Direct-Expansion Water-Source Evaporator Temperature

[microHydro]

I'm extracting heat from a 37°F stream flowing at 220 gpm. I plan to use copper coils immersed in the open flow as an evaporator. Conventional wisdom suggests that I keep the evaporator temperature high enough to prevent icing. However, the limiting convection heat transfer coefficient is on the refrigerant side [23.4 BTU/(hr•ft^2•°R)]. The water side of the evaporator is much better[758 BTU/(hr•ft^2•°R)]. So, I'm thinking that I could run the evaporator much colder, say 0°F. My guess is that ice would form until an equilibrium between ice formation and ice melting occurs. Thus I would see much higher heat transfer with the larger delta T (37°F versus 6°F). Does anyone have ideas or experience with this?

[mad fridgie]

Are you sure about your heat transfer co-efficients, check again, remember that you are changing state!
Are you trying to cool the water ? Or are you trying to remove enregy from the water so this energy can be else where "heat pump"?
Second I guess, by droping your evap temp you are just reducing the efficiency of the heat pump. Sure reducing the evap temp will increase the heat transfer through the evap, meaning less coils, but at the detrement of the system

[microHydro]

I used Bo Pierre's correlation for the evaporator refrigerant coefficient. It was about half of the Dittus-Boelter value. The water side coefficient is a Dittus-Boelter value for a 4" tube in tube. I figure it is just a ballpark value.

Yes, this is a heat pump.

Hmm. I see what you are saying. At 0°F my suction pressure is half what it is at 31°F, and thus the evaporator Q is about half.

The problem 'm trying to solve is that to collect 25,000 BTU/hr from the creek, I need 175 sf of HX area, about 1400 ft of 3/8 copper tube. The pressure drop would be about 7 psi. This length seems excessive. Ideas?

[microHydro]

I tried 20F for the evaporator, and found my system capacity reduced to 20,000 BTU/hr, but the evaporator coil length also reduced to 550 feet, which would work. So, back to the original question, Will ice formation on the evaporator coils in the creek flow reach a steady state?

[mad fridgie]

For your 25000BTU, i think you need about 32 sqF, with a 9F TD, if you use 3/8 five circuits, do not know the velocity of the coils assume around 2m/s, maybe 6 * 50ft coils (standard coil length ?), a long time since working imperial

[microHydro]

Thanks for your replies, mad fridgie!

A 9°F TD is 28°F evaporator temperature. I'll put 30°F in my model:
System Capacity: 25,000 BTU
Convection heat transfer coefficient is on the refrigerant side: 25 BTU/(hr•ft^2•°R)
Refrigerant velocity: 1.4 f/sec
Pipe Size: 3/8"
Evaporator Length: 1365 ft

It's about the same as initial 31°F evaporator stats.

My velocity is much lower than 2 m/sec you suggest. The heat exchange coefficient is very dependent on velocity - maybe 1/4 " pipe would work better. Is 2 m/s optimal for liquid velocity in the evaporator?

[microHydro]

I tried 1/4 inch:

System Capacity: 25,000 BTU
Convection heat transfer coefficient is on the refrigerant side: 43 BTU/(hr•ft^2•°R)
Refrigerant velocity: 2.6 f/sec
Exchanger area is down from 150 sf to 87 sf
Evaporator Length: 1054 ft - still long because the smaller tube has less surface area
Pressure Drop: 22 psi - way too much.

[mad fridgie]

Your model is wrong!
The refrigerant going through the pipe is not a liquid,it is a liquid/vapour mix, try again

[microHydro]

I could well be wrong. Please help me see where.

I'm using Bo Pierre's correlation for the refrigerant side heat transfer coefficient. As the text and graph below show, it does take into account a reduced h as the refrigerant mixture becomes mostly gas.

http://www.refrigeration-engineer.com/forums/attachment.php?attachmentid=6542&d=1301678843


The formula in the text below below includes a load factor Kf, which does not appear to be in dimensionless units. I assumed that Kf was in metric units. Is this right?

http://www.refrigeration-engineer.com/forums/attachment.php?attachmentid=6543&d=1301678843

Delta x is the change in fraction for the evaporator. I choose .6 as there will be a second evaporator following this one. It will be a coil with 40°F spring water flowing over it, and about 15,000 BTU available.

Notice that the text says the Nusselt and Reynolds numbers are from liquid refrigerant.

This is what I'm working from, any suggestions?

[mad fridgie]

Instead of trying to achieve your load via a single coil, split the load (multiple coils/circuits)

[Bigfreeze]

Instead of trying to achieve your load via a single coil, split the load (multiple coils/circuits)

Exactly. The original premise is a flawed one, extracting too much energy from too small an area which a) leads to freezing and b) leads to the conclusion that you should be dropping your evap temp in order to retrieve more energy. This will affect your compression ratio and drop your COP.
In my estimation you shouldn't be looking to retrieve more than 40W per linear metre of pipe in this installation and fit mutiple circuits as suggested by MF

[Peter_1]

We have some heatpumps running where we extract the heat from a big water source.
Guide: 30m/100 ft of 1/2" extracts +/- 1000 W at Te= 1°C and water temperature around 10°C.
Don't make your circuits longer than 100ft, otherwise your pressure drop will be too high.
So 7 kW will be +/- 7 coils of 1/2"@100 ft/coil

[ah fai]

halo everyone, i think i have the same problem on my system.
do have some formula for this?
what is the result for pressure drop too high?
my system discharge = 200psi
suction = only 30psi
water temp in tank = 22cel