# Heat Transfer question, estimation of coefficients

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1. Nov 29, 2014

### timmayy

Hi, ive recently been given a series of questions on heat transfer to do and have done most of them with general ease, but this one question ive been stuck on for ages and i cant seem to figure out:

"A double pipe heat exchanger is made up from a length of 25mm i.d. steel pipe of 2.5mm thickness, and a length of 75mm i.d. pipe. The fluid in both pipes is water. Estimate the heat transfer coefficients at the outside and inside of the inner pipe for the following conditions.

Outer pipe flow-rate 5kg/s @ 23˚ C
Inner pipe flow-rate 0.5kg/s @ 51˚C

Use the correlation for turbulent flow:
Nu = 0.023 Re0.8 Pr0.33 (μ/μw)0.14
The physical properties of water are:
Cp of water @ 23˚C = 4.187kJ/kgK
Cp of water @ 51˚C = 4.184kJ/kgK
k of water @ 23˚C = 0.670W/mK
k of water @ 51˚C = 0.635W/mK"

Included in the question is also a density and viscosity to relevant temperatures table
Ive been working through it and have calculated all cross sectional areas, but then i cant seem to put it into any equations (nusselts etc), as i am unable to find the in temperature (and so relevent density/viscosity) - i tried finding log means of the temperatures, densities and viscosities but nothing seems to work...
any help would be great thanks :)
Tm

2. Nov 29, 2014

### Staff: Mentor

Show us more details on what you did. Tell us if it is counterflow or cocurrent flow. Can you calculate Re and Pr at least for the inlet conditions?

Chet

3. Nov 30, 2014

### timmayy

Hi, whether it was counter or co current wasnt given in the question, we also dont have inlet or outlet temperatures to work it out with (im not entirely sure what the temperatures given are), all we could do was work out cross sectional areas and we tried taking log mean values for k, viscosity and density to work out reynolds number, but the values calculated was too small for turbulent flow:
found CSA of inner pipe = 4.9087 x10-4 m
CSA of outer pipe = 3.711 x10-3m
μ log mean = 0.71156 mNs/m2
ρ log mean (ρlm) = 992.3646 kg/m3
using U=G/ρlm CSA = 11.36
so, RE = ρUd/μ
RE = 992.3646 x 11.36 x 0.0025 / 0.71156x10-3
= 396075.35
then:
Klm=0.65234 w/mk
Q = 2πr Klm(T1-T2)/(r1-r2)
so we took log mean of temprature and radius to give:
Q = 166.54 W
but all these values seem wrong and we cant piece together how to work it out, everything weve tried involves far too many log means as no exact values are given?
Thanks
Tim

4. Nov 30, 2014

### Staff: Mentor

The input data describes the local conditions at a specific point within the heat exchanger. You need to determine the overall heat transfer coefficient, including inside the tube, within the annulus, and through the steel wall. You will need to determine the inside wall temperature and the outside wall temperature. This is going to involve a little trial and error.

For the first iteration, assume that the inside wall temperature is 51C and the outside wall temperature is 23C, so that the temperature is uniform within the tube and within the annulus. These temperatures will be corrected in subsequent iterations. Calculate the Nussult number for the inner flow and for the outer flow. Then get the heat transfer coefficients inside and outside. Now you will need to get the heat flow and the new inside and outside wall temperatures, taking into account the three resistances in series. You now have new values for the temperatures on the inside and outside walls to start the second iteration.

Chet

5. Nov 30, 2014

### timmayy

ahh ok, thats great thanks :)
Tim