Answer:
isobaric expansion = 281.09 Btu
isothermal compression= 72 Btu
Explanation:
The first law of thermodynamics is:
[tex]Q_{AB}=W_{AB}+deltaU_{AB}[/tex]
where:
Q=heat transferred
W= work
U=internal energy
[tex]W_{AB}=P*(V_{B}-V_{A})[/tex]
[tex]U_{AB}=n*C_{v}(T_{2}-T_{1})[/tex]
P=pressure, V= volume, T= temperature, n = moles, Cv= specific heat at constant volume.
In a isobaric process heat transferred is:
[tex]Q=P*(V_{B}-V_{A})+n*C_{v}(T_{2}-T_{1})[/tex]
For an isothermal process (T2-T1 = 0) so
[tex]Q=P*(V_{B}-V_{A})= W_{AB}[/tex]
From the data we know that the energy transferred to the system in the isothermal compression by work was 72 Btu that is the heat transferred to the system.
For the first process
[tex]Q=P*(V_{B}-V_{A})+n*C_{v}(T_{2}-T_{1})[/tex]
we have to properties at the beginning of the process : temperature (350°F) and specific volume (V/mass)
[tex]specific-volume=\frac{71.7 ft^{3}}{3Lb}=23.9\frac{ft^{3}}{Lb}[/tex]
we use this information in the appropriate unit to find the pressure in thermodynamic tables.
T1= 176°C
v1= 1.49 m^3/kg
P=1.37 bar
in the second state we have
P=1.37 bar =137000Pa
[tex]v_{2}=\frac{85.38ft^{3}}{3Lb}= 28.46\frac{ft^{3}}{Lb}[/tex]
with thee properties we check in the thermodynamic tables
T2= 255°C
[tex]n=mass/Mw = 3Lb*\frac {1kg}{2.2Lb}*\frac{1000gr}{1kg}*\frac{1mol}{18gr}=75.75 mol[/tex]
we usually find Cp on tables for water but from the Mayer relation we have:
[tex]C_{v}=C_{p}+R[/tex]
Cp for water vapor is: 33.12 J/mol*K
R=8.314 J/mol*K
Cv= 41.434 J/mol*K
replacing in the equation for Q
[tex]Q=137000 Pa*(2.41m^{3}-2.030m^{3})+75.75mol*41.434\frac{ J}{mol*K}*(528.15-449.81 K)=296569J[/tex]
296569J =281.09 Btu
