Question

A 500 L rigid tank contains a saturated water mixture at 200 C as shown: the mixture is 40% liquid and 60% vapour by volume (State 1). A valve on top of the tank is opened, and saturated vapor is slowly withdrawn from the tank. Heat transfer occurs during this process such that the temperature in the tank remains constant. The valve is closed when 50 % of the initial mass is withdrawn from the tank (State 2)

(a) Determine the amount of saturated water withdrawn from the tank (in kg).

(b) Determine the quality of the mixture in State 2.

(c) Determine the amount of heat transferred during the process (in

MJ), and clearly indicate the direction of the heat flow.

Answer 1

Volume of tank= 500L, volume of liquid = 40%. Hence volume of liquid = 500*0.4= 200L and Volume of vapor =500*0.6= 300L

Specific volume at saturated conditions of 200 deg.c ( from steam tables)

: Liquid =1.126 cm³/g and gas= 126 cm³/gm

Density , liquid = 1/1.26 g/cc=0.793g/cc=793 g/L and gas= 1/126 gm/cc=0.0079g/cc=7.9 g/L

Masses : liquid = 793g/L* 200L=158600 gm = 158.6 kg and gas= 7.9* 300 gm=2370 gm= 2.370 kg

Total mass of the contents in the tank = 158.6+2.370= 169.97 kg

50% of the mass is withdrawn from the tank, mass of tank =160.97*0.5= 80.485 kg

All the mass of vapor is withdrawn from the tank which is 2.370 kg

liquid withdrawn from the tank = 80.485-2.370=78.115 kg

Liquid remaining in the tank = 158.6-78.115 =80.485 kg

Entire vapor is removed. Hence the entire tank will be containing liquid only after 50% of the mass is withdrawn.

Enthalpy of initial mass = total enthalpy of liquid + enthalpy of vapor

= 158.6kg* 852.4 KJ/Kg+ 2.370*1938.6 kj/kg ( the specific enthalpy data is obtained from steam tables)=139785 KJ

Enthlpy of liquid in the tank = 852.4*80.485=68605 KJ

Change in enthalpy =final enthalpy-initial enthalpy = 68605-139785 Kj=-71180 Kj =-71.18 MJ. Heat is transferred from tank to surroundings.