5 replies to this topic

[ChaosTamed]

Hey friends.

I'm sure this is a basic chemical engineering calculation but I've been out of engineering for 6 years now and I don't have my books around me. I've been scratching my head for a couple of days so hope you can help.

If a tank of propane, 80% full, is at 834kPa at 20C and is then heated to 55oC (by an external fire say) so that the pressure becomes 1900kPa, how do I determine the vapour mass fraction within the tank?

Hoping you can help.

Regards
M

[chenblue]

Hi,

Ur case is a standard promblem for "phase equilibrium". Because it is a closed system, and you have known the P and T of both initial state and final state, what u want to do is just two steps. The first step is to find the state equation of gas phase, and the second step is use the Clapeyron equation (http://en.wikipedia....peyron_relation) to calculate the delta V (how many propane liquid becomes gas) in Clapeyron equation. To simplify this problem, u can use ideal gas to approah it. But u should know that it may be not a good approximation, because the pressure is a bit high.

chenblue

[Art Montemayor]

M:

Chenblue is in a different ballpark; you don't have an Equation of State situation and Clapyron is not required. The problem is far simpler but you need to furnish additional data.

You state you’re an engineering graduate, but yet you submit a typical Student Homework problem in the Student Forum. What is all the background to this query? It helps to have ALL the facts.

Your stated “problem” asserts certain conditions that fix the state of the Propane. And in the real-life scenario, one is not normally interested in the “vapor mass fraction within the tank”. Therefore, I suspect there is a longer story behind this query.

For example, if one goes to the NIST free on-line Database, one learns that:

At 20 oC, saturated propane is at 836.54 kPaA

At 55 oC, saturated propane is at 1,907.7 kPaA

A professional engineer would give the units of the pressures he is employing; you fail to do so. I have to assume you are a student and the pressures are those from a text book and, as such, in absolute terms. Therefore, the conditions you have given us for your propane identify them both as being SATURATED (an important fact you also failed to state). This is the normal way Propane is stored in real-life.

You also failed to state the real-life information regarding your propane tank: its dimensions and its water capacity (or total internal volume). With that information and the thermodynamic information you obtain from the NIST Database, you have a quick and accurate answer to your “problem”.

[Andree]

my (third) proposal to solve this problem is:

1) p1*(20%*V)=n1*R*T
2) p2*(x%*V)=n2*R*T
3) n2-n1=(x%*V-20%*V)*ro_liquid/M
4) n=n1+(V-20%*V)*ro_liquid/M
5) n=n2+(V-x%*V)*ro_liquid/M

5 eqns and 5 unknown variables: n1, n2, n, V, x% - solve for x%

n1,n2 - no. of moles of vapour propane @ state 1 and 2, respectively
n - total no. of moles of propane (vapour + liquid)
ro_liquid - density of liquid propane (assumed that it is constant in given pressure range = incompressible liquid)
M - molar mass of propane
V - total volume of the system
p1,p2 - pressures
R - universal gas constant
T - temperature in [K]!

I think it can be solved without knowledge of total volume of the system (vessel or whatever it is)

[Andree]

x% - is volume fraction of vapour propane

[chenblue]

Thanks for Art, I forgot this is student forum, and did a simple homework just like a textbook did= =.