MAE 204 Thermodynamics Assignment: HW #5 Solutions and Analysis

This assignment solution addresses three thermodynamics problems from MAE 204. The first problem involves an isothermal expansion of air, calculating initial and final specific volumes, work done, and heat transfer. The second problem analyzes a rigid container with R-134a undergoing cooling, determining heat transfer, the final phase, and temperature. The third problem explores the properties of a substance using the ideal gas equation, steam tables, and a compressibility chart, calculating and comparing specific volumes and determining the percentage error of the estimates. The solution demonstrates the application of thermodynamic principles and equations to solve practical engineering problems, including the use of T-V and P-V diagrams.

Thermodynamics
Thermodynamics
SEPTEMBER 30, 2019

Question: 1
P1 = 2 MPa = 2000 KPa
T1 = 350 °C = 350+273 = 623K
P2 = 500KPa
T2 = T1 = 623 K
P1 V1 = mRT1
Specific volume V1 = V 1
m
P1 V1 = RT1
V1 = RT 1
P1 = 287 x 623
2 X 106
V1 =0.0894 m^3/Kg
V2 = RT 2
P2 = (287 x 623)/(500x10^3) m^3/Kg
Initial specific volume = V1 = 0.0894 m^3 /Kg
Final specific volume = V2 = 0.3576 m^3 /Kg
T-V diagram
P-V diagram
T
V
P
1 2

Work done = P1V1ln ( P 1
P 2 )
= 2x10^6 x 0.0894ln ( 2
0.5 )
= 2.479 x 10^6 Nm/Kg
Heat transfer for isothermal process = work done
δQ = δW = 2.479 X 10^6 J/Kg
δQ = 2.479 MJ/Kg
Question: 2
M = 5 Kg
P1 = 800 KPa
T1 = 70 °C = 70+273 = 343 K
P2 = 400 KPa
Since container is rigid
V1 = V2
P1/ T1 = P2/ T2
800/ 343 = 400/ T2
T2 = 171.5 – 273 = -101.5° - 273 = -101.5°C
Since boiling temp of R134a is -26.1 °C and T2 < boiling temp therefore R-134a has changed its phase
from vapor state to liquid or solid state
CV = 0.23901 KJ/KgK
Q = ∫
T 1
T 2
CvdT (at constant volume)
Q = mCV (T2 – T1)
Q = 5 X 0.2390 X (171.5 -343) KJ
Q = 204.951 KJ
Negative sign shows heat transfer takes place outside the system

T-V diagram
Question: 3
P = 15 MPa = 150 bar
T = 370 °C = 370 + 273 = 643 K
By ideal gas equation:
PV = mRT
P V = mRT
P V/m = RT
V = RT/P = (287X643)/(15 X 10^6) = 0.0123 m^3/Kg
By steam table
V(at P= 15MPa, T = 350 °C) = 0.010338
V(at P= 15MPa, T = 400 °C) = 0.015649
By interpolation
(370-350)/(400-350) = (v-0.010338)/(0.015649-0.010338)
20/50 = (v-0.010338)/0.00531
V = 0.0124624 m^3/Kg
From compressibility chart
At P = 150 bar & T = 370 °C
T
V
1
2

V =?
V = 0.01 m^3 / Kg
For (a & c)
True value = 0.01 m^3 /Kg
Estimate value = 0.0123 m^3 /Kg
% error = 100 X ((0.0123-0.01)/0.01)
=23%
For b & c
True value = 0.01 m^3 /Kg
Estimate value = 0.0124624 m^3 /Kg
% error = 100 X ((0.0124624-0.01)/0.01)
=24.624%