Wednesday, May 10, 2006

HW #9, P#7 - Performance of a Geothermal Heat Pump - 6 pts

A heat pump with refrigerant-134a as the working fluid is used to keep a space at 25°C by absorbing heat from geothermal water that enters the evaporator at 50°C at a rate of 0.065 kg/s and leaves at 40°C. The refrigerant enters the evaporator at 20°C with a quality of 23 percent and leaves at the inlet pressure as saturated vapor. The refrigerant loses 300 W of heat to the surroundings as it flows through the compressor and the refrigerant leaves the compressor at 1.4 MPa at the same entropy as the inlet. Determine...
a.) The degrees of subcooling of the refrigerant in the condenser
b.) The mass flow rate of the refrigerant
c.) The heating load and the COP of the heat pump
d.) The theoretical minimum power input to the compressor for the same heating load


Graham said...

Again, I'm not sure if this is in ThermoCD, and I'm from Castner's class so we didn't cover this in lecture. What are "degrees of subcooling" and "heating load." And a more precise question, if I'm working on table for this cycle, I've got states 1-3 but am completely stuck on state 4, except for it has the same enthalpy as state 1. Where do I go from here?

Stuck said...

I'm having difficulty determining the COPr.
I determined Qh and Qc using:
Qh = -m(H4-H3) ~3.078 KW
Qc = m(H2-H1) ~2.721 KW
(using stream numbering found in the book)
My incorrect COPr of 8.6 was found using COPr = Qh/(Qh-Qc)

Any ideas?

Also, to the person above me, you know the pressure of state 4 because it doesn't change in the condenser.

stuck said...

I didn't account for the heat loss of the compressor. I got it.


brendan said...

so i got a-c but i dunno what to do for d work,min i'm lost at exactly what this is after.

Graham said...

ok thanks to stuck for his comment, I resolved my issues with this problem, except I'm unsure how to account for the 300 W of lost heat. I solved for Qh using Mdot(H4-H3) + 300W and by first finding Wcomp from the first law and adding it to Qc. Both ways I got Qh equal to 3.378 and COPhp=5.14.

Yoda said...

Like brendan, I too am stuck on part d; not quite sure what about Carnot HP we need to analyze.

Anonymous said...

I'm having trouble determining the mass flow rate. Can someone please give me some hints about this?

Anonymous said...

nevermind about mass flow rate. It was something really simple I was missing :-)

Claire said...

Maybe Im reading this incorrectly, but can someone tell me how the compressor loses heat, but is still isentropic (has same entropy).

Graham said...

Now I'm not sure why this works but heres what I did. For part c) Qc= heat that enters in the evaporator from 1 to 2 MINUS the 300 watts lost in the compressor. For part d) The COPrev=1/(1-(Tc/Th)). Assuming Th to be 50C or 323 K and Tc to be 25C or 298K, we get a COPrev. Equate this to Qh/Wcycle and solve for Wcycle to find the minimum work.

Dr. B, this sounds completely wrong to me but somehow it yields the right answer. I thought that the 300 watts would be ADDED to Qh not subtracted from Qc, and that the Hot resevoir would be at 25C and the cold at 50C, but both of these are completely reversed, can you please explain?

phoot said...

Hey Dr. B- to find mass flow rate i was kind of stuck so i used an estimate of the heat capacity of water we used for another class, 4.18 in order to calculate delta H of the geothermal water. Using this estimation i got a mass flow rate of 19.36 g/s, is that an ok assumption to make?

Dr. B said...

Graham 3:44 PM
Degrees of subcooling = Tsat - Tactual
How much cooler is the subcooled liquid than the saturated liquid at the same P.
Heating load is QH, as stated inthe hints for the problem.
Which is state 4 ? Leaving the condenser ? If so, it is a subcooled liquid. The valve is isenthalpic, H4 = H1 !

Dr. B said...

Stuck 5:06 PM
Thanks for helping outfor Graham.
You forgot that the compressor is not adiabatic.
COPHP = QH / Wcycle
But Wcycle is not = Qh-Qc because there is also Qcompressor !

Dr. B said...

Stuck 5:38 PM
Good work !

Dr. B said...

Brendan 7:51 PM
Think of a Carnot HP operating between these same 2 thermal reservoirs...

Dr. B said...

Graham 10:32 PM
the 300 W lost from the compressor is not part of Qh. QH = Mdot*(H2-H1).

Apply the 1st Law to the whole cycle to see how the 300 W lost at the compressor sneaks into the COP.

Remember that COPHP = QH/Wcycle.

Dr. B said...

Yoda 12:47 AM
Just use the T's to compute the COP of a Carnot HP operating between the same two thermal reservoirs.

Dr. B said...

Anon 12:49 and 12:55 AM
Good work.

Dr. B said...

Claire 10:05 AM
Mdot deltaShat = INT{dQ/T) + Sgen
Q < 0
Sgen > 0
deltaS = 0
All the entropy generated by irreversibilities in the compressor is removed in the form of heat loss.

Dr. B said...

Graham 2:13 PM
Wow, I have no idea what you did for part (c) or why it "worked". See my comments, above.
You are correct for part (d).
OK, here is the 1st Law for the cycle.
Q=W (with sign convention)
QC-QH-Qcomp=-Wcomp (without sign convention, all terms positive)
This is true because Qc and Wcomp go INTO the cycle and QH and Qcomp go OUT of the cycle.
COPHP = QH/Wcycle
COPHP = QH /(QH+Qcomp-Qc)

Dr. B said...

Phoot 4:07 PM
It is fine to use Cp = 4.18 kJ/kg-K for the heat capacity of the cooling water.
As an alternative, you can assume the cooling water is saturated at the inlet and outlet.
Basically, these are equivalent if the water is incompressible (H not a fxn of P).
Good work.

Anonymous said...

umm guys can you please tell me how to get T actual to calculate the delta T subcooled?

Dr. B said...

Anon 10:44am
I think you are asking me how you can calculate the T of the stream leaving the expansion valve. You know P = 1.4 MPa. If you assume the valve is isenthalpic, then the outlet specific enthalp is the same as the inlet. Determine H at the inlet from the given x and P. Then, knowing P and H of the valve effluent, you can determine T by interpolation on the R-134a tables. It will be easier to determine T if you use the NIST Webbook.

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