Heat pump efficiency calculations - Compressor power issue

  • #1
The heat pump comprises of the 4 components: evaporator, compressor, condenser and expansion valve.

Thermal power required to heat the building: 12.1 kW at condensing temperature tc = 44.3 deg C
For the evaporator: vaporizing temperature tv = -7 deg C
Subcooling temperature for the heat pump Δ tsc = 5 deg C
adiabatic coefficient: k = 1.16
vaporising pressure: pv =t(tv) =paspiration =4bar
condensing pressure: pc = t(tc) = pdischarge =17bar
vapor density: ρv=17.0 kg/m3
specific volume for aspiration vaspiration = 1/ρv = 1/17kg/m3 = 0.058823529 m3/kg
internal compressor efficiency: ηi=0.84
h1= hv(tv) = 402.56 kJ/kg
h3 =hl(tsc,pc) =248.6 kJ/kg

Now I tried finding out the following:
a)Thermal power Q0 at the evaporator
b)Compressor power
c)efficiency of the heat pump (both COP and refrigeration cycle efficiency ε )

a)Q0=q0*mrf, where q0 = specific heat load, and mrf=mass floware for the refrigerant

q0 = h1-h4 = h1-h3; because h3=h4
so q0 = 402.56 kJ/kg - 248.6 kJ/kg => q0 = 153.96kJ/kg

compressor work: compressor_work = (k/((k-1)*ηi))*paspiration*vaspiration*((pdischarge/paspiration)^((k-1)/k)-1)*100
compressor_work = 44.85735668 kJ/kg

enthaply in the 2nd point: h2 = h1+compressor_work = 402.56 kJ/kg+44.85735668 kJ/kg
h2 = 447.4173567 kJ/kg

specific heat load at the condenser: qc = h2-h3 = 447.4173567 kJ/kg - 248.6 kJ/kg
qc =198.8173567 kJ/kg

so: mrf = Qc/qc = 12.1 kW/198.8173567 kJ/kg
mrf = 0.060859878 kg/s (refrigerant mass flowrate)

Thermal power Q0 at the evaporator: Q0=q0*mrf
Q0 = 153.96kJ/kg*0.060859878 kg/s
Q0 = 9.369986761 kW

b)Compressor_power = (mass_flowrate_refrigerant * compressor_work) / (ηmotor*ηelectrical)
Problem arises here because I was given just the internal efficiency ηi=0.84, but the formula uses the efficiency of the motor ηmotor and the electrical efficiency ηelectrical...so how can I calculate this? (as I don't have the ηmotor*ηelectrical), I don't want to guess though...

c)COP =qc/work_compressor = 198.8173567 kJ/kg / 44.85735668 kJ/kg
COP = 4.432
ε = heat load evaporator/work done = COP - 1 = 3.432

What do you think of the calculations, do they make any sense? I'd really appreciate a peer review.

My actual question is, how can I calculate the compressor power at b) as I don't have those 2 efficiencies (ηmotor and ηelectrical), but just the ηi which is the internal efficiency of the compressor?
 

Answers and Replies

  • #2
russ_watters
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My actual question is, how can I calculate the compressor power at b) as I don't have those 2 efficiencies (ηmotor and ηelectrical), but just the ηi which is the internal efficiency of the compressor?
I don't have time to check the calcs right now, but based on what was given, they are probably asking for the mechanical power, not electrical.
 
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  • #3
Hi,
Thanks for the reply, I was given just the actual internal efficiency for the compressor.
Is there any way to calulate the compressor power with just the data I have, without having mechanical and electrical efficiencies?
 
  • #4
Please find below a schematic diagram if it is of any help.
The heat exchangers have plates and are using water and refrigerant.
 

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  • #5
21,054
4,641
This pertains to the work done on the working fluid in the compressor.

If I use the isentropic equations for k = 1.16, I get a temperature of 52.4 C exiting the compressor and an enthalpy increase of 37.7 kJ/kg in the compressor. If I use the thermodynamic tables for R22, I get a temperature of 67.9 C exiting the compressor and an enthalpy increase of 36.5 kJ/kg in the compressor. So, either way, the work done on the refrigerant is about 37 kJ/kg in the compressor. Assuming an isentropic efficiency of 0.84, doing it the first way, I confirm your value of 44.8 kJ/kg for the enthalpy change and the work.

I also confirm your determination of the mass flow rate.
 
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  • #7
Thank you!
All right, in this case, how can those be calculated then with the data I was given? I have my attempt in post#1, but still can't do the compressor power because I don't have the motor and electrical efficiencies, just the internal compressor efficiency which was used to calculate the work for the compressor.
 
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  • #8
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The compressor power is the specific enthalpy change of the compressor times the mass flow rate, right?
 
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  • #9
Hi,
Right, okay. So can we say that compressor power = mass flowrate of refrigerant * specific work for the compressor?
 
  • #10
If I'm asked to plot a log p-h diagram for the following statement:

"A building needs a thermal power for heating Q = 12 [kW] at with a condensing temperature tc = 44.3 deg C, the heating being performed with heat pump. The heat source is represented by the soil. The installation works with a refrigerant whose vaporization temperature is tvap = -7 deg C. The degree of subcooling of the installation ΔtSC = 5 deg C."

So for this case on the log p-h diagram I should plot just the sub-cooling part, right? I wasn't told anything about superheating.

The question is...can we have sub-cooling without super-heating though?
So should I plot just the sub-cooling process on the log p-h diagram? Or should I plot both sub-cooling and superheating on the log p-h diagram?

Should the log p-h diagram be like Fig.1 or like Fig.2? (please see below)
 

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  • #12
Hi, thanks for the quick reply, I do appreciate it.

The problem is that I have to plot the p-h diagram for the heat pump myself, but not sure if I should have just the subcooling process since I was given just the subcooling degree "It is known that the degree of subcooling of the heat pump ΔtSC = 5 deg C."

Or should I have both sub-cooling and super-heating, though the word superheating was not specified within the text....that got me confused, is it possible to have just subcooling without superheating for the heat pump?

"A building needs a thermal power for heating Q = 12 [kW] at with a condensing temperature tc = 44.3 deg C, the heating being performed with heat pump. The heat source is represented by the soil. The installation works with a refrigerant whose vaporization temperature is tvap = -7 deg C. The degree of subcooling of the heat pump is known such as ΔtSC = 5 deg C."

Don't know if to plot the p-h diagram just with the sub-cooling or with both sub-cooling and superheating, based on the text provided above.

Anyway, for the coolant (R22) I assume the log p-h diagram is something like this:
1619113237312.png
 
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  • #13
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I found a P-H diagram. In my judgment, the correct diagram should be Fig 1. Although they don't say there is superheating, its seems obvious that the vapor exiting the compressor should be superheated at 17 bars. Just follow an isentrop upward from a saturated vapor at 4 bars.
 
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  • #14
Thank you for your logical response.

I've been given some additional data:

internal efficiency for the compressor: 0.84

pv(vaporizing pressure) = t(tv) = 4bar

pc(condensing pressure)=t(tc)=17 bar

ρv=17 kg/m3

h1=hv(tv)=402.56 kJ/kg, where tv = vaporizing temperature, yet it doesn't say anything about tsh (superheating temperature)

h3=h1(tsc,pc)=248.6 kJ/kg, where tsc=subcooling temperature, pc=condensing pressure

k=1.16 (adiabatic index)

Yet again, superheating is not specified, so I guess, it's just subcooling, right?
 
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  • #15
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Thank you for your logical response.

I've been given some additional data:

internal efficiency for the compressor: 0.84

pv(vaporizing pressure) = t(tv) = 4bar

pc(condensing pressure)=t(tc)=17 bar

ρv=17 kg/m3

h1=hv(tv)=402.56 kJ/kg, where tv = vaporizing temperature, yet it doesn't say anything about tsh (superheating temperature)

h3=h1(tsc,pc)=248.6 kJ/kg, where tsc=subcooling temperature, pc=condensing pressure

k=1.16 (adiabatic index)

Yet again, superheating is not specified, so I guess, it's just subcooling, right?
When you compress the saturated vapor that enters the compressor, the vapor is compressed isentropically and exits the compressor in a superheated state at 17 bars and about 70 C. So it enters the condenser superheated and exits sub-cooled.
 
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  • #16
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Dear All ,
how can i post a new question
regards
Go to the forum you are interested in posting to, and click on the button POST THREAD.
 
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  • #17
When you compress the saturated vapor that enters the compressor, the vapor is compressed isentropically and exits the compressor in a superheated state at 17 bars and about 70 C. So it enters the condenser superheated and exits sub-cooled.
Right, okay. So, now with the actual given data if I'm asked to plot a log p-h diagram for the entire heat pump, it should resemble the one like in Fig.1, right? Just the subcooling part basically, because superheating (before entering the compressor 1-1') was not stated anywhere within the text. When I am speaking about superheating I am refering to the one before entering the compressor. I currently don't know based on the provided data if the p-h diagram should include the superheated process right before entering the compressor, given that the input data (text) hasn't specified anything about superheating.

So how should the log p-h diagram look like for the heat pump?
 
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  • #18
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I would assume it is saturated vapor before entering the compressor. It exits the compressor as a superheated vapor at 17 bars and on the order of 70 C. This is consistent with your Fig.1. So it enters the condenser as a superheated vapor and exits as a sub cooled liquid.
 
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  • #19
Hi

Thanks for your reply.
Like I should not have superheated process 1-1' before entering the compressor.

I am trying to understand why I was not given any information on purpose regarding the superheated process before entering the compressor.

So, the p-g diagram, should look something like this, right? (please see below)
 

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  • #20
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Hi

Thanks for your reply.
Like I should not have superheated process 1-1' before entering the compressor.

I am trying to understand why I was not given any information on purpose regarding the superheated process before entering the compressor.

So, the p-g diagram, should look something like this, right? (please see below)
I don't know what 1-1' means. But, in any case, Fig. 4, not Fig. 3
 
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  • #21
Hi,

Thanks for the reply.

Please see Fig.5 and Fig.6 below, where 1->2 compressor, 2->3 condenser, 3->4 expansion valve, 4->1 evaporator.

Which one would you say that is now the correct p-h diagram for the above given data for the heat pump?
 

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  • #22
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Hi,

Thanks for the reply.

Please see Fig.5 and Fig.6 below, where 1->2 compressor, 2->3 condenser, 3->4 expansion valve, 4->1 evaporator.

Which one would you say that is now the correct p-h diagram for the above given data for the heat pump?
Figure 5
 
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  • #23
Thank you for your quick reply, really much appreciated!

Yeah right, absolutely, it makes sense to enter as saturated vapor in the compressor, yet I am afraid that if it won't be superheated the compressor might corrode due to liquid droplets, but then again, I was not told anything about superheated vapors.
 
  • #24
Can we say that the efficiency for the refrigeration cycle for the heat pump = COP -1 ?

I know that usually the refrigeration cycle efficiency = Qevaporator/Work done by compressor
 

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