I'm working on the following problem:

Use a stainless-steel pipe with an internal diameter of 0.016m and a length of 2.5 m to conduct a convective heat transfer experiment. During the experiment, the stainless-steel tube is directly heated by direct current with a voltage of 5V and current of 900A, and the inlet temperature of the water is 20°C. The flow velocity is 0.5 m/s, the pipe is insulated externally, and the heat loss is ignored. Find the surface heat transfer coefficient and the temperature difference of the convective heat transfer in the tube.

Since there is no width of the pipe, I can't use the specific heat of stainless steel to figure out the temperature of the pipe. I resorted to calculating the added temperature through the water's specific heat, and got 30.7 °C as the outlet temperature, but I'm not confident in this solution at all.

How should I go about it? Thanks in advance.

Edited to add my calculations: I used the Joule heating formula . Then, I divided Q by the specific heat capacity per the mass of water, so . The result was 10.69°C, so it would be 30.69°C for the final temperature.

I wasn't sure if this was the correct way to go about this, since the problem seems to stress on the stainless-steel bit, so I was wondering if there is another way to do this more accurately.

  • 1
    $\begingroup$ It's been a long time since I've done anything like this, but ... electrical power, P=IV. In your case P= 5(900)=4500 W. As a starting point I'd assume this would be the heat energy provided by the electrical heater & that this gets added to the 20C inlet water temperature. Finish it off. $\endgroup$
    – Fred
    Commented Jan 2, 2023 at 8:15
  • $\begingroup$ Can you Edit to show your calculations? I'm getting more like a 10°C temperature rise. $\endgroup$
    – Transistor
    Commented Jan 2, 2023 at 9:14
  • $\begingroup$ @Fred, thank you, I did go that route with my solution, just wasn't quite sure about it $\endgroup$
    – Miya
    Commented Jan 2, 2023 at 14:00
  • $\begingroup$ @Transistor Edited, sorry I might've been a bit unclear, the rise I got was 10°C as well, 30°C was the overall outlet water temperature so 10° + the inlet 20° $\endgroup$
    – Miya
    Commented Jan 2, 2023 at 14:01
  • $\begingroup$ Attach a thermocouple. $\endgroup$ Commented Jan 2, 2023 at 15:41

1 Answer 1


I agree with your calculations. The temperature rise should be about 10.7 °C.

The stainless steel aspect is mostly irrelevant other than it should ease any worries about rust, etc., affecting the result.

Now, what's the answer to the other bits? I'd have to do some more thinking and research before I could answer confidently.

  • $\begingroup$ As for the rest, my thinking process is that since the DC current and water inlet temperature are constants, the pipe would be heated up to 20°C by the water and up an additional 10.7°C by the constant current. Got confused when I tried to use these values to get the LMTD, so I tried to use the heat transfer coefficient formula using heat flux over temperature difference between the pipe and the fluid. So the heat flux is 22500/(pi*0.016*2.5) = 179049.31 W/m^2, and that divided by the temp difference 10.7 C is 16744 W/m^2 C. $\endgroup$
    – Miya
    Commented Jan 2, 2023 at 15:14
  • $\begingroup$ For the second part of the question, I think I should use the heat transfer formula Q = hALMTD. I’m not quite sure if the “temperature difference of the convective heat transfer in the pipe” part of the problem is referring to the 10.7 C bit or to the LMTD though. $\endgroup$
    – Miya
    Commented Jan 2, 2023 at 15:14

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