A 10 m3 tank contained saturated liquid water at 1 bar. First step with the fixe

Question

A 10 m3 tank contained saturated liquid water at 1 bar. First step with the fixed volume if the tank, it is heated until the water reaches to 10 bar and the water remains as saturated liquid. second step, the heating source is removed, and the tank is expanding adiabatically to 1 atm. No shaft work is involved in the process. 1. What is the final temperature of the tank contents? What is the quality of the final stream? (Hint, enthalpy change of the second step is O) How much work has the tank done to the environment? a. b. c.

Explanation / Answer

Volume of tank = 10 m3

Here we will make use of steam table

For saturated liquid at 1 atm

Tsat = 100 oC

H1sat =417.436 KJ/Kg

Now heat is being supplied and the water is pressurized to 10 bar and yet the water is maintained as saturated liquid

From steam table at P = 10 bar

T2sat = 180 oC

H2sat = 762.68 KJ/Kg

Now the steam is adiabatically expanded to 1 bar

Hence during adiabatic expansion there is no loss of heat

Hence final enthalpy = 762.68 KJ/Kg and P = 1 bar

From steam table

at P = 1 bar

Hl = 417.4 KJ/Kg

Hs = 2505.4 KJ/Kg

Let he quality of steam be x

H =x*Hs + (1-x)*Hl

762.68 = x*2505.4 + (1-x)*417.4

solving the above equation for x we get

x = 0.168

b) Hence the final state is a mixture with quality of steam being 16.8 %

a) Final temperature will be nothing but the saturation temperature Tsat = 100 oC

c) Here there were two process. In first process heat was given to the tank from the environment and in second one there was no heat exchange between system and surrounding.

hence tank has not done any work on the environment . The environment has some work on the system ( H2-H1 =762.68 - 417.4 = 345.28 KJ/Kg)

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