Question

Two adiabatic rigid tanks are connected by a valve. The left tank is filled with 4 ft³ of air initially at 28. 5 psia and 89. 4°F. The right tank is filled with 13. 4 ft3 of air initially at 19. 7 psia and 107. 8°F. The valve is opened and the system reaches equilibrium. Assume that the specific heat is constant and can be taken at a temperature of 80°F. A) What is the final temperature of the system after the valve is opened? °F b) What is the final pressure of the air after the valve is opened? psia

Answer 1

To find the final temperature and pressure of the air in the system after the valve is opened and the two tanks reach equilibrium, we can apply the principles of adiabatic and isentropic processes for an ideal gas. We will use the following equations:

For adiabatic process: P1 * V1^γ = P2 * V2^γ, where γ is the specific heat ratio (Cp / Cv) for the gas.

For isentropic process: T1 * V1^(γ-1) = T2 * V2^(γ-1), where T1 and T2 are temperatures in absolute units (Rankine).

We are given:
Left tank: V1 = 4 ft³, P1 = 28.5 psia, T1 = 89.4°F
Right tank: V1 = 13.4 ft³, P1 = 19.7 psia, T1 = 107.8°F

Step 1: Convert temperatures to Rankine (°R):
T1_left = 459.67 + 89.4 = 549.07 °R
T1_right = 459.67 + 107.8 = 567.47 °R

Step 2: Calculate the final temperature (T2) when the two tanks reach equilibrium.

Using the isentropic process equation for both tanks:
T1_left * V1_left^(γ-1) = T2 * V2_left^(γ-1)
T1_right * V1_right^(γ-1) = T2 * V2_right^(γ-1)

Since both tanks have the same final temperature (T2), we can set the two equations equal to each other:

T1_left * V1_left^(γ-1) = T1_right * V1_right^(γ-1)

Step 3: Calculate the final pressure (P2) when the two tanks reach equilibrium.

Using the adiabatic process equation for both tanks:
P1_left * V1_left^γ = P2 * V2_left^γ
P1_right * V1_right^γ = P2 * V2_right^γ

Since both tanks have the same final pressure (P2), we can set the two equations equal to each other:

P1_left * V1_left^γ = P1_right * V1_right^γ

Step 4: Substitute the given values and solve for T2 and P2.

For T2:
(T1_left / T1_right) = (V1_right / V1_left)^(γ-1)
T2 = T1_left / ((V1_right / V1_left)^(γ-1))
T2 = 549.07 / ((13.4 / 4)^(1.4-1))
T2 ≈ 549.07 / (3.35^0.4)
T2 ≈ 549.07 / 1.766
T2 ≈ 310.95 °R

Now, for P2:
P1_left * V1_left^γ = P2 * V2_left^γ
P2 = P1_left * (V1_left / V1_right)^γ
P2 = 28.5 * (4 / 13.4)^1.4
P2 ≈ 28.5 * 0.3824
P2 ≈ 10.91 psia

Step 5: Convert the final temperature back to Fahrenheit.
Final temperature (T2) = 310.95 - 459.67 ≈ -148.72°F

So, the final temperature of the system after the valve is opened is approximately -148.72°F, and the final pressure of the air is approximately 10.91 psia.