Convective heat transfer coefficient

AI Thread Summary
The discussion revolves around calculating the convective heat transfer coefficient for butanol flowing through a tube. Initial calculations were incorrect due to mistakes in the Reynolds and Prandtl numbers, which led to an unrealistically low Nusselt number. After correcting the diameter and ensuring proper scientific notation, the user recalculated the Reynolds number, indicating turbulent flow, and applied the Dittus-Boelter correlation for the Nusselt number. The final calculation yielded a convective heat transfer coefficient of 6,859.7 W/m²K, which was confirmed as correct by other participants. The conversation highlights the importance of accurate arithmetic and notation in thermal calculations.
Melvinamarie
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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
 
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Melvinamarie said:
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)
 
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
 
Melvinamarie said:
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.
 
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
 
Melvinamarie said:
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.
 
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.
 
Melvinamarie said:
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.
 
Brilliant, thankyou
 
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