Convective heat transfer coefficient

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Discussion Overview

The discussion revolves around calculating the convective heat transfer coefficient for butanol flowing through a tube. Participants explore the necessary equations and properties of butanol, addressing issues related to scientific notation, arithmetic errors, and the application of relevant correlations for turbulent flow.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Initial calculations of Reynolds number, Prandtl number, and Nusselt number are presented, with some participants expressing concern over the small values obtained.
  • Participants identify arithmetic mistakes in the calculations, particularly regarding the diameter of the tube and the use of scientific notation.
  • Corrections are suggested for the Reynolds number and Prandtl number calculations, leading to a revised understanding of flow conditions.
  • A later participant proposes using the Dittus-Boelter correlation for the Nusselt number based on the corrected Reynolds number indicating turbulent flow.
  • Final calculations for the heat transfer coefficient are presented, with participants confirming the correctness of the results and expressing gratitude for assistance.

Areas of Agreement / Disagreement

Participants generally agree on the need for corrections in initial calculations, but there is no consensus on the correctness of the initial results before corrections were made. The discussion evolves from confusion to clarity as participants refine their calculations.

Contextual Notes

Limitations include potential misunderstandings of scientific notation and arithmetic, as well as the initial misinterpretation of the tube diameter affecting the Reynolds number calculation.

Who May Find This Useful

Students and professionals interested in fluid dynamics, heat transfer, and practical applications of thermodynamics may find this discussion beneficial.

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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