Question

An insulated 40 ft3 rigid tank contains air at 50 lbf/in2 and 120oF. A valve connected to the tank is now opened, and air is allowed to escape until the pressure inside drops to 25 lbf/in2. The air temperature within the tank and at the exit during this process is kept constant at 120oF by an electric heater placed in the tank. Determine the electrical work [Btu] for the process. Assume that the air behaves like an ideal gas.

Answer 1

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:

$m_{1}=\frac{P_{1}V }{RT_{1} }$

Where P1=50 psig

V=40 ft^3

T1=120°F=580 R

$m_{1}=\frac{50*40}{0.3704*580} =9.3 lbm$

The final mass m2 is:

$m_{2} =\frac{P_{2}V }{RT_{2} }$

P2=25 psig

V=40 ft^3

T2=580 R

$m_{2}=\frac{25*40}{0.3704*580} =4.66 lbm$

me=9.3-4.66= 4.64 lbm

$W_{e}=(4.64*138.66)+(4.66*98.9)-(9.3*98.9)=184.49 Btu$

Answer 2

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