Convective heat transfer coefficient

In summary, after correcting for diameter and other minor errors, the convective heat transfer coefficient of butanol flowing through a 100mm diameter tube at a temperature of 28oC and a velocity of 14ms-1, with a wall temperature of 90oC, is calculated to be approximately 6,859.7Wm-2K-1. The Dittus-Boelter correlation was used due to the Reynolds number indicating turbulent flow.
  • #1
Melvinamarie
5
0
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
 
Engineering news on Phys.org
  • #2
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)
 
  • #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
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.
 
  • #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
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.
 
  • #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
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.
 
  • #9
Brilliant, thankyou
 

Related to Convective heat transfer coefficient

1. What is the convective heat transfer coefficient?

The convective heat transfer coefficient is a measure of the rate at which heat is transferred between a solid surface and a fluid in motion. It reflects the efficiency of the convective heat transfer process and is typically expressed in units of watts per square meter per degree Celsius (W/m^2·°C).

2. How is the convective heat transfer coefficient calculated?

The convective heat transfer coefficient is calculated by dividing the convective heat transfer rate by the product of the surface area and the temperature difference between the solid surface and the fluid. It can also be determined experimentally through laboratory tests or simulations.

3. What factors affect the convective heat transfer coefficient?

The convective heat transfer coefficient is influenced by several factors, including the fluid properties (such as density and viscosity), the fluid flow characteristics (such as velocity and turbulence), and the geometry and surface properties of the solid surface.

4. How does the convective heat transfer coefficient impact heat transfer?

The convective heat transfer coefficient plays a crucial role in determining the rate of heat transfer between a solid surface and a fluid. A higher convective heat transfer coefficient means a more efficient heat transfer process, resulting in a faster rate of heat transfer. It is an important parameter to consider in various industrial and engineering applications.

5. How can the convective heat transfer coefficient be improved?

To improve the convective heat transfer coefficient, one can increase the surface area of the solid surface, increase the velocity or turbulence of the fluid, or use a fluid with higher thermal conductivity. Additionally, optimizing the geometry and surface properties of the solid surface can also improve the convective heat transfer coefficient.

Similar threads

Convective heat transfer coefficients of fuel oil?
    • Mechanical Engineering
Replies
2
Views
3K
Engineering Average heat transfer coefficient (forced convection)
    • Engineering and Comp Sci Homework Help
Replies
6
Views
2K
I Heat transfer coefficient and convection
    • Differential Equations
Replies
3
Views
2K
Two-phase flow heat transfer coefficient
    • General Engineering
Replies
34
Views
3K
Is a Convection Coefficient of 37 Indicative of Free Convection?
    • Mechanical Engineering
Replies
5
Views
4K
How do I get the Convection Coefficient for steam in a jacketed vessel?
    • Materials and Chemical Engineering
Replies
3
Views
1K
Heat Transfer Coefficient to Air Flow Rate
    • Mechanical Engineering
Replies
7
Views
13K
Evaluating heat transfer coefficient for a cylinder
    • Mechanical Engineering
Replies
1
Views
2K
Logical explanations of factors, Heat transfer coefficient
    • Thermodynamics
Replies
22
Views
2K
Overall heat transfer coefficient for free convection exchanger
    • General Engineering
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top