enthalpy of steam water mixture

[dwene]

Hello all,

I'll be taking the Mechanical PE exam next week. I'm studying hard and am stuck on a very simple problem. Can you please help?
"Water(100lbm/min, 60F) and steam (100lbm/min, 600F) both at atmospheric pressure enter an insualted chamber through separate inlets. After thorough mixing, 200lbm/min of the product is withdrawn at atmpospheric pressure. What is the quality of the product stream.

It shows the solution -
hmix = 1/2(hwater + hsteam) From the tables I get hwater=28.08 and hsteam= 1335.2btu/lbm. This first equation gives 681.64 btu/lbm. The next step is where I lose them. They show h=hf + xhfg = 681.64=180.1+x(970.4) therefore x = .516

Where did they get 180.1 and 970.4 from? This is driving me nuts.

Thanks in advance!

 

[chicopee]

The first equation s/b
hmix= (Mwater*hwater+Msteam*hsteam)/(Mwater+Msteam)
assuming no accumulation and M is mass rate

The second equation would be dealing with saturated steam from which you can calculate saturated steam enthalpy or quality.
The values that you are inquiring about are from the steam tables.

 

[TBP]

Water freezes at 32*F, and boils at 212*F, right?

212 - 32 = 180. 1

BTU raises the temp of 1 lb of water 1*F, so 1 lb of water at the boiling point (at atmospheric pressure) has 180 BTU.

970 is the latent heat of steam. That is the energy it takes to turn 1 lb of water at 212* in to 1 lb of steam at 212*, at atmospheric pressure. Note that the temp doesn't increase, this heat is what is required to effect the change of state.

 

[TBP]

Oops - I should have said "water at 32*F".

You can also have ice at 32*F. The difference between water at 32* and ice at 32*, is 144 BTU/lb, and is the latent heat of fusion. But you don't care about that for this problem.