Convective heat transfer coefficient

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SUMMARY

The forum discussion centers on calculating the convective heat transfer coefficient for butanol flowing through a 100 mm diameter tube at a temperature of 28°C and a wall temperature of 90°C. The initial calculations presented by the user contained errors in the Reynolds number and Prandtl number due to incorrect unit conversions and arithmetic mistakes. After corrections, the user successfully applied the Dittus-Boelter equation to determine the Nusselt number, leading to a final convective heat transfer coefficient of 6,859.7 W/m²K.

PREREQUISITES
  • Understanding of fluid dynamics concepts such as Reynolds number and Prandtl number.
  • Familiarity with heat transfer principles, specifically convective heat transfer.
  • Knowledge of the Dittus-Boelter equation for calculating Nusselt number.
  • Proficiency in scientific notation and unit conversions in engineering calculations.
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  • Study the Dittus-Boelter equation for Nusselt number in turbulent flow conditions.
  • Learn about the significance of Reynolds and Prandtl numbers in heat transfer applications.
  • Explore advanced topics in convective heat transfer, including laminar vs. turbulent flow analysis.
  • Review common mistakes in fluid dynamics calculations and how to avoid them.
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Mechanical engineers, thermal system designers, and students studying heat transfer principles will benefit from this discussion, particularly those focusing on convective heat transfer in fluid systems.

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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