Effect of altering cooling water flow rate

[scarecrow75]

Hi

I'm new to this and am still trying to get my head around some parts. So I was shown an experimental procedure for investigating the performance of a refrigeration plant (R134a) incorporating a water cooled condnser. I varied evaporator loads 1400W, 1200W and 1000W and adjusted the water flow rate across the condenser between 0.01kg/s, 0.02kg/s and 0.03kg/s with results that I interpret that as the flowrate of water across the condenser increases the cooling water differential decreases. But I thought that if Q=mcdt then surely if I increase water flowrate, my temperature differential should increase. What am I missing here? See my recorded results below.

Thanks and sorry if post is in wrong section

Barry

[frank]

Hi Barry
Just think that as the flow rate increases then the water passes through the condenser quicker so it picks up less heat. In effect, with increased flow, we have a shorter 'dwell' time.

A good example is to picture an electric shower. To get the shower water hotter, you restrict the flow through the heat exchanger - increased dwell time, to make the water cooler, you increase the flow - reduce the dwell time.

[scarecrow75]

Hi Barry
Just think that as the flow rate increases then the water passes through the condenser quicker so it picks up less heat. In effect, with increased flow, we have a shorter 'dwell' time.

A good example is to picture an electric shower. To get the shower water hotter, you restrict the flow through the heat exchanger - increased dwell time, to make the water cooler, you increase the flow - reduce the dwell time.

Thanks Frank. Maybe I'm having a "dumb" moment and I agree with everything above BUT if heat transfer follows the guide of Q(kW) = m(kg/s) x Cp(kJ/kgK) x dt('C) then the increase in flowrate should return an increase in Output?

Something I'm not grasping right here?

Any thoughts?

Barry

[cduque]

Hello Scarecrow75,
You can increase the flow, decrease Dt and still increase heat transfer, it's only a matter of having a flow increase bigger then the Dt decrease, right?

Q=m.cp.dt
example:
Q1 = 10 x 4 x 10 = 400
Q2 = 20 x 4 x 8 = 640

The question is that when you increase the flow you also increase the heat transfer coefficient, so your Dt will be lower but still increasing the total heat transfer.

CDuque

[Brian_UK]

You are performing your calculation using the heat rejection flow rate when you should be using the heat input side I think.