A free TFS practice problem (maybe interesting for HVACR folks too)

A rigid tank with an internal volume of 100 gallons contains saturated water at 50 psia and 50% of the mass of water inside the tank is liquid. The tank is fitted with a pressure relief valve (PRV) set to open if the pressure ever reaches 150 psia. In one occasion, a process upset caused the tank to be heated and the PRV was activated. The average temperature (ºF) inside the tank when the PRV opened was most nearly:

(A) 615

(B) 635

(C) 715

(D) 805

Initial State:

P = 50 psia, quality = 0.5 (sat'd mixture),

two states fix specific volume (v_i) on steam table, use quality to find initial specific volume,  vf < v_i < vg. 

Rigid Tank implies Constant Volume Process,  so v_i = v_f = 4.267 ft3/lbm.

Final State - At instant when PRV was opened:

P = 150 psia, v_f = 4.267 ft3/lbm, two states fix temperature = 635 F

I got Ans B

Thank you for the problem!.

ICM said:

Initial State:

P = 50 psia, quality = 0.5 (sat'd mixture),

two states fix specific volume (v_i) on steam table, use quality to find initial specific volume,  vf < v_i < vg. 

Rigid Tank implies Constant Volume Process,  so v_i = v_f = 4.267 ft3/lbm.

Final State - At instant when PRV was opened:

P = 150 psia, v_f = 4.267 ft3/lbm, two states fix temperature = 635 F

I got Ans B

Thank you for the problem!.

Click to expand...

Nicely done!