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01092016, 01:39 PM #1
Energy balance in evaporator. Thermal load and heat transfer rate
Hi,
I have a question regarding energy balance in an evaporator. We have an experimental setup of a single stage vapor compression system. The evaporator is a B15THx20 SWEP heat exchanger, and the fluid to cool is a glycol solution in a small tank where we have a variable resistance to control the thermal load. The glycol solution is pumped constantly from the tank to the evaporator at a rate of about 12 l/min.
I want to calculate the heat transfer rate. To my knowledge, provided the system reach a stable point, the energy balance should be the following:
Q = m'_ref * deltaH = m'_glycol * c * delta_T = resistancePower
where:
Q = heat transfer rate (kW)
m'_ref = mass flowrate of the refrigerant (kg/s)
deltaH = specific enthalpy difference between the outlet and inlet of the evaporator (refrigerant side) (kJ/kg)
m'_glycol = mass flowrate of the refrigerant (kg/s)
c = specific heat capacity (kJ/kg/K)
delta_T = temperature difference between inlet and outlet of the evaporator (glycol side) (K)
resistancePower =imposed thermal load (kW)
I have done several (a lot) of tests by analyzing a lot of stable points, modifying the thermal load, the evaporating temperature, condensing pressures, etc.
Given the test I did a data logging and I calculated Q based on the previous equations. I was expecting to get similar values to the imposed thermal load, but that does not happen. For example, if I impose 1 kW, I was expecting the term m'_ref * deltaH to be about 1 kW, and the same for the glycol side. However, I always get about 80100% greater value of the calculated heat transfer rate, i.e, if I impose a thermal load of 1 kW I get about 2 kW from the refrigerant side calculation (m'_ref * deltaH). Regarding the glicol side the measurement is more noisy (my delta_T is between 1 and 3 ºC and I have NTC probes which can give a +0.5 ºC ) but I also get higher values.
Thus, There is always a positive offset in the calculation of the heat transfer rate in the evaporator (refrigerant and glycol side) with respect to the imposed thermal load. The calculated heat transfer rate seems to be bigger than the imposed thermal load. What really intrigues me is that the aforementioned offset happens in both sides of the evaporator, and I am pretty sure we are controlling well the imposed thermal load in the resistance since we have measured the voltage and current across it (it is a threephase resistance controlled by a 420 mA signal). What am i missing?
Attached is a figure where I show the problem.
EnergyBalance.PNG
As you can see in the figure, and as I said before, it seems there is a constant offset. The question is, provided there are no errors in the measurement (ideal case) am I right in the energy balance equation or am I missing something else? I could expect some losses but not twice the imposed thermal load.
Looking forward to your comments.
PS: we have a direct measurement of the refrigerant flow rate and enthalpies are calculated by using pressure probes at the evaporator input and condenser output.
PS2: The hydraulic pump that recirculates the glycol solution has an electrical power of 0.2 kW. I suppose the heat it transfers due to friction to the glycol solution is negligible.Last edited by mrr; 01092016 at 10:27 PM. Reason: clarification

01092016, 08:27 PM #2
Re: Energy balance in evaporator. Thermal load and heat transfer rate
glycol pump power?

01092016, 10:26 PM #3
Re: Energy balance in evaporator. Thermal load and heat transfer rate
Hi Ranger,
The hydraulic pump that recirculates the glycol solution has an active power (electrical) of about 0.20.25 kW. A small fraction of this power due to friction is transformed in heat, and would be added to the thermal load we impose with the electrical resistance. However, to my understanding this contribution is negligible. Anyways I have edited the post for claritication, thanks!!

19092016, 10:37 AM #4
Re: Energy balance in evaporator. Thermal load and heat transfer rate
As a rule of thumb, in chillers:
kWr = mass flow(m3/h) * deltaT
which in your case is:
12 l/min = 0,72 m3/h > kWr = 0,72 * 2 = 1,44 kW
According to the original formula
m'*c*deltaT > 12/60*1*4,2(cal to J)*2= 1,76 kW
As you might think, with so small deltaT, probe's error can become up to 1C (i.e 50% what you're trying to measure), so you need to calibrate the probes or raise the deltaT(by lowering flow), but i don't think you could raise the deltaT up to say 10C so the error could be much more reasonable.
Assuming +0,5C, the value of heat transfer is between:
12/60*1*4,2*(3,50,5)= 2,52
and
12/60*1*4,2*(2,51,5)= 0,84
If you were working with a greater delta:
0,84*(10,52,5)= 6,72
0,84*(9,53,5)= 5,04
Same total difference best/worst case but much smaller %.
Also, there's the +0,5C error from the probe itself, but the testbed may also add some error.
The point is, if you want to get correspondence from theory to practice, you gotta focus on the measurement precision.
About the offset, that's probably energy losses, i.e. electric motor(the pump will dissipate heat through the glycol), heat transmission on the pipings/tank, etc...
That all sums up.

29092016, 02:31 PM #5
Re: Energy balance in evaporator. Thermal load and heat transfer rate
Hi el_donks,
I did not read your answer. Thank you, it makes sense to me. Actually I have thought about the possible error measurement. However my theoretial doubts are:
1 If i put the heater at 2 kW shouldnt I get 2 kW of frigorific power? (theoretically).
2 How can i calculate the heat that the hydraulic pump introduces in the glycol solution? I am wondering also what the hydraulic energy (related with the flowrate value) has to do in the energy balance.
Thanks!
Regards,

29092016, 09:09 PM #6
Re: Energy balance in evaporator. Thermal load and heat transfer rate

30092016, 10:25 AM #7
Re: Energy balance in evaporator. Thermal load and heat transfer rate
That's correct. Beside the language error (I meant volume flow), due to water's density close to 1, and also the specific heat of water, you "can" round it to the given rule of thumb (term which I understand is a quick and not 100% exact rule to calculate something). Of course chillers work with water+glycol usually, so mix's density isn't 1, neither is the specific heat. So you can use this rule of thumb with fluids whose characteristics are similar to water's.

01102016, 01:09 PM #8
Re: Energy balance in evaporator. Thermal load and heat transfer rate
Thank you el_donks.
2 How can i calculate the heat that the hydraulic pump introduces in the glycol solution? I am wondering also what the hydraulic energy (related with the flowrate value) has to do in the energy balance.
I'm not pretty sure about this. If you have a given cooling power, you could calculate losses as the difference between given cooling power and the cooling effect over the load(what you are cooling). This way you would get the total thermal losses, but i don't know how to calculate thermal losses due to friction or transmission.
Regarding the rule of thumb, just for clarification for frank. The point is that the units should be (and this is the difference with the rule of thumb, where the volume flow is applied in m^3/h):
Q [kW] = density [kg/m^3] * volume flow [m^3/s] * specificheat [kJ/kg*K] * dT[K]
kW = 1000 kg/m^3 * m^3/h * 1h / 3600 s * 4.2 kJ/(kg * K) * K.
So a conversion factor of 4.2*1000/3600 = 1.16 is not applied in the rule of thumb, but you get an approximation as el_donks said.Last edited by mrr; 01102016 at 05:11 PM.

03102016, 08:32 AM #9
Re: Energy balance in evaporator. Thermal load and heat transfer rate
One question, in the first post, i see differences much higher than 40% difference. Have you improved this? Because a difference of 40%, given the measurement error you indicated previously, isn't crazy off. Because its not only the probes error. Let's say the probe isn't in the best location, or isn't well insulated, it could also add more error(i'm assuming this is not the case, but is something to think about).
I assume you're still working with the same thermal probes. If so, try using a higher dT with lower flow.
Remember, you're trying to get theory results with real system. You gotta identify all the "not ideal" things happening. I would start checking accuracy of the measuring tools. Do you actually have a flow meter for the glycol side? How do you measure mass flow on the evaporator? Thermal insulation on the system?
I'd assume that the electrical load is the most accurate parameter you're getting, since resistors have almost 100% efficiency with virtually a power factor of 1(are you using any kind of inverter to control power?). Starting from this, i'd go reverse, i.e. from resistor side to glycol side and on.
Again, assuming you have a decent thermal insulation, on the glycol side you should get slightly higher cooling power(couldn't say a number, 510% maybe?) Let's say you have 20% total thermal losses. Anything higher than that are measurement errors, one would think. Then try to improve measurements and check if you are reducing the gap.
When I say improve measurements, I don't necessary say that you need to change the whole measuring system, just need to reduce the gap due to the possible measurement error. Like i said, increasing dT for example.Last edited by el_donks; 03102016 at 08:36 AM.

04102016, 12:38 PM #10
Re: Energy balance in evaporator. Thermal load and heat transfer rate
Hi el_donks,
Thank for your interest. I lowered the glycol flowrate from about 12 l/min (about 0.22 kg/s) to 2.16 l/min (0.038 kg/s). dT changed from about 2.7 ºC to 8 ºC. Here you can see now the calculated cooling power from both sides from a experiment of about 5 hours.
figure.png
In the picture a transient can be observed when the flowrate was changed. All the external conditions (temperature setpoint, condensing pressure, superheat) remained constant. I could observe that the compressor increased its % from about 77% to 95%. This is OK for me, decreasing the glycol flowrate makes the evaporating pressure to decrease from about 1 barg (relative pressure!) to 0.66 barg.
As you can see from the picture, now the black and blue curves are quite similar, but I do not understand why the cooling power calculated from the refrigerant side (black curve) has a great value. The black curve should have remained constant (except in the transient).
Let me answer your questions now:
 I have a flowrate meter in the glycol side and another one in the refrigerant side (liquid line). They measure volume flow (m^3/h), so I calculate mass flow rate starting from densities and specific heats.
 The system was insulated but now it isnt completely. Especially at the input and output of the heat exchanger (glycol side). Ice and snow are formed in the pipes. I tried to isolate the probes with armaflex. So this can be a possible error source (also, one of the probes is not located exactly at the output of the heat exchanger).
 Regarding the resistor, it is a threephase resistance which provides up to 5.4 kW. It is regulated with a threephase solid state relay. The % power is controlled by a 420 mA signal. Basically, the solid state relay chops the voltage. I measured the three phases and with a clamp meter and I calculated that a little bit less of power is being provided. This is due to the voltage grid which is 395 V instead of 400. For example, when 0.7 kW is set, about 0.51 kW are, If I made the calculation properly, being given.
So now it seems that probe error was determinant as you said, and as I thought, but still I dont understand why thermal losses seem to be very high. In addition, the glycol circuit is very small, there is just the pump and the electric resistance.
Thanks!

04102016, 05:56 PM #11
Re: Energy balance in evaporator. Thermal load and heat transfer rate
As long as i understand, what provides power to the resistor is a 3 phase PWM. Depending on the filters implemented on it, it will generate harmonics which could eventually make clamp meter readings not reliable, and if you are controlling resistor power based on the clamp meter, there could be the issue.
Difference between blue and black curve is about 1520%, which doesn't seem too odd for thermal losses given the circumstances.