Convective heat transfer coefficient

  • #1
Not sure if I'm getting this right, my answer seems very small, can anyone advise?

Question- butanol at a temperature of 28oC is pumped at a velocity of 14ms-1 through a 100mm diameter tube kept at a wall temp of 90oC. The properties of butanol are below. Determine the convective heat transfer coefficient.
Data:
Specific heat capacity - 2.142kJkg-1K-1
ρ 950kgm-1
Dynamic viscosity at 28oC - 2.9x10-3kgm-1s-1
Dynamic viscosity at 90oC- 1.2x10-3kgm-1s-1
Thermal conductivity - 2.4x10-4kWm-1K-1

Attempt:
Re=pUd/μ
Pr=Cpμ/K
Nu=0.332Re0.5Pr0.33
hx=Nu(k/x)

Re=950*14*0.1x10-3/2.9x10-3 458.62x10-4
Pr=2.142x2.9x-3/2.41x10-4 2.588
Nu=0.332x(458.62x10-4)0.5x(2.588)0.33 0.0973
Hx=0.0973*2.41x10-4/0.1 2.3352x10-4kWm-2K-1
Answer: 0.2335 Wm-2K-1
 

Answers and Replies

  • #2
SteamKing
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Not sure if I'm getting this right, my answer seems very small, can anyone advise?

Question- butanol at a temperature of 28oC is pumped at a velocity of 14ms-1 through a 100mm diameter tube kept at a wall temp of 90oC. The properties of butanol are below. Determine the convective heat transfer coefficient.
Data:
Specific heat capacity - 2.142kJkg-1K-1
ρ 950kgm-1
Dynamic viscosity at 28oC - 2.9x10-3kgm-1s-1
Dynamic viscosity at 90oC- 1.2x10-3kgm-1s-1
Thermal conductivity - 2.4x10-4kWm-1K-1

Attempt:
Re=pUd/μ
Pr=Cpμ/K
Nu=0.332Re0.5Pr0.33
hx=Nu(k/x)

Re=950*14*0.1x10-3/2.9x10-3 458.62x10-4
Pr=2.142x2.9x-3/2.41x10-4 2.588
Nu=0.332x(458.62x10-4)0.5x(2.588)0.33 0.0973
Hx=0.0973*2.41x10-4/0.1 2.3352x10-4kWm-2K-1
Answer: 0.2335 Wm-2K-1
Your calculations are confusing. Use '=' to separate the calculation from the final result.

As to your calculation of the Reynolds number, apparently you are saying that d = 0.1*10-3m, which is equal to 0.1 mm. A tube diameter of 100 mm = 0.1 m.
The calculation you have shown contains arithmetic mistakes (a Re < 1 is unusual)
 
  • #3
Thank you SteamKing. I missed the diameter mistake. I have corrected this but still get a low result.

Re=950*14*0.1/2.9x10-3 =0.4586
Pr=2.142x2.9x10-32.41x10-4 =2.588x10-7
Nu=0.332*(0.4586)0.5*(2.588x10-7)0.33 =0.00151
Hx=0.00151*(2.41x10-4/0.1) =3.6391x10-6kWm- 2K-1
 
  • #4
SteamKing
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Thank you SteamKing. I missed the diameter mistake. I have corrected this but still get a low result.

Re=950*14*0.1/2.9x10-3 =0.4586
You still have a basic arithmetic problem here. You're taking a relatively sizable number and dividing by a small number and getting a small number. Are you sure you know how scientific notation works?

BTW, this equation should be written: Re = 950*14*0.1 / (2.9x10-3) = ?

Pr=2.142x2.9x10-32.41x10-4 =2.588x10-7
This equation should read: Pr = 2.142 * (2.9x10-3) / (2.41x10-4) = ?

Try not bunching your terms together. Use additional spaces and parentheses to make calculations clear.

You also missed out the division by K in your version, making your calculation of the Prandtl No. incorrect.

Use the 'Preview' Button to check how your work will appear before posting. I know I use it a lot.

Nu=0.332*(0.4586)0.5*(2.588x10-7)0.33 =0.00151
You are using an incorrect Reynolds number and an incorrect Prandtl number to calculate the Nusselt number.

Hx=0.00151*(2.41x10-4/0.1) =3.6391x10-6kWm- 2K-1
We'll worry about the calculation of Hx after you have fixed the previous calculations.
 
  • #5
Ok, reviewed scientific notation, and am now getting the below. Am I on track?

Re = 950*14*0.1 / (2.9x10-3) = 458621
Pr = 2.142 * (2.9x10-3) / (2.4x10-4) = 25.8825

Re now indicates turbulent flow so dittus-bottler correlation for nusselt number to be used:
Nu= 0.023 *(458621)0.8 * (25.8825)0.4 = 2858.206
 
  • #6
SteamKing
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Ok, reviewed scientific notation, and am now getting the below. Am I on track?

Re = 950*14*0.1 / (2.9x10-3) = 458621
Pr = 2.142 * (2.9x10-3) / (2.4x10-4) = 25.8825

Re now indicates turbulent flow so dittus-bottler correlation for nusselt number to be used:
Nu= 0.023 *(458621)0.8 * (25.8825)0.4 = 2858.206
These calculations look good.
 
  • #7
Great, so I'm now calculating hx using hx=Nu(k/x).

Hx = 2858.206 * ((2.4x10-4) / 0.1 ) = 6.8597kWm-2K-1

= 6,859.7Wm-2K-1

Does this look along the right lines? Thank you for all your help.
 
  • #8
SteamKing
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Great, so I'm now calculating hx using hx=Nu(k/x).

Hx = 2858.206 * ((2.4x10-4) / 0.1 ) = 6.8597kWm-2K-1

= 6,859.7Wm-2K-1

Does this look along the right lines? Thank you for all your help.
This calculation looks OK.
 
  • #9
Brilliant, thankyou
 

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