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# Flow rate heat transfer confusion

## Flow rate heat transfer confusion

(OP)
Hello,

I try to use equation to calculate how much flow rate needs to be to removed heat from apparatus, but I confuse myself. I thought more flow means more heat taken away?

I take this equation:

Qdot = Vdot * cp * ρ * ΔT.

I rearrange to find Vdot:

Vdot = Qdot / cp * ρ * ΔT

If Qdot kept same, but I increase value of ΔT, Vdot gets smaller. Why?

I understand bigger number on bottom of fraction means smaller number total, but for concept I do not understand. Why smaller temperature difference has bigger flow rate?

Thank you all,

Miran
Replies continue below

### RE: Flow rate heat transfer confusion

What do the terms of the equation mean?

How do they interact with each other?

### RE: Flow rate heat transfer confusion

(OP)

#### Quote (MintJulep)

What do the terms of the equation mean?

How do they interact with each other?

Qdot is rate heat transfer (Watts). cp is specific heat capacity of fluid (kJ/kg C), water in this case. ρ is fluid density (kg/m^3). ΔT is temperture difference (°C). Vdot is volume flow rate (m^3/s).

When arrange like this, Qdot = Vdot * cp * ρ * ΔT, I see that more flow is more heat. If double flow rate heat is double as well. If ΔT is double then heat is double. But when arranged for Vdot and keeping heat keep constant, increasing ΔT means less flow.

I imagine two apparatus. I want flow rate of both. Apparatus 1 has Qdot and ΔT, appartus 2 has Qdot but 2ΔT. With same Qdot, but 2ΔT apparatus 2 has half flow rate. Mathematically makes sense, but not undertsnad physical concept.

Miran

### RE: Flow rate heat transfer confusion

You have kept the heat flow the same but increased the delta T.

That is only possible if you reduce the flow rate.

So in your example you want the same heat flow to two systems with different DT.

So volume flow must change as well.

Think of say a delivery of parcels.
You want the same number of parcels in a day (Q), but you have two different trucks. One can take say twice the number of parcels as the other (DT). So to get the same number of parcels per day you only need half the number of big trucks (V)

Did that help?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

### RE: Flow rate heat transfer confusion

Or drive faster.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

### RE: Flow rate heat transfer confusion

Miran,

You are missing the other "half" of the equation. Your equation simply determines how much energy is transferred by raising a certain mass flow of material by a particular temperature. It gives no insight as to what the actual dT is. That dT is governed by your design.

Q = U * A * dT

U = overall heat transfer coefficient
A = surface area for heat transfer
dT = average temperature difference between hot and cold sides

Calculating U is the difficult part, as it is the reciprocal of the sum of the thermal resistances, each of which has its own calculation involving Reynolds numbers, Prandtl/Grashoff Numbers (multiplied to get Rayleigh number), thermal conductivity, and physical dimensions.

Increasing flow of either hot or cold side will served to increase both average dT as well as decrease that respective thermal boundary layer heat transfer resistance, which impacts U.

The fact that you don't appear to know what an equation is (evidenced by your confusion over why Vdot decreases when dT increases at constant Q), makes me think the above information is probably more than you will ever use.

### RE: Flow rate heat transfer confusion

#### Quote (Qdot = Vdot * cp * ρ * ΔT. I rearrange to find Vdot: Vdot = Qdot / cp * ρ * ΔT If Qdot kept same, but I increase value of ΔT, Vdot gets smaller. Why?[/quote)

Your equations have to do with the carrying capacity of the cooling fluid. If you have a certain amount of heat to carry, then it's the product of the flow rate and deltaT that matter.

[quote ]I thought more flow means more heat taken away?

It is, but you forced the heat flow to be constant. If you increase velocity, then the thermal mass flow is higher, assuming deltaT does not decrease.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
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### RE: Flow rate heat transfer confusion

(OP)

#### Quote (LittleInch)

Did that help?

Yes thank you very much. Think I was just tired and thinking about what I was doing at the time too much and confuse myself.

Miran

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