# Constitutive Equation for required flow to maintain temperature

1. Jan 4, 2013

### Trespaser5

I have an endothermic gas generator which maintains a heat of 850 °C while not generating endothermic gas. The generator maintains it's temperature through a burner control system, however coolant still runs through a water jacket around the outlet pipe (which in this state has no gas running through it) the average flow out is 13.5 litres per minute. When the flow is stopped some heat is picked up from the generator and the coolant steadily increases temperature, however the temperature inside the generator does not appear to change due to the control system controlling the gas burner, however I presume that the coolant will carry some heat away from the generator. When full flow is resumed there is currently no difference in inlet and outlet coolant temperature (that is one I can measure with a standard thermometer).

Can anyone think of an constitutive equation that will determine the required flow to maintain a specified cooling water temperature assuming say a pipe temperature of 200°C ?

2. Jan 4, 2013

### CWatters

If there is no temperature difference then there is no heat being removed by the coolant.

Most likely the temperature difference is very small so the measurement is error prone. You need to find a better way to measure it. Perhaps reduce the flow to deliberatly increase the temperature uplift. Then you can work out the power that needs/is being removed. From that you can calculate the flow rate required for any particular coolant temperature increase.

or another way...

If you know the volume/mass of coolant, it's specific heat capacity and the rate of temperature increase then you can work out how much power the coolant is absorbing. From that you can calculate the flow rate required for any particular coolant temperature increase.

3. Jan 4, 2013

### CWatters

Can I check that when you say..

They are both stable as well?

4. Jan 4, 2013

### CWatters

For info..

Assuming:

1) the cooling medium is water
2) the flow rate is 13.5 L/min
3) the temperature rise less than 1C (insert lower figure if required)
4) Input and output temperatures are stable

then the power is less than..

Flow rate in L/S = 13.5/60 = 0.225 L/S = 0.225 Kg/S

Power = SCH * Δt * flow rate

= 4180 * 1 * 0.225
= 940W

Lets call it 1kW.

How much do you think your generator is producing? If less than say 500W that might explain why the temperature rise is hard to measure on an ordinary thermometer.