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moideen
14-10-2012, 08:31 AM
dears,

I have a indoor swimming pool 300cubic mtr volume and its heater 120 kw electrical output. Has any formula/relation for electrical KW to energy KW for swimming pools.:cool:

Tesla
14-10-2012, 08:52 AM
You will need a heat pump that will produce 120kW heating under the ambient conditions it is used. It should use less than a third of the power of the electric heater. The difference here is the heat pump is transporting heat from one area to another where the electrical heater is transforming electrical energy to heat - transforming / transporting. The formula is dependent on what the COP is and the ambient temp. The conversion formula is 1 J = 1 W/second. Incorporate the efficiency or COP or output divided by input of heat pump.

moideen
14-10-2012, 01:14 PM
yes tesla, 120 kw is not output.it is electrical input.To find out the cop, I should get the existing flow rate but there are no any records. As you said I am planning to fix new heat pump by replacing heater box. Please check my calculation:

300*4.18*17*1000= 21318000Kj
Kj to Kw =21318000/3600
= 9921 Kw/48 hr.
=123KW. (NO ADDED HEAT LOSS)
Ambient temp can expect 15c, TD is 17C.

Tesla
14-10-2012, 10:46 PM
Sorry moideen
A little confusion here and I did not explain very clearly. Electric heat output from resistance heater is the same as input so we don't need to convert anything here nor do we need any formula. Simply we need a heatpump which has an out put of 120kW. That is the heatpump can add 120kW to the water. The heatpump will need to be able to do 120kw at 15degC and the flow rate range of heatpump will need to match your water pump flow rate. That is if it is working (heating enough) with the existing electric heater.

As for the heatload calc - I think I answered another post in much detail for a swimming pool application.
Q = M x SHC x TD
Where
M = Mass flow rate of water per second
SHC = Specific heat capacity of water = about 4.2kJ/s x K
TD = the temp difference between water in and water out
As a double check you could measure the in/out temps and the flow rate of the water (not always easy). With the calc above would tell us how much heat is going into the water in kW as the J/s component is in the calc.

Magoo
14-10-2012, 11:14 PM
The flow rate through existing HX will be approx 1.736 litres per second, for a 123 Kw heater.
( 1.736 L/sec x 4.18 Kj/Kg x 17'KTD = 123.36 Kw , or 123 Kw / (4.18Kj/Kg x 17'KTD) = 1.73 L/sec )

Magoo
15-10-2012, 04:50 AM
The flow rate ties up with your stated 48 hour turn around of pool volume. To me the flow rate is low and temp rise is high considering surface loses and general loses. If applying a heat pump I would be looking at a higher flow rate and temp rise across HX at around 7.0 'C., which would give a better pool volume turn around.

moideen
15-10-2012, 07:46 AM
Sorry moideen
A little confusion here and I did not explain very clearly. Electric heat output from resistance heater is the same as input so we don't need to convert anything here nor do we need any formula. Simply we need a heatpump which has an out put of 120kW. That is the heatpump can add 120kW to the water. The heatpump will need to be able to do 120kw at 15degC and the flow rate range of heatpump will need to match your water pump flow rate. That is if it is working (heating enough) with the existing electric heater.

As for the heatload calc - I think I answered another post in much detail for a swimming pool application.
Q = M x SHC x TD
Where
M = Mass flow rate of water per second
SHC = Specific heat capacity of water = about 4.2kJ/s x K
TD = the temp difference between water in and water out
As a double check you could measure the in/out temps and the flow rate of the water (not always easy). With the calc above would tell us how much heat is going into the water in kW as the J/s component is in the calc.
Thanks tesla. Can you clear 4.2kj/s*k(reading of “K” whether delta T. by this formula ,I think can calculate running capacity.


The flow rate through existing HX will be approx 1.736 litres per second, for a 123 Kw heater.
( 1.736 L/sec x 4.18 Kj/Kg x 17'KTD = 123.36 Kw , or 123 Kw / (4.18Kj/Kg x 17'KTD) = 1.73 L/sec )
thanks magoo,What is the ideal flow rate of heatpump like we consider 2.4 gpm as per 10F delta T for chilled water A/c system?

Tesla
15-10-2012, 10:58 AM
Cool to clear it is not the ambient temp diff. It is the water temp diff. The confusion has come in with the ambient temps - this is just to ensure the heatpump might be rated at say 120kW for an ambient of 22degC which may be lacking for an ambient of 15 degC

mad fridgie
15-10-2012, 09:21 PM
You maybe approaching this from the wrong direction.
How many times do you heat the pool up from 15C to 32C, maybe once a year. If this takes 2 or 5 days does it really matter!
Your heat pump should be sized to meet the losses over a 24 hour period divided by the normally run time of the pump filtration system (normally 8-10hrs per day)

The flow rate is very high and the split is very low around the 2-3C mark. (which may be higher during start up)

To increase the heat pump duty and COP (when in normal use), you could look at introducing the air extracted from the pool area into the air inlet stream of the heat pump (HIGH LATENT HEAT)

desA
16-10-2012, 04:56 AM
MF:
To increase the heat pump duty and COP (when in normal use), you could look at introducing the air extracted from the pool area into the air inlet stream of the heat pump (HIGH LATENT HEAT)

A nice touch.

Please however take great care in the following:
1. Fin material selection (to minimise galvanic corrosion);
2. Compressor low-side protection to remain within compressor operating window.

Humidity has a way of raising Te,sat dramatically. Be prepared.

moideen
16-10-2012, 05:38 AM
A nice touch.Please however take great care in the following:1. Fin material selection (to minimise galvanic corrosion);2. Compressor low-side protection to remain within compressor operating window.Humidity has a way of raising Te,sat dramatically. Be prepared.on heating mode,15C ambient+humidity,will it increase or decrease the evaporator temperature!?please .can you explain

desA
16-10-2012, 05:59 AM
Increased humidity onto the evap coil will increase Te,sat - improving COP.