# Material Energy Balance / Thermodynamics Question

1. Mar 17, 2007

### twiztidmxcn

I have this problem as a homework question for my MEB/Thermodynamics class and have been having some trouble with it.

A 10.0 m^3 tank contains steam at 275°C and 15.0 bar. The tank and its contents are cooled until the pressure drops to 1.2 bar. Some of the steam condenses in the process.

a) How much heat was transferred from the tank?
b) What is the final temperature of the tank contents?
c) How much steam condensed (kg)?

I started by doing the energy balance on the closed, transient system:

mi(hi+v2/2+gzi) - mo(ho+v2/2+gzo) + Q + W = ∆U

where v2 = v^2, i is subscript in, o is subscript out

Since the system is closed, mi = mo = 0, Ep = 0 (no height change), Ek = 0 (no movement), W = 0 (no work)

So I'm left with Q = ∆U = U2 - U1

I assume for part A that I am looking for Q, the heat transferred.

I started out knowing that at 275 C and 15 bar, we have a superheated steam. So using steam tables, I interpolated a value for U1 being 2739.25 kJ/kg.

My problem is with finding U2. Using the pressure of 1.2 bars, I used steam tables to find a value for U2, but since the tank's final contents are both steam and liquid, I am not quite sure which value to use.

In any case, U2 for liquid water is 439.2 kJ/kg and U2 for steam is 2512.1 kJ/kg.

Pretty much, I need to find out how to use this information to get a reasonable value and find the heat transferred.

As for B and C....any pointers in the right direction would be much appreciated.

Any help at all is appreciated (as usual)

Much thanks
-twiztidmxcn

2. Mar 17, 2007

### chaoseverlasting

Ok. For the first part, since you have the pressure, volume and temperature, you can find the moles of steam using pv=nrt. Next, you know final pressure, final volume, but not the final moles or temperature.

Lets assume n1 moles were condensed during the process. If the latent heat of vapourisation is x per mol, then the total heat taken away from the system is n1*x=dQ. Since dQ=dU, then $$U_i=nfRt_i/2, U_f=n'fRt_f/2, dU=\frac{fR}{2}(nt_i-n't_f$$ and n=n1+n'.
Where f is the degree of freedom of steam.

Two variables and two equations, solve for them and you have final temperature and number of moles condensed. From there the final part is easy.

3. Mar 17, 2007

### twiztidmxcn

Woah....I'll take a look at it and get back if I need any more help