Heat Ex-changer Calculations

Refeek Maher

Hi all, I am trying to work out a few calculations on a plate heat ex-changer. Any help would be greatly appreciated.

On the primary side there is 90 degrees Celsius going in and 70 degrees Celsius going out at 30.1 m3/h

The secondary side I have 65 degrees going in at 17.8 m3/hr.

The question is what temperature will I have coming out of the secondary side and kW will have been exchanged?

What is the correct formula to use for this? Is the size of the Heat ex-changer needed?

Ports are DN 80

Need this to find out a figure that is missing in a schematic of a CHP system.

Thanks!

Pete67

q = h / ((4.2 kJ/kgoC) (1000 kg/m3) dt)

= h / (4200 dt)

where

q = water flow rate (m3/s)

h = heat flow rate (kW or kJ/s)

dt = temperature difference (oC)

Dtp1979

Xyz(65) = 2134
A23 (e45) @ 566788
Flow 45y = rt% (2134)
Return 45y = thr% (rt%4567)

So simple really. 18 degrees.

Refeek Maher

Thanks Pete 67! But what about the flow rate? there are two different flow rates.

Thanks as well Dtp1979, but finding it hard to follow your method.

Dtp1979

Refeek Maher wrote: »

Thanks Pete 67! But what about the flow rate? there are two different flow rates.

Thanks as well Dtp1979, but finding it hard to follow your method.

Madness was my method. I was only kidding around mate. Maths isn't my strong suit

Pete67

Flowrate is important, and it does not have to be the same on the primary and secondary sides. Basically the heat input on the primary side should equal the heat output on the secondary side. You need the flowrate in each case to calculate this. Assuming the fluid in both cases is water, from a quick calculation you have just over 700 kW of heat coming in on the primary side, however there is a problem with the figures on the secondary side - adding heat at a rate of 700kW to the 65 degree water circulating at 17.8 m3/hr would increase the temperature to above that of the incoming hot water on the primary side. Obviously this is not possible so you need to either increase the flow on the secondary side, or lower the inlet temperature.

Refeek Maher

That is correct. This is actually based on a real 700 kW Plate heat ex-changer used in a CHP system.

How were you able to calculate it at 700 kW?

The 65 degrees input on the secondary side is actually variable, it can be more or less then 65 degrees.

If the flow rate on the secondary side needs to be increased, how is this calculated?

So the heat output on the secondary should technically be 90 degrees?

Thanks very much "Pete67" for the info, and taking your time to help me out with this, greatly appreciated!

Pete67

OK, worked example.

q = h / ((4.2 kJ/kgoC) (1000 kg/m3) dt)

or simplifying the above: q = h / (4200 dt)

where

q = water flow rate (m3/s)

h = heat flow rate (kW or kJ/s)

dt = temperature difference (oC)

if q = h / (4200 dt)

then h = q (4200 dt)

so considering the primary side and putting numbers into the formula:

q = 30.1 m3/hr but we need m3/sec so divide by 3600, q = 0.008361 m3/sec (which is 8.361 l/s)

dt = difference between inlet and outlet temps, so 90 - 70 = 20

so h (in kW) = 0.008361 * 4200 * 20 = 702.33 kW


I'm sure you can do the same calculation for the secondary side.