Temperature/Pressure relationship of a fluid in a sealed chamber

Question

What will be the change of pressure of water in a completely filled and sealed vessel per degree fahrenheit temperature change. ie: 10 gallons of 65˚f water at 125psi raises in temperature to 75˚f

Answer 1

This an interesting problem that you don't see everyday and I don't have nice big Thermodynamic charts so I went to the internet for Thermodynamic table calculator for sub-cooled water.

Starting with 65 F and 125 psi or 140 psia, I get

Density = 62.4 lbs/ft³

Volume = .016 ft³/lb

Then when you heat it to 75 F, you get

Density = 62.3 lbs/ft³

Volume = .0161 ft³/lb

Temp = 75 F

Pressure = 140 psia

Then when you compress it back to .016 ft³/lb & keep the temperature constant, you get

Temp = 75 F

Density = 62.4 lb/ft³

Volume = .016 ft³/lb

But the Pressure is now = 732 psia

if h = enthalpy, T = temperature, v = specific volume, c_p = heat capacity & P = pressure and I use the governing equation (2 = final state & 1 = initial state)

h₂ -h₁ = c_p(T₂ -T₁) + v(P₂-P₁) and see if the first two terms equals the last term after plugging in the enthalpies and internal energies (not included here), I get a slight discrepancy of about .153 Btu/lb. This tells me that the answer is very close to 732 psia (absolute pressure). I calculated everything on a per lb basis.

So the difference is pressure going from 65 F to 75 F is about 732-140 psia = 592 psia

Answer 2

Hi Fred

This problem exercises the ideal gas laws. The volume, number of moles and gas constant are the same for both cases (PV=nRT) ' so this question reduces to P1/T1 = P2/T2.

The only hitch here is that, for the temperatures, you can't use Farenheit degrees in the temperature calculation. you need a scale that has an absolute zero point. So you need to use Rankine degrees.

Zero Rankine = - 460 degrees F. you should be able to do the calculation now.