Heat transfer between water tank and pipe system

In summary, the water circulation rate needs to be high enough to remove heat from the office, while being slow enough so that the fan coil unit can operate.
  • #1
Leo Lee
4
0
Hello All,
I have a question for heat exchange calculation and hope someone can help. I would like to recover the "cooling energy" from a ice melting pool for air-conditioning for a small office. I plan to install three fan coil units with 800CFM for the office with closed loop water pipe to circulate water from fan coil units to the cooling coil immersed in the ice melting water tank and then back to fan coil units. Hence, heat from the office will transfer to the ice melting water.
Ice melting tank effective size : 4m x 2.2m x 1m(H)
Ice melting water temperature : 3 ~ 5*C as measured
Cooling coil : OD : 9.52mm OD; Length : 32.4m ; Material : copper
Fan coil unit : each cooling capacity 7.92kW (assume water in at 7*C and leave at 12.5*C)
What is the water circulating flow rate for the above system ?
How long will the ice water support for the system (as the water temperature will rise up) ? Ice will be refilled for every 2 to 3 hours and cannot support air-conditioning at 7*C.
Any assumption can be made. I only need rough estimation for reference. I have a normal air-conditioning system for the office as backup. This project is for environmental education purpose. Thanks !
 
  • #3
Anyone can help ?
 
  • #4
Hello Leo, :welcome:

It is a bit unclear how we can help here. Are you asking simple things, like
Leo Lee said:
What is the water circulating flow rate for the above system ?
with as a naive answer$$ {dQ\over dt} = \rho \,c_p \, \Delta T\, F_v \quad \Rightarrow \quad F_v = { 7.92\; kJ/s \over 1000 \; kg/m^3 \ 4.2 \;kJ/(kg\cdot \mathstrut^\circ\! C) \ 5.5 ^\circ \!C} \ \ ? $$So some 1.2 m3/h per coil.

Similarly for the melting rate of ice:
Leo Lee said:
How long will the ice water support the system
$$

{dM\over dt} = {dQ\over dt} \Delta H_f \quad \Rightarrow \quad T = \rho \ V \ \Delta H_f {dQ\over dt} = {593\; kg/m^3 \ 8.8 \;m^3 \ 334\; kJ/kg \over 7.92\; kJ/s}$$or some 20 hours if all three are at full capacity (and the ice tank is perfectly isolated :rolleyes: (*) )

(## \; 593\; kg/m^3\ ## is the density of dry crushed ice)But there are some things in your account I can't follow: ice in contact with water should yield a temperature of ##0 \; ^\circ \!C\ ##, so how come you measure 3 - 5 ? And use 7 degrees for the coil input ?

How did you determine the length of your cooling coil ? Isn't it a design parameter to be established ? You have ##\displaystyle {dQ\over dt} = U\; A\; \Delta T## with ##A## the contact area and ##U## the over-all heat transfer coefficient (which you probably don't know yet).
 
  • #5
BvU said:
Hello Leo, :welcome:

It is a bit unclear how we can help here. Are you asking simple things, like
with as a naive answer$$ {dQ\over dt} = \rho \,c_p \, \Delta T\, F_v \quad \Rightarrow \quad F_v = { 7.92\; kJ/s \over 1000 \; kg/m^3 \ 4.2 \;kJ/(kg\cdot \mathstrut^\circ\! C) \ 5.5 ^\circ \!C} \ \ ? $$So some 1.2 m3/h per coil.

Similarly for the melting rate of ice:
$$

{dM\over dt} = {dQ\over dt} \Delta H_f \quad \Rightarrow \quad T = \rho \ V \ \Delta H_f {dQ\over dt} = {593\; kg/m^3 \ 8.8 \;m^3 \ 334\; kJ/kg \over 7.92\; kJ/s}$$or some 20 hours if all three are at full capacity (and the ice tank is perfectly isolated :rolleyes: (*) )

(## \; 593\; kg/m^3\ ## is the density of dry crushed ice)But there are some things in your account I can't follow: ice in contact with water should yield a temperature of ##0 \; ^\circ \!C\ ##, so how come you measure 3 - 5 ? And use 7 degrees for the coil input ?

How did you determine the length of your cooling coil ? Isn't it a design parameter to be established ? You have ##\displaystyle {dQ\over dt} = U\; A\; \Delta T## with ##A## the contact area and ##U## the over-all heat transfer coefficient (which you probably don't know yet).

Thanks for your reply. Sorry for confusion in my question. I would like to determine the water circulation rate inside the pipe system. That is critical to determine the water temperature entering and leaving the ice melting pool, as well as entering and leaving fan coil units. If water circulation rate is too fast, there will be no sufficient time for heat transfer to the ice melting pool. If it is too slow, the fan coil unit will not have sufficient water flow to remove heat from the office. How can I determine a suitable water circulation rate ?

I measured the water temperature at the ice melting pool by infrared temperature measurement sensor and the reading is 3~5 degree. Is it not suitable for the measurement? On the other hand, the pool has hot water supply for ice melting. I am not sure any hot water mixed in the pool before my measurement.

Regarding the cooling coil length, I will use an abandoned 1600 CFM fan coil unit's coil and the measured length is 32.4m.

I don't know the U-value of the cooling coil in the ice melting pool, any ball-part figure can be used ?

Thank you very much.

Best Regards

Leo
 
  • #6
Hello, anyone can help ? Thanks.
 
  • #7
Leo Lee said:
Thanks for your reply. Sorry for confusion in my question. I would like to determine the water circulation rate inside the pipe system. That is critical to determine the water temperature entering and leaving the ice melting pool, as well as entering and leaving fan coil units. If water circulation rate is too fast, there will be no sufficient time for heat transfer to the ice melting pool. If it is too slow, the fan coil unit will not have sufficient water flow to remove heat from the office. How can I determine a suitable water circulation rate ?
Since you specified an inlet and outlet temperature and power, that leads directly to only one possible water circulation rate, as calculated by @BvU
I measured the water temperature at the ice melting pool by infrared temperature measurement sensor and the reading is 3~5 degree. Is it not suitable for the measurement?
It is a bit surprising, but I suppose it is possible in a dynamic situation.
Regarding the cooling coil length, I will use an abandoned 1600 CFM fan coil unit's coil and the measured length is 32.4m.

I don't know the U-value of the cooling coil in the ice melting pool, any ball-part figure can be used ?
Since this is real-world, that makes life easier. You can calculate it based on the actual testing you've done, but I don't think I would bother. With a 3-5C melting pool temperature and 7C supply to the FCU, I'd say your main problem is too small of a tank and/or not enough ice. Ice storage can be a useful solution to some HVAC problems, but as you are finding out, you need a lot of ice. In the USA, we have a unit of refrigeration called a "ton", which refers to the melting of a US ton (900 kg) of ice in 24 hours. Each of your FCUs has a capacity of 2 tons.
 
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1. How does heat transfer occur between a water tank and pipe system?

Heat transfer occurs through conduction, convection, and radiation. Conduction involves the transfer of heat through direct contact between the tank and pipes. Convection occurs as the water flows through the pipes, carrying heat away from the tank. Radiation occurs when heat energy is transferred through electromagnetic waves.

2. What factors affect the rate of heat transfer between a water tank and pipe system?

The rate of heat transfer is affected by the temperature difference between the tank and pipes, the material of the tank and pipes, the surface area of contact, and the flow rate of the water through the pipes.

3. How can heat transfer be maximized between a water tank and pipe system?

To maximize heat transfer, the temperature difference between the tank and pipes should be increased, the surface area of contact should be increased, and the flow rate of the water should be increased.

4. What are the potential consequences of poor heat transfer between a water tank and pipe system?

Poor heat transfer can lead to inefficient heating or cooling of the water, resulting in increased energy costs. It can also lead to uneven heating or cooling, which can cause damage to the tank or pipes.

5. How can heat transfer between a water tank and pipe system be improved?

To improve heat transfer, insulation can be added to the pipes and tank to reduce heat loss. Increasing the surface area of contact between the tank and pipes can also improve heat transfer. Additionally, using materials with high thermal conductivity can help to improve heat transfer.

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