I have a tank of oil at 55 degrees c. I plan to run a copper pipe 8mm in diameter (1mm thickness) into a coil 15m long inside the tank. For all purposes of assumption, the copper pipe is perfectly transferring heat across to the oil and vice-versa.

I have a pump of 10l/min that circulates this oil around the tank, so we will just assume that the 55degree oil is always well distributed.


I want to cool this down as much as possible, and was wondering if I was running tap water (28degrees), cold water (10degrees), how much would the coil be able to bring the temperature down by. We will assume that water is running through the coil at 5l/min.

Specifically I'm looking to convert the heat removal capacity into temperature in degrees. So that I can decide whether I want to

i) Use cold or room temperature water ii) Whether this concept works at all


  • $\begingroup$ You will cool more with colder water, that's for sure, but it is not easy to theoretically compute the efficiency of your coil. But if you run it and measure the temperature going in and out, a global heat transfer factor would be easy to calculate. Also, the heat removal capacity cannot be converted into degrees, unless you list the total volume of oil, as well as its physical properties, specifically its heat capacity. $\endgroup$
    – Jaime
    Nov 6, 2012 at 19:00

1 Answer 1


Suppose you have a mass $M$ of oil at a temperature $T$ and specific heat capacity $C_{oil}$. You pump water at a temperature of $T_0$ and flow rate of $f$ kg/sec, and we'll assume as you say that the heating of the water by the oil is effectively instant. The heat absorbed by the water per second is:

$$ \frac{dU}{dt} = f \space \left(T - T_0 \right) \space C_w $$

where $C_w$ is the specific heat of water. The corresponding temperature drop per second in the oil is:

$$ \frac{dT}{dt} = \frac{1}{M \space C_{oil}} \frac {dU}{dt} = - \frac{f \space C_w}{M \space C_{oil}} \left(T - T_0 \right) $$

and one quick integration later we get:

$$ T - T_0 = T_{init} \space \text{exp} \left( - \frac{f \space C_w}{M \space C_{oil}} t \right) $$

where $T_{init}$ is the initial temperature in your tank i.e. 55C.

You give values for $f$ = 0.0833 kg/sec and $T_{init}$ = 55C. The specific heat of water, $C_w$ is 4.2 kJ/kg and oil is around 2 kJ/kg depending on the exact type of oil. However you don't give a value for M, so it's not possible to give an actual figure for the cooling rate. However if you plug in the mass of oil in your tank you can calculate the temperature as a function of time.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.