Question

A 10.0-m3 tank contains steam at 275°C and 15.0 bar. The tank and its contents are cooled until the pressure drops to 1.8 bar. Some of the steam condenses in the process.
(a) How much heat was transferred from the tank?
(b) What is the final temperature of the tank contents?

(c) How much steam condensed (kg)?

Answer 1

Solution som n moles steam(W), 2755, 15 bax = lom nmoles of water (+4) 1.8 bar 10.0 m H₂O(v) 10.om Tank min kg 2756, 15 bar → mu [kg H₂O(v) me [kg H₂04) 1.8 bar, saturated fig- Schematic of process final temp of tank contents at 1.8 bay from table B6, T₂ = 116.9 C ② Steam Condensed (kg) kg Total mass of water = min = 10m x 0.1818 m3 55 kg 10.omi Mass balance mrt ml=55 volume additivity & Vrtke my (1428 milky) + me (0.001048 m (kg) my = 7.0kg , me = 48 kg condensed © Heat transferred from tank, from table Bf at P=15 bar T = 275°c = 2739.2 KJ/Kg, Vin = 0.1818 molkg from table B6, Ud=439.2 KJ/kg, Va= 0.001048 m ů = 25121 kJ /kg, W = 1.428 m²/kg Energy balance Q = SU = myúrt me te min ûin Q=TGX2512.1) + (48x439.2) -(55 x 273.9.2)) =-1.12 x 105 kJ Ein- m 1kg