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# The power required for heating a circulating fluid

## The power required for heating a circulating fluid

(OP)
the equation to use to work out the power required for heating a circulating fluid:
Pch = (Qm × Cp ×(t2 − t1) × 1,2) ÷ 860
-Heating power : Pch (kW)
– Mass flow rate : Qm (kg/h)
– Specific heat of fluid : Cp (kcal/kg × °C)
– Inlet temperature : t1 (°C)
– Required outlet temperature : t2 (°C)
– 1,2 : Safety coefficient linked to our manufacturing tolerances and variations in network power

however the system im using is a closed loop so whatever leaves t2 will come back into t1 therefore my t1 temperature will always be changing so how would i calculate it with the constant change ?

### RE: The power required for heating a circulating fluid

Your delta T should be the average temperature of your fluid vs. the average temperature of the ambient to which it loses heat. THAT is the steady state case.

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### RE: The power required for heating a circulating fluid

Well power in watts ( joules /second) needs a mass flow rate in hour to be divided by 3600, not 860

If you're talking about heating up then you need to use a transient program or do this in a series of small steps.

As noted in your previous post Cp changes with temperature.

This is a transient event with many things changing as the temperature rises (including any heat losses in your circuit). It is not possible to find a simple equation which can calculate this.

If you do the bulk energy method and then divide by a factor of 1.5 you won't be far away providing you heat losses aren't too big.

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### RE: The power required for heating a circulating fluid

(OP)
ok thank you
why would i need to divide the bulk energy by a factor of 1.5?

Also when calculating a object heating up time do i factor in the heat loss which will increase the time or would i put a tolerence of +/- a suitable number to make up for the heat loss ?

### RE: The power required for heating a circulating fluid

The 860 is because they have kCal/kgC for the specific heat but power is in kW. 3600 / 4.184 = 860.

"Also when calculating a object heating up time do i factor in the heat loss which will increase the time or would i put a tolerence of +/- a suitable number to make up for the heat loss ?
"

If you are heating it up to ambient? Heat loss will be quite small. Heating it up to 500 degC? A lot more heat loss to ambient...

### RE: The power required for heating a circulating fluid

You should be able to design the system so that a pump is not needed, and you would use gravity to cause the heated fluid to circulate "naturally". You would need to know how the fluid's density changes with temperature, and the other physical properties needed to calculate the fluid friction etc.

"...when logic, and proportion, have fallen, sloppy dead..." Grace Slick

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