Initially, an insulated rigid tank contains 3.94 kg of water at 240 o C and 1 MP
ID: 1561661 • Letter: I
Question
Initially, an insulated rigid tank contains 3.94 kg of water at 240oC and 1 MPa. A heater in the wall of the tank maintains the wall temperature at 600oC. After a period of time, the pressure in the tank rises to 1.5 MPa. Determine the final volume (m3) of the water in the tank. Assume variable specific heats (use the property tables for water).
Initially, an insulated rigid tank contains 3.94 kg of water at 240oC and 1 MPa. A heater in the wall of the tank maintains the wall temperature at 600oC. After a period of time, the pressure in the tank rises to 1.5 MPa. Determine the final volume (m3) of the water in the tank. Assume variable specific heats (use the property tables for water).
Explanation / Answer
The specific gravity of water at 240oC with respect to water at 15 degree celsius is 0.818. Hence the density of water at 240 degree C is 818 kg/m^3.
Mass of the water at this temperature is 3.94 kg. Hence the volume of the water is 3.94/818 = 0.00482 m^3. This is the volume of the water at 240o celsius at 1MPA.
From Boyle's law,
P1V1=P2V2. Now the pressure in the tank is increased to 1.5MPA,i.e. P2=1.5MPA. Hence the new volume of water is V2=1*10^6*0.00482/(1.5*10^6)=0.003211 m^3.