§ME 231 Homework 4

Sep 25, 2025

§3.62

Determine the volume, in m3, occupied by 2 kg of H2O at 100 bar, 400°C, usinga. data from the compressibility chart.b. data from the steam tables.Compare the results of parts (a) and (b) and discuss.

m=2kg

P=100bar

T=400\degree C=673.15K

§a.

Critical temperature from the end of table A-2:

T_C=374.14\degree C=647.29K

Critical temperature from the end of table A-3:

P_C=220.9bar

Now we get the reduced temperature and pressure:

T_R=\frac{T}{T_C}=\frac{673.15K}{647.29K}=1.04

P_R=\frac{P}{P_C}=\frac{100bar}{220.9bar}=0.453

Since

T_R<2

and

P_R>0.2

z

cannot be safely assumed to be

1

. I must refer to the generalized compressibility chart.

Figure 3.12 from Fundamentals of Engineering Thermodynamics

T_R\approx 1.00

Z\approx 0.85

Now we use a variation of the ideal gas equation with compressibility effects in mind:

Pv=ZRT

Solving for

v

v=\frac{ZRT}{P}

I will need

R

so let's get that real quick:

R=\frac{\bar{R}}{M}=\frac{8.3144598\frac{J}{mol*K}}{18.01528 \frac{g}{mol}}=0.4615227\frac{J}{g*K}

v=\frac{ZRT}{P}=\frac{0.85 * 0.4615227\frac{J}{g*K} * 673.15K}{100bar}=0.026407\frac{m^3}{kg}

V=mv=2kg*0.026407\frac{m^3}{kg}=\boxed{0.0528m^3}

§b.

From table A-5 where

P=100bar=10.0MPa

and

T=400\degree C

v=0.02641m^3/kg

V=mv=2kg*0.02641m^3/kg=\boxed{0.0528m^3}

§Discuss

The steam tables hide the finer details about compressibility effects and just gives your easy specific values (less work = happier me). They both lead to the same results which should not be a surprise.

§3.65

Determine the temperature, in °C, of air at 30 bar and a specific volume of 0.013 m3kg.

P=30bar

v=0.013m^3/kg

R=287\frac{J}{kg*K}

Pv=RT

T=\frac{Pv}{R}=\frac{30bar * 0.013m^3/kg}{287\frac{J}{kg*K}}=136K=\boxed{-137\degree C}

§3.77

Carbon dioxide (CO2) contained in a piston–cylinder arrangement, initially at 6 bar and 400 K, undergoes an expansion to a final temperature of 298 K, during which the pressure–volume relationship is pV1.2 = constant. Assuming the ideal gas model for the CO2, determine the final pressure, in bar, and the work and heat transfer, each in kJ/kg.

P_1=6bar

T_1=400K

T_2=298K

pV^{1.2}=\text{const}=C

n=1.2

R=0.1889\frac{kJ}{kg*K}

Because it's an ideal gas,

P_2

can be evaluated easily:

\frac{P_1}{P_2}=\left(\frac{T_1}{T_2}\right)^\frac{n}{n-1}

P_2=\frac{P_1}{\left(\frac{T_1}{T_2}\right)^\frac{n}{n-1}}=\frac{6bar}{\left(\frac{400K}{298K}\right)^\frac{1.2}{1.2-1}}=\boxed{1.026bar}

Work also has a friendly equation:

W=\frac{mR(T_2-T_1)}{1-n}

We don't want the total work, just the specific work in

kJ/kg

\frac{W}{m}=\frac{R(T_2-T_1)}{1-n}=\frac{0.1889\frac{kJ}{kg*K} * (298K - 400K)}{1 - 1.2}=\boxed{96.34kJ/kg}

Q/m

cannot be evaluated directly so I will need

\Delta E/m=\Delta U/m

c_v

from table A-20 where

T=300K

c_v=0.657\frac{kJ}{kg*K}

\Delta U=mc_v(T_2-T_1)

\frac{\Delta U}{m}=c_v(T_2-T_1)=0.657\frac{kJ}{kg*K}*(298K - 400K)=-67.01kJ/kg

\Delta E/ m=\Delta U/m=Q/m-W/m

Q/m=\Delta U/m+W/m=-67.01kJ/kg+96.34kJ/kg=\boxed{29.33kJ/kg}

§3.83

Argon contained in a closed, rigid tank, initially at 50°C, 2 bar, and a volume of 2 m3, is heated to a final pressure of 8 bar. Assuming the ideal gas model with k = 1.67 for the argon, determine the final temperature, in °C, and the heat transfer, in kJ.

T_1=50\degree C=323.2K

P_1=2bar

P_2=8bar

V=2m^3

k=1.67

PV=mRT

Here

m

and

R

are constant so we can rewrite it in terms of constants:

\frac{PV}{T}=mR

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

Abd since it's a rigid tank:

V_2=V_1

\frac{P_1\cancel{V_1}}{T_1}=\frac{P_2\cancel{V_1}}{T_2}

\frac{P_1}{T_1}=\frac{P_2}{T_2}

Now we can solve for

T_2

T_2=\frac{P_2T_1}{P_1}=\frac{8bar * 323.2K}{2bar}=1292.8K=\boxed{1020\degree C}

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

The gas does not do work on the environment so

W=0

Q=m(u_2-u_1)

From table A-21 in the "Monoatomic gases" row:

\frac{C_P}{R}=\frac{\bar{C_P}}{\bar{R}}=2.5

For ideal gases

u

(internal energy) and

h

(enthalpy) are functions of

T

. So a change is

u

is some scalar of a change in

T

u_2-u_1=c_v(T_2-T_1)

Q=mc_v(T_2-T_1)

Still need the mass:

PV=mRT

P_1V=mRT_1

m=\frac{P_1V}{RT_1}

R=\frac{\bar{R}}{M}=0.2081\frac{kJ}{kg*K}

m=\frac{P_1V}{RT_1}=\frac{2bar * 2m^3}{0.2081\frac{kJ}{kg*K} * 323.2K}=5.9473kg

Now time for

c_v

:(I don't need this but keeping it here for memory)

c_p=\frac{kR}{k-1}

c_v=\frac{R}{k-1}=\frac{0.2081\frac{kJ}{kg*K}}{1.67-1}=0.3106\frac{kJ}{kg*K}

Q=5.9473kg * 0.3106\frac{kJ}{kg*K} * (1292.8K - 323.2K)=\boxed{1791kJ}

§3.91

Carbon dioxide (CO2) is compressed in a piston–cylinder assembly from p1 = 0.7 bar, T1 = 280 K to p2 = 11 bar. The initial volume is 0.262 m3. The process is described by pV1.25 = constant. Assuming ideal gas behavior and neglecting kinetic and potential energy effects, determine the work and heat transfer for the process, each in kJ, using (a) constant specific heats evaluated at 300 K, and (b) data from Table A-23. Compare the results and discuss.

P_1=0.7bar

T_1=280K

P_2=11bar

V_1=0.262m^3

pV^{1.25}=const=C

n=1.25

§(a)

From table A-20 with

T=300K

as instructed in the question:

c_v=0.657\frac{kJ}{kg*K}

\Delta u=u_2-u_1=c_v(T_2-T_1)

T_1

was given but

T_2

is also needed to get

\Delta u

. Because we're assuming ideal gas, the following holds true:

\frac{P_1}{P_2}=\left(\frac{T_1}{T_2}\right)^\frac{n}{n-1}

Now solving for

T_2

\left(\frac{P_1}{P_2}\right)^\frac{n-1}{n}=\frac{T_1}{T_2}

T_2=\frac{T_1}{\left(\frac{P_1}{P_2}\right)^\frac{n-1}{n}}=\frac{280K}{\left(\frac{0.7bar}{11bar}\right)^\frac{1.25-1}{1.25}}=485.8K

\Delta u=0.657\frac{kJ}{kg*K}*(485.8K-280K)=135.21kJ/kg

It would help to have the mass to turn

\Delta u

into

\Delta U

R=0.1889\frac{kJ}{kg*K}

PV=mRT

P_1V_1=mRT_1

m=\frac{P_1V_1}{RT_1}=\frac{0.7bar * 0.262m^3}{0.1889\frac{kJ}{kg*K} * 280K}=0.347kg

\Delta U=m\Delta u=0.347kg * 135.21kJ/kg=46.92kJ

Change in internal energy is great but what we really need is

Q

and

W

W

can be evaluated easily:

W=\frac{mR(T_2-T_1)}{1-n}=\frac{0.347kg * 0.1889\frac{kJ}{kg*K} * (485.8K - 280K)}{1-1.25}=\boxed{-53.96kJ}

And

Q

can now be evaluated:

\Delta E=\Delta U=Q-W

Q=\Delta U+W=46.92kJ-53.96kJ=\boxed{-7.04kJ}

§(b)

From table A-23 where

T=300K

\bar{u_1}=6939\frac{kJ}{kmol}

From table A-23 where

T=485.8K\approx 490K

\bar{u_2}=13158\frac{kJ}{kmol}

These values aren't very helpful when normalized by moles. Converting them to kilograms:

M=44.01g/mol

u_1=\frac{\bar{u_1}}{M}=\frac{6939\frac{kJ}{kmol}}{44.01g/mol}=157.67kJ/kg

u_1=\frac{13158\frac{kJ}{kmol}}{44.01g/mol}=298.98kJ/kg

\Delta U=m(u_2-u_1)=0.347kg*(298.98kJ/kg-157.67kJ/kg)=49kJ

That's pretty close!

W

would be evaluated in the same way as in part (a) so let's jump to

Q

W=\frac{mR(T_2-T_1)}{1-n}=\boxed{-53.96kJ}

Q=\Delta U+W=49kJ-53.96kJ=\boxed{-4.96kJ}