Calculating Min Water Required to Condense Steam from 5 bar to 4 bar

[matt222]

Homework Statement

If we have a containment sprayed by a subcooled water to condense the steam, the pressure decreased from 5 bar to 4 bar, calculate the amount of sprayed water required as a minimum to decrease this pressure?
Heat of vaporization 2550 kJ/Kg, steam at saturated condition, subcooling 70c and C=4kJ/kg-k which is constant over pressure and temperature.

Homework Equations

The Attempt at a Solution

at 1 kg we have Q=2550kJ ,

Q=mC(T2-T1)------> m=2550kJ/((4kJ/kgk)x(100-70)K)=21.25 Kg!

do you agree with me?

 

[mark726]

I would first like to clarify the problem statement. It is not clear what is meant by "containment" and how it is related to the subcooled water and steam. Can you please provide more information or a diagram to better understand the setup?

Assuming that the containment is a closed vessel and the subcooled water is being sprayed onto the steam inside the containment, then your calculation for the amount of sprayed water seems correct. However, it is important to note that the subcooling of 70°C is not constant over pressure and temperature. It may be constant for the specific conditions given in the problem, but it will vary for different pressures and temperatures.

Also, it would be helpful to know the initial temperature and pressure of the steam before it is sprayed with the subcooled water. This information is necessary to accurately calculate the amount of water required to decrease the pressure from 5 bar to 4 bar.