Question

Assignment 3 Question: A 1 m3 rigid tank has propane at 100 kPa and 300 K. The tank is connected to another 0.5 m3 rigid tank which has propane at 250 kPa and 400 K by a ball valve. The valve is opened and both tanks come to a uniform state at 325 K. 1- State all assumptions you need to solve the problem. 2- Calculate the final pressure of propane in both tanks? 3- Calculate the mass of propane before and after opening the valve in each tank. 4- Sketch the process on a P-V diagram. 5- Find the compressibility factor of propane in each tank before opening the valve. 6- Calculate the change in internal energy and enthalpy of propane in each tank.

Answer 1

Ok I will answer throught question 5, cause I have a little doubt with the expressions to use in part 6.

For tank 1 and Tank 2 we need to calculate the moles of propane, so we can calculate the mass of propane in part 3, or we can use the derivate of the ideal gas equation with mass, so we can know the mass of propane, and then the final pressure:

From the ideal gas equation modified:

PV = mRT Where R for propane = 0.1886 kJ/ kg K

m1 = PV / RT

m1 = 100 * 1 / 0.1886 * 300 = 1.7674 kg

m2 = 250 * 0.5 / 0.1886 * 400 = 1.6569 kg

Mass of propane after opening the valve: 1.7674 + 1.6569 = 3.4243 kg

V = 1 + 0.5 = 1.5 m³

P = 3.4243 * 0.1886 * 325 / 1.5

P = 139.93 kPa is the final pressure.

I have a doubt with compressibility factor calculations, however I look online a table for this and found that for the first tank at those conditions Z = 0.99 while for tank 2 Z = 0.98

For part 6, I have a doubt cause I couldn't find the interna energy for propane at those conditions, but I do know that the equation to use is :

U = u*m

H = h*m

where u and h are coefficientes of internal energy, but I couldn't find those values. If you do have them, use it in these equations.

If you need something else, something to be fixed, or is there anything you don't understand of what I did, tell me in a comment and I'll help you out.

Hope this helps