Answer:
The electrical work for the process is 256.54 Btu.
Explanation:
From the ideal gas equation:
n = PV/RT
n is the number of moles of air in the tank
P is initial pressure of air = 50 lbf/in^2 = 50 lbf/in^2 × 4.4482 N/1 lbf × (1 in/0.0254m)^2 = 344736.2 N/m^2
V is volume of the tank = 40 ft^3 = 40 ft^3 × (1 m/3.2808 ft)^3 = 1.133 m^3
T is initial temperature of air = 120 °F = (120-32)/1.8 + 273 = 321.9 K
R is gas constant = 8.314 J/mol.k
n = 344736.2×1.133/8.314×321.9 = 145.94 mol
The thermodynamic process is an isothermal process because the temperature is kept constant.
W = nRTln(P1/P2) = 145.94×8.314×321.9×ln(50/25) = 145.94×8.314×321.9×0.693 = 270669 J = 270669 J × 1 Btu/1055.06 J = 256.54 Btu
Answer:
The electrical work is 184.49 Btu
Explanation:
T=120°F=580R
From table of properties of air the enthalpy of air at 580R is he=138.66 Btu/lbm, internal energy at initial and final state is u=98.9 Btu/lbm
The mass balance is:
me=m1-m2
And the energy balance assuming the tank as the steady flow is:
We=mehe+m2u2-m1u1
The initial mass m1 is:
[tex]m_{1}=\frac{P_{1}V }{RT_{1} }[/tex]
Where P1=50 psig
V=40 ft^3
T1=120°F=580 R
[tex]m_{1}=\frac{50*40}{0.3704*580} =9.3 lbm[/tex]
The final mass m2 is:
[tex]m_{2} =\frac{P_{2}V }{RT_{2} }[/tex]
P2=25 psig
V=40 ft^3
T2=580 R
[tex]m_{2}=\frac{25*40}{0.3704*580} =4.66 lbm[/tex]
me=9.3-4.66= 4.64 lbm
[tex]W_{e}=(4.64*138.66)+(4.66*98.9)-(9.3*98.9)=184.49 Btu[/tex]
