

Q=\Delta U=U_2-U_1=m(u_2-u_1)

Q/m=u_2-u_1=237.91kJ/kg-309kJ/kg=\boxed{-71.1kJ/kg}

§3.40

A piston–cylinder assembly contains water, initially a saturated vapor at 200°C. The water is cooled at constant temperature to saturated liquid. Kinetic and potential energy effects are negligible.a. For the water as a closed system, determine the work per unit mass of water, in kJ/kg.b. If the energy transfer by heat for the process is −1200 kJ, determine the mass of the water, in kg.

T_1=T_2=200\degree C

From table A-2 with

T_1=200\degree C

u_1=u_{1g}=2595.3kJ/kg

u_2=u_{2f}=850.65kJ/kg

Q/m=u_2-u_1=850.65kJ/kg-2595.3kJ/kg=\boxed{-1745kJ/kg}

(a)

Q/m=-1745kJ/kg

m=\frac{Q}{-1745kJ/kg}=\frac{−1200kJ}{-1745kJ/kg}=\boxed{0.688kg}

(b)

§3.43

A closed, rigid tank filled with water, initially at 20 bar, a quality of 80%, and a volume of 0.5 m3, is cooled until the pressure is 4 bar. Show the process of the water on a sketch of the T–υ diagram and evaluate the heat transfer, in kJ.

P_1=20bar

x_1=80\%

V_1=0.5m^3

P_2=4bar

From table A-3 with

P=20bar

v_{1f}=1.1767*10^{-3}m^3/kg

v_{1g}=0.09963m^3/kg

u_{1f}=906.44kJ/kg

u_{1g}=2600.3kJ/kg

v_1=v_f+x(v_g-v_f)

v_1=1.1767*10^{-3}m^3/kg+(80\%)(0.09963m^3/kg-1.1767*10^{-3}m^3/kg)

v_1=0.079939m^3/kg

u_1=906.44kJ/kg+(80\%)(2600.3kJ/kg-906.44kJ/kg)

u_1=2261.5kJ/kg

Now solving for

m

which I will use later to multiply with specific heat to get heat just in

kJ

v_1=V_1/m

m=\frac{V_1}{v_1}=\frac{0.5m^3}{0.079939m^3/kg}=6.25kg

Now I need

u_2

and because it's a rigid tank, the specific volume is constant.

v_2=v_1=0.079939m^3/kg

From table A-3 with

P=4bar

v_{2f}=1.0836*10^{-3}m^3/kg=0.0010836m^3/kg

v_{2g}=0.4625m^3/kg

u_{2f}=604.31kJ/kg

u_{2g}=2553.6kJ/kg

v_{2f}<v_2<v_{2g}

The second state of the liquid is partially saturated. This should allow me to calculate the quality and then the later the specific energy for state 2 after which I will have access to specific energies of both states and also the mass from above.

x_2=\frac{v_2-v_{2f}}{v_{2g}-v_{2f}}*100\%=\frac{0.079939m^3/kg-1.0836*10^{-3}m^3/kg}{0.4625m^3/kg-1.0836*10^{-3}m^3/kg}*100\%=17.09\%

u_2=u_{2f}+x_2(u_{2g}-u_{2f})

u_2=604.31kJ/kg+(17.09\%)(2553.6kJ/kg-604.31kJ/kg)=937.4kJ/kg

And now everything comes together.

Q=\Delta U+\cancel{W}=\Delta U=U_2-U_1=m(u_2-u_1)

Q=6.25kg*(937.4kJ/kg-2261.5kJ/kg)=\boxed{-8280kJ}

I may be able to write pretty now, but drawing... that's a different story.

§3.56

A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes a process to a final pressure of 4 bar, during which the pressure–volume relationship is pV1.1 = constant. For the propane, evaluate the work and heat transfer, each in kJ. Kinetic and potential energy effects can be ignored.

T_1=27\degree C

p_1=1bar

V_1=0.2m^3

p_2=4bar

pV^{1.1}=\text{const}

It's a polytropic process.

n=1.1

p_1V_1^{1.1}=p_2V_2^{1.1}

\frac{p_1V_1^{1.1}}{p_2}=V_2^{1.1}

V_2=\left(\frac{p_1V_1^{1.1}}{p_2}\right)^{\frac{1}{1.1}}=\left(\frac{1bar*(0.2m^3)^{1.1}}{4bar}\right)^{\frac{1}{1.1}}=0.05672m^3

W=\frac{p_2V_2-p_1V_1}{1-n}=\frac{4bar*0.05672m^3-1bar*0.2m^3}{1-1.1}=\boxed{-26.86kJ}

Now for the heat transfer.

\Delta U=Q-W

Q=\Delta U+W=m(\Delta u)+W=m(u_2-u_1)+W

W

is known and I can get specific energy for both states. So what I really need now is mass. Assuming ideal gas law, I should be able to get

m

R=\frac{\bar{R}}{M}=\frac{8.3144598J/(K*mol)}{44.09kg/kmol}=188.58J/(kg*K)

PV=mRT

P_1V_1=mRT_1

m=\frac{P_1V_1}{R_1T_1}=\frac{1bar*0.2m^3}{188.58J/(kg*K)*27\degree C}=0.353kg

Specific internal energies of state 1 from table A-18 with

p=1bar

around my target

T=27\degree C

$T$	$u$
$20\degree C$	$463.3kJ/kg$
$30\degree C$	$478.2kJ/kg$

u_1=463.3kJ/kg+(478.2kJ/kg-463.3kJ/kg)\frac{30\degree C-27\degree C}{30\degree C-20\degree C}=467.77kJ/kg

Before I can get the specific internal temperature of state 2, I need the temperature of state 2.

P_2V_2=mRT_2

T_2=\frac{P_2V_2}{mR}=\frac{4bar*0.05672m^3}{0.353kg*188.58J/(kg*K)}=340.8K=67.65\degree C

Specific internal energies of state 2 from table A-18 with

p=4bar

around my target

T=67.65\degree C

$T$	$u$
$60\degree C$	$521.1kJ/kg$
$70\degree C$	$538.1kJ/kg$

u_2=521.1kJ/kg+(538.1kJ/kg-521.1kJ/kg)\frac{70\degree C-67.65\degree C}{70\degree C-60\degree C}=525.1kJ/kg

Now it all comes together.

Q=m(u_2-u_1)+W=0.353kg*(525.1kJ/kg-467.77kJ/kg)+(-26.86kJ)=\boxed{-6.62kJ}