Two tanks of the same volume are connected by a pipe containing a valve. Initially, the valve is closed, one tank is evacuated, and the other tank contains 40 g of steam at 15 bars and 280 C. The valve is opened and the steam flows into the evacuated tank until pressure equilibrium is reached. If the steam temperature at the end of the process is 440C in both tanks,determine
a. the final equilibrium pressure, in bars
b. the heat transfer, in kj
c. the entropy change, in kj / K, of the steam
d. the total entropy change, in kj / K

(a) The final equilibrium pressure in bars = 10 bars (b) The heat transfer in Kj=14.264 KJ (c) The entropy change inkj/k of the steam = 0.03KJ/K

(d) The total entropy change, in kj/K

How to Calculate the equilibrium and Entropy change?

We should understand that insulated tanks deal with systems that has change with respect to the amount of heat transferred through its surroundings.

The given parameters are

h₁=2993.4KJ/kg

S₁=6.8402KJ/kh.K

V₁=0.163cm³/kg

All these values are drawn from the steam table

The tank volume can be determined first using the formula

V=mx

Volume = 0.00652m³ while the mass balance = 0.5m₁=mₙ

This implies that the specific balance is V₂=V/mₙ

V₂=0.3350KJ/kg

S₂=7.59KJ/kg.k

(a) The final equilibrium pressure in bars = P₂=10 bars

(b) The heat transfer, in kj is

m₂h₁+Q=m₂h₂

This implies that Q when made the subject is 14.264KJ

(c) The entropy change in kj/K of the steam is

Δs=0.03KJ/K

(d) The total entropy change, in kj/k steam is determined by

Δs=m(s₂-s₁) +Q/T

This implies that Δs=0.05KJ/K

In conclusion, (a) The final equilibrium pressure in bars = 10 bars (b) The heat transfer in Kj=14.264 KJ (c) The entropy change inkj/k of the steam = 0.03KJ/K

(d) The total entropy change, in kj/K

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