Condenser calculations - uncertainty

  • #1
Summary:
Hi
I got a condenser, which is using a refrigerant that will be condensed using water ( a water cooled plate condenser). I would really appreciate it if someone could peer review my calculations before submitting them for work.
A water cooled plate condenser, condensing refrigerant.

Heat exchanger (condenser with plates) characteristics:
Descriptionsymbolvaluevalue (converted)
number of platesnp28
number of channelsnc27
number of water channels
nwc
14
number of flowing channels for refrigerantncrf13
plates thickness
δp
0.8 mm0.0008 m
width of a platew0.07 m

length of a plate

L
0.2m
height of flowing channelH02.2mm0.0022 m
equivalent hydraulic diameterdh = 2*H04.4 mm0.0044 m

thermal conductivity of the material of the plates (steel)

λp

20 W/m*K
area of one plateSp = w*L
0.014 m2
plates area
S = Sp*(np-2)

0.364 m2
Enthalpy at the inlet of condenserh2448.8154408 kJ/kg

Enthalpy at the outlet of condenser
h3248.6436269 kJ/kg

Please note that I am using the C system which is basically the SI system using ° C instead of K.

For the refrigerant:
refrigerant mass flowrate mrf=0.06kg/s

For water:
water mass flowrate = mw=0.7kg/s
outlet water temperature Twoutlet=35°C

Calculations for the condenser:

1) Heat flux qc=h2−h3 = 448.8154408 kJ/kg - 248.6436269 kJ/kg = 200.171814 kJ/kg
qc=200.171814kJ/kg

2) Heat Qc=mrf∗qc = 0.06 kg/s * 200.171814 kJ/kg = 12.01030884 kJ
Qc=12.01030884kJ

3) Inlet water temperature in the condenser twinlet=twateroutlet−Qc/(mw∗cpw) = 35 ° C - 12.01030884 kJ / (0.7 kg/s*4.18 kJ/kg*K)
hence twinlet=30.89531482°C

4)mean water temperature twm=(twinlet+twoutlet)/2 = (35 ° C + 30.89531482 ° C) /2 = 32.94765741 ° C
twm=32.94765741°C

5)Water properties at mean temperature twm Calculations (calculated by ForTran Refprop)

a)Water thermal conductivity

λw = thermal conductivity (water, "PT", "C" 0.101325, twm; mean water temperature) / 1000 = 0.6187661 W/m*K
where: PT = as a function of pressure and temperature, pressure was used as atmospheric in MPa, and temperature was the mean water temp, divided by 1000 to get W from mW. Alternatively instead of PT I could have used TQ, such as mean water temperature twm and Q (quality, such as x = 0, liquid, but at 35 ° C we are far from saturation, so that's why I used PT, atmospheric pressure and mean water temperature).

λw=0.6187661W/(m∗K)

b)Water density
ρw = density(water, "PT", "C" 0.101325, mean water temperature) = 994.721891 kg/m3
ρw=994.721891 kg/m3

c)Water dynamic viscosity
μw = viscosity(water, "PT", "C" 0.101325, mean water temperature)*0.000001, so *10-6 was used to convert μ Pa*s to kg/m*s

μw=0.000749615kg/(m∗s)

d)Water kinematic viscosity
νw=μw/ρw= (0.000749615kg/m*s) / (994.721891 kg/m3) = 7.53592E-07 m2/s

νw=7.53592E−07m2/s
e) Prandtl
Pr = Prandtl (water, "PT", "C" 0.101325, mean water temperature)
Pr=5.063195386

f) Water flowing section
Scw=nwc∗w∗H0 = 14 * 0.07 m * 0.0022 m = 0.002156 m2
Scw=0.002156 m2

g) Water flowing velocity
uw=mw/(ρw∗Scw) = 0.7 kg/s / (994.721891 kg/m3*0.002156 m2)
uw=0.326398089m/s

h)Reynolds
Re=(uw∗dh)/νw = (0.326398089 m/s*0.0044 m)/7.53592E-07 m2/s
Re=1905.740592

i)Nusselt
Nu=0.212∗Re0.638∗Pr0.33
Nu = 0.212*1905.7405920.638*5.0631953860.33
Nu=44.82050053

j)Water convection coefficient
αw=(Nu∗λw)/dh = (44.82050053*0.6187661 W/m K)/0.0044 m

αw=6303.046887 [W/(m2*K)]

I would really appreciate it if someone could peer review my calculations from 1-5 (a-j), and let me know if they are correct.
 
Last edited:

Answers and Replies

  • #2
21,054
4,641
Here are a few comments:

Please provide some diagrams of the layout, including internal channels. I am particularly interested in the channel cross sections.

Where did you get your Nu correlation from? The flow is laminar (close to transition), so the channel length should matter. What range of Re is your correlation supposed to apply to?

What about the heat transfer resistance on the refrigerant side? Are you assuming that the heat transfer coefficient on the refrigerant side is very larger compared to the water side?

What is the refrigerant temperature?
 
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  • #3
Here are a few comments:

Please provide some diagrams of the layout, including internal channels. I am particularly interested in the channel cross sections.
Hi

Thanks for the quick reply
Please see attached below a sample figure for the condenser (on the right-hand side) with plates its shape would be rectangular, and it will be used to condense the refrigerant.


Where did you get your Nu correlation from? The flow is laminar (close to transition), so the channel length should matter. What range of Re is your correlation supposed to apply to?

I looked online for Nu correlations based for laminar flow based on the calculated Re.

The channel length L = 0.2m is used just to calculate the surface of a plate, and then the surface for all the plates, but other than that it was not used. However its width w = 0.07 m was used to calculate the Water flowing section (please see above).


What about the heat transfer resistance on the refrigerant side? Are you assuming that the heat transfer coefficient on the refrigerant side is very larger compared to the water side?

What is the refrigerant temperature?
I was not given any heat transfer resistance, just the thermal conductivity of the steel for the condenser, perhaps it's just simple calculations, not going too much into details for the moment. Not sure if my calculations are correct, though.

I have a heat pump, so 2 heat exchangers (evaporator + condenser), 1 compressor and 1 expansion valve.

The refrigerant is: chlorodifluoromethane, this is going to be condensed within the condenser by using water.

Regarding the temperatures:
For the compressor: condensing temp 44.28 ° C, vapourising temp: -6.55 ° C, superheating temp: -4.55 ° C, subcooling temp: 39.28 ° C
 

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  • #4
21,054
4,641
This is basically laminar flow between two parallel plates (neglecting the ends of the high aspect ratio channel). There are definitely going to be analytic solutions in the literature for laminar flow heat transfer between two parallel plates, and the heat transfer coefficients are going to be functions of the channel length. See Bird, Stewart, and Lightfoot, Transport Phenomena, page 431. For a very long channel, where the thermal boundary layer is easily able to penetrate to the center of the channel, the lower bound to the Nussult number is 7.5 (based on twice the channel height, 0.44 cm). For your channel, I don't think that the Nu is going to be much higher than this. But it will depend on the channel length, according to the analytic solution in Bird. Run some numbers on their analytic solution and see what you get. Right now, in my judgment, your estimate of the Nu is high.

After you get a value for the overall heat transfer coefficient, you need to determine whether the heat load that you have calculated so far is consistent with the heat load that you calculate using the overall heat transfer coefficient, the temperature differences between hot and cold sides, and the heat transfer area.
 
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  • #5
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Better yet, see the graph in Fig.14.2 on page 429. From this graph for your conditions, I estimate an average Nussult number of about 15 (based on twice the channel height).
 
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  • #6
Hi
I'd like to thank you for both of your answers. I'm fully with you..

I've been given this (please see below) correlation based on rectangular plates for the condenser.

Nu = 0.212*Re(0.638) *Pr(0.33)

The thing is that my Re =1905 which indicates laminar, and seems quite right for a water velocity of ~.3m/s. So my personal opinion was that either Re is wrong or that Nu is not quite right...Seems weird to have laminar flow and Nu = 44...

I started all calculations from rock bottom by hand...I couldn't really spot any mistakes...with either the calculations or conversions to SI....
 
  • #7
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The difference between your rectangular channel with a height of 0.22 cm and a width of 7.0 cm (30:1 aspect ratio) and two infinite parallel plates separated by a distance of 0.22 cm (in terms of heat transfer) is insignificant. I don't know where you got that correlation, but in my judgment (based on abundant practical experience), it is inappropriate.

It looks to me like your arithmetic is correct. But with an inappropriate heat transfer correlation, it can't be an accurate quantitative representation of your system behavior. Try checking to see if the predicted heat load based on the temperature differences and heat transfer coefficients are consistent with the heat load that you have already calculated based on the enthalpy change. I'm betting the Nussult number correlation grossly overstates the heat load.
 
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  • #8
Hi

You're quite right there, guess you won the bet. My Nu of 44ish seems to greatly overstate the heat load. So it all comes down to the Nu correlation, like everything else is correct, but the correlation itself is faulty, right? I looked up for more correlation on laminar flow for a rectangular condenser but couldn't really get one that would fit my input data.

I am just curious how the estimation is about 15 for the Nu? Indeed I read that the Nu for laminar flow shouldn't exceed 7.5 based on BSL book, so now I'm a bit confused, as I thought my Nu should be no more than 7.5, but now it seems to be double the upper limit of 7.5 such that Nu = 15.
 
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  • #9
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The 7.5 is a lower bound for the Nussult Number. If you plug in your specific numbers for the parameters on the graph on page 429, you get a local Nussult Number for the exit length of the channel (z =. 0.2 m) of about 12 (maybe a little higher if you do it accurately with a ruler). Because the Nussult number is decreasing inversely with the cube root of the channel length for most of the channel, the average nussult number is about 3/2 the local Nu at the exit. So that would make your result, based on my crude reading from the graph 12 x 1.5 = 18 for the average Nu.
 
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  • #10
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An average Nu of 18 is consistent with the given heat load and with a water temperature entering at 29 C and exiting at 33 C.
 
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  • #11
Hi

You're absolutely right, a Nu of 18 seems quite correct, perhaps the correlation I initially used was faulty. Thank you for your detailed explanation, much appreciated!

I guess this thread can be marked as solved.
 

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