PPI Mech Practice Exam 3rd Edition - HVAC+R Problem 50

Anyone out there work the PPI Mechanical Practice Examination (3rd edition) HVAC+R practice problems?  I have an issue with problem #50 on page 14.

I am calculating a crawl space temp of 30 F.  The solution says it is 11.9 F. 

I calculated my solution two different ways and suspect the solution is bogus.  Anyone out there come to the same conclusion?

Thanks,

Ryan

Ahhhhhhhhhhhhhhhh!!!

Now I see.

The ventilation rate is given in ft^3/hr, not CFM.

You got me, PPI... you got me.

Yeah I'm working through these problems right now too. I remember this one, that would be a very leaky crawlspace if that was CFM, not CFH.

Let me know when you get to the rankine cycle problem and determining the efficiency. I think its like #61? I don't understand their efficiency equations, or at least I don't see why it is different from what's in the MERM.

namod65 said:

...Let me know when you get to the rankine cycle problem and determining the efficiency. I think its like #61? I don't understand their efficiency equations, or at least I don't see why it is different from what's in the MERM.

Click to expand...

Hi namod65

It is #63. They just neglected the pump work.

Let's see with detail what's going on. Define the following states: 

1 = condenser discharge = pump suction

2 = pump discharge = boiler inlet

3 = boiler discharge = turbine inlet

4 = turbine discharge = condenser inlet

Then, the thermal efficiency would be


where Wpump= h2 - h1. In terms of enthalpies, this equation is:


Neglecting the pump work means saying that  h2 = h1. If you insert that in the above equation you get what they used in the solution.

Here's why we can neglect the pump work: Remember that for compressed liquids, the change in enthalpy is usually approximated as:


so we can actually calculate it as:

h2 - h1 = (0.01623 ft3/lbm) x (700 - 2) lbf/in2 x [144in2/ft2] x [32.2 lbm ft/lbf/s2] x [ 1 (Btu/lbm)/ (25037 ft2/s2) ] = 2.1 Btu/lbm

So if h1 =94.02 Btu/lbm then h2 = 96.12 Btu/lbm. (they are indeed very close)

If we use the original equation to calculate the efficiency we get:


In the basic Rankine cycle, it is usually a safe bet to neglect the pump work. You would have to be dealing with a HUGE pressure rise across the pump for it to make a difference. That is in fact the point of vapor power cycles: performing the work consuming step (isentropic compression) when the fluid is liquid, which requires much less energy per unit mass than what the work producing step (turbine expansion) produces per unit mass for the same pressure levels.