Butanol at a temperature of 28 C is pumped at a velocity of....

[Tiberious]

Homework Statement

Butanol at a temperature of 28 C is pumped at a velocity of 14 m s-1 through a 100 mm diameter tube kept at a wall temperature of 90°C. The properties of butanol are given below.
Determine the convective heat transfer coefficient (you will find the appropriate correlation in the lessons).
Data:
ρ=950 kg m^(-3)
c_p=2.142 kj kg^(-1) K^(-1)
μ=2.9∙10^(-3) kg m^(-1) s^(-1) at 28°C
μ=1.2∙10^(-3) kg m^(-1) s^(-1) at 90°C
k=2.4∙10^(-4) kW m^(-1) K^(-1)

Homework Equations

Relevant equation

R_e= ρux/μ

P_r= (c_p μ)/k

h_x= N_u (k/x)

The Attempt at a Solution

Determining the Reynolds number,

R_e= pUd/μ
Inputting our values,

R_e= ((950)∙(14)∙(0.1))/((2.9∙10^(-3)))
Yields,

R_e= 458 621For R_e > 2300. Apply the below equation.

N_u=0.023 〖R_e〗^0.8 〖P_r〗^0.33

Determining the Prandtl number,

P_r= (c_p μ)/k

Inputting our values,

P_r= ((2.142)∙(2.9∙10^(-3)))/(2.4∙10^(-4) )

Yields,

P_r=25.8825Hence,

N_u=0.023 〖(458 621)〗^0.8 〖(25.8825)〗^0.33

Yields,

N_u= 2276.06

Am I applying the correct N_u equation ? Or, should I be applying the below as the Re > 10 000

N_u=0.027 〖R_e〗^0.8 〖P_r〗^0.33 (μ/μ_w )^0.14

 

[Chestermiller]

The two equations you are considering give very similar results, right? So, choosing between them is more of a judgment call. However, you should not be using the bulk viscosity in calculating the Re and Pr. You should be using the viscosity at the arithmetic mean temperature between the wall temperature and the bulk temperature, or, in this case, the arithmetic mean viscosity. The final equation you have for Nu includes an additional correction to Nu that is supposed to apply for extreme variations in viscosity, but can also be used here.

 

[Tiberious]

"However, you should not be using the bulk viscosity in calculating the Re and Pr. You should be using the viscosity at the arithmetic mean temperature between the wall temperature and the bulk temperature" - with respect to this, what would be the equation ?

 

[Chestermiller]

"However, you should not be using the bulk viscosity in calculating the Re and Pr. You should be using the viscosity at the arithmetic mean temperature between the wall temperature and the bulk temperature" - with respect to this, what would be the equation ?

$$Re=\frac{\rho v D}{\bar{\mu}}$$
where $$\bar{\mu}=\frac{\mu(28)+\mu(90)}{2}$$

 

[Tiberious]

How does the below look ?

Determining the Reynolds number,

R_e= pUd/μInputting our values,

R_e= ((950)∙(14)∙(0.1))/((2.9∙10^(-3)))
Yields,

R_e= 458 621

This is in the range of Reynolds numbers for which the correlation below applies.
N_u=0.027 〖R_e〗^0.8 〖P_r〗^0.33 (μ/μ_w )^0.14
Determining the Prandtl number,

P_r= (c_p μ)/k

Inputting our values,

P_r= ((2.142)∙(2.9∙10^(-3)))/(2.4∙10^(-4) )

Yields,

P_r=25.8825Hence,

N_u=0.027 〖(458 621)〗^0.8 〖(25.8825)〗^0.33 ((2.9∙10^(-3))/(1.2∙10^(-3) ))^0.14

Yields,

N_u=502.07

But,

h_x= N_u (k/x)

So, we obtain,

h_x=502.07((2.4∙10^(-4))/100)

Answer:

h_x=1.205 ∙10^(-3)

 

[Chestermiller]

You should be using the average of the wall viscosity and the bulk viscosity in calculating the Reynolds number and the Prantdl number. That would be $2.05\times 10^{-3}\ Pa.s$. So the correct Reynolds number to use would be:
$$Re=\frac{(950)(14)(0.1)}{2.05\times 10^{-3}}=6.5\times 10^5$$ and the Prantdl number should be $$Pr=\frac{(2.142)(2.05\times 10^{-3})}{2.4\times 10^{-4}}=18.3$$So, $$Nu=\frac{hD}{k}=0.027(6.5\times 10^5)^{0.8}(18.3)^{0.33}(2.9/1.2)^{0.14}=3600$$
So the convective heat transfer coefficient estimate is: $$h=\frac{(3600)(2.4\times 10^{-4})}{0.1}=8.63\ \frac{kW}{m^2.K}$$

So,
1. you should have used the arithmetic mean viscosity in calculating Re and Pr
2. your arithmetic was faulty in determining the Nu
3. your units were messed up in determining the h

 

[Tiberious]

Slightly confused how you've obtained 3600. Using the above equation I get 3564.13. (I assume rounding)

I've had a further work on this today and another look over the module notes and they all seem to agree with R_e being 458 621. Why in this case would we require the average ?

 

[Chestermiller]

Slightly confused how you've obtained 3600. Using the above equation I get 3564.13. (I assume rounding)

I've had a further work on this today and another look over the module notes and they all seem to agree with R_e being 458 621. Why in this case would we require the average ?

See Bird, Stewart, and Lightfoot, Transport Phenomena, Section 14.2. For better accuracy, they recommend using the viscosity at the average temperature between the wall temperature and the bulk temperature. This is because, for heating, the higher temperature and lower viscosity near the wall results in a steeper velocity gradient, and this translates into a steeper radial temperature gradient near the wall and a higher heat transfer coefficient.