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I'm trying to design a liquid fueled rocket engine, with copper walls cooled by kerosene.

While designing it, the following formula comes out:

$$Q=qA=wc(T-T(0))$$ Where:

  • Q = total heat transferred

  • q = average heat transfer rate

  • A = heat transfer area

  • w = coolant flow rate

  • c = specific heat of coolant

  • T = temperature of coolant leaving the engine

  • T(i) = temperature of coolant entering the engine

This formula gives the total heat transferred from the combustion chamber to the cooling liquid.

To proceed with my calculations, I need to know the value of "q", known as "overall heat transfer coefficient" or "average heat transfer rate", between a copper chamber and the coolant (kerosene).

I need this number to find w:

$$w =\frac{Q}{c-(T-T(i))}$$

"A", from the first formula, is known: "q" is needed to find the value of Q and finally find the actual cooling liquid flow rate

I'm currently working with Btu, lb, sec, in as units

Thank you in advance

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  • $\begingroup$ I'd like to help but can you describe the set-up a little better? Ta. $\endgroup$ – Gert Jan 22 at 17:59
  • $\begingroup$ @Gert I need to find the value of "q" which should be called "average heat transfer rate" as I need it to find the proper dimension of the kerosene flow to cool down my rocket engine. I don't really know where to find it tho: how do I calculate the average heat transfer rate between a solid wall and a liquid? $\endgroup$ – Nicop.dev Jan 22 at 18:02
  • $\begingroup$ Why do you have a $T(0)$ and a $T(i)$? $\endgroup$ – Gert Jan 22 at 18:03
  • $\begingroup$ I do have the difference between the two, which is 40F $\endgroup$ – Nicop.dev Jan 22 at 18:04
  • $\begingroup$ What's the geometry of the copper walled engine? $\endgroup$ – Gert Jan 22 at 18:09
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First off, it looks like your formula should be $\frac{dQ}{dt}$ equals the rest; it doesn't make sense to set TOTAL heat transferred equal to RATE of heat transferred, which is what your other terms give. The standard formula for total rate of heat flow is as follows: $$\frac{\Delta Q}{\Delta t}=\frac{-kA\Delta T}{\Delta x}$$ Where $k$ is the thermal conductivity of the material (I don't remember what $k$ is for copper; I'm sure you can look it up), $\Delta T$ is the difference between the temperature on either side of the copper (based on your setup, difference between coolant outside and fuel inside?), and $\Delta x $ is the thickness of the copper. Of course, $\Delta $ can be sent to $d$ when integrating over time.

It looks like your paramter $q$ should then be equal to $\frac{-k\Delta T}{\Delta x}$. All you have to do is look up $k$ for copper, figure out thickness of the copper, and figure out the temperature of liquid kerosene and of the interior of the engine. I hope this helps.

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    $\begingroup$ You are assuming no convective resistance at all, only conductive. That is unrealistic. $\endgroup$ – Gert Jan 22 at 21:43
  • $\begingroup$ I see, I completely neglected the effect of the coolant's flow rate. Thank you for your comment. $\endgroup$ – Shura Zeryck Jan 23 at 22:53
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Engine cooling

The go-to equation here is Newton's Law of Cooling:

$$\dot{q}=-uA(T_E-T_C)$$

where:

  • $\dot{q}$ is the heat flux (heat carried off by the coolant, per unit of time),
  • $u$ is the heat transfer coefficient,
  • $A$ is the total surface area separating coolant and engine,
  • $T_C$ and $T_E$ the coolant and engine temperatures respectively.

The trouble is that neither $T_C$ and $T_E$ are constant and this makes calculating a total heat flux rather difficult (except for the simplest of geometries). For that reason average temperatures are often used, as an approximation:

$$T_C=\frac{T_{C,i}+T_{C,o}}{2}$$

and:

$$T_E=\frac{T_{E,i}+T_{E,o}}{2}$$

Bar any losses, we also know that:

$$\dot{q}=wc(T_{C,i}-T_{C,o})$$

The only remaining unknown is $u$. Usually a value is chosen from an engineering resource like this one. Alternatively a value can be determined empirically.

Also, I know the temperature inside the cylinder and outside it,

Strictly speaking that is not possible, as these temperatures aren't constant by definition. Something that is cooled is lowered in temperature, no matter how little it might be.

and I also know the thickness of the cylinder.

In that case $u$ has to be calculated as a thermal resistance:

$$\frac1u=\frac{1}{h_i}+\frac{\tau}{k}+\frac{1}{h_o}$$

where:

  • $h_i$ the 'inside' (convection) coefficient of heat transfer,
  • $\tau$ the wall thickness and $k$ copper's heat conductivity,
  • $h_o$ the 'outside' (convection) coefficient of heat transfer.
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  • $\begingroup$ How can I choose a value from those charts? $\endgroup$ – Nicop.dev Jan 22 at 19:24
  • $\begingroup$ Also, I know the temperature inside the cylinder and outside it, and I also know the thickness of the cylinder. I don't know if they'd be any useful tho $\endgroup$ – Nicop.dev Jan 22 at 19:27
  • $\begingroup$ Will make an edit in about 30''. $\endgroup$ – Gert Jan 22 at 19:39
  • $\begingroup$ Great! Thank you! $\endgroup$ – Nicop.dev Jan 22 at 20:59
  • $\begingroup$ How can I obtain the h1 and h0 values? $\endgroup$ – Nicop.dev Jan 22 at 21:45

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