Estimate the local heat transfer flux

[electr]

Homework Statement

Liquid ammonia is heated as it flows at a mean velocity of 2 m s–1 through
a circular pipe. The pipe, which has an internal diameter of
75 mm, is at a uniform temperature of 27°C, and the ammonia at a section
1.2 m from the inlet to the pipe has a temperature of –23°C. Use the
following information to estimate the local heat transfer flux at l = 1.2 m.
Note, the properties of ammonia liquid have been taken at –23°C, except
where stated.
Liquid ammonia properties:
Density = 600 kg m–3
Specific heat capacity = 4.86 kJ kg–1 K–1
Dynamic viscosity (at 27°C) = 1.19 × 10–4 kg m–1 s–1
Dynamic viscosity = 2.05 × 10–4 kg m–1 s–1
Thermal conductivity = 5.11 × 10–4 kW m–1 s–1
Heat transfer correlations:

Nu=1.86Re^1/3 Pr^1/3 (d/l)^1/3 (μ/μw)^0.14 for laminar flow
Nu= 0.023Re^0.8Pr^0.14 for turbulent flow

Homework Equations

The Attempt at a Solution

Re =600 x 2 x 0.075/2.05 x10^-4 = 4.39 x 10^5 which is > 2000 so turbulent flow
Pr=2.05 x 10 ^-4 x 4.86 x 10^3 / 5.11 x 10 ^-4 x 10^3 = 1,949[/B]
Nu=0.023 (4,39 x 10^5)^0.8 x (1.949)^0.33 = 936.132

Nu=hd/k => h=936.132 x 5.11 x 10^-4 x 10 ^3 / 0.075 =6378.183 w/m^2k

Q=h x π χ d x l x (Ts - To) =>6378.183 x π χ 0.075 x 1.2 x (27 - (-23)) = 90.16 kw

is this correct?if not i would like your help,thank you

 

[haruspex]

Homework Statement

Liquid ammonia is heated as it flows at a mean velocity of 2 m s–1 through
a circular pipe. The pipe, which has an internal diameter of
75 mm, is at a uniform temperature of 27°C, and the ammonia at a section
1.2 m from the inlet to the pipe has a temperature of –23°C. Use the
following information to estimate the local heat transfer flux at l = 1.2 m.
Note, the properties of ammonia liquid have been taken at –23°C, except
where stated.
Liquid ammonia properties:
Density = 600 kg m–3
Specific heat capacity = 4.86 kJ kg–1 K–1
Dynamic viscosity (at 27°C) = 1.19 × 10–4 kg m–1 s–1
Dynamic viscosity = 2.05 × 10–4 kg m–1 s–1
Thermal conductivity = 5.11 × 10–4 kW m–1 s–1
Heat transfer correlations:

Nu=1.86Re^1/3 Pr^1/3 (d/l)^1/3 (μ/μw)^0.14 for laminar flow
Nu= 0.023Re^0.8Pr^0.14 for turbulent flow

Homework Equations

The Attempt at a Solution

Re =600 x 2 x 0.075/2.05 x10^-4 = 4.39 x 10^5 which is > 2000 so turbulent flow
Pr=2.05 x 10 ^-4 x 4.86 x 10^3 / 5.11 x 10 ^-4 x 10^3 = 1,949[/B]
Nu=0.023 (4,39 x 10^5)^0.8 x (1.949)^0.33 = 936.132

Nu=hd/k => h=936.132 x 5.11 x 10^-4 x 10 ^3 / 0.075 =6378.183 w/m^2k

Q=h x π χ d x l x (Ts - To) =>6378.183 x π χ 0.075 x 1.2 x (27 - (-23)) = 90.16 kw

is this correct?if not i would like your help,thank you

Your answer strikes me as much too high. I can't check it, though, because I am not an expert in this subject matter and you have not quoted the Relevant Equations you are depending on.
What is Pr? Working from the variables you seem to have used to calculate it (except the 1000 at the end... I assume that's some conversion between kilo and non-kilo units) it appears to have units of s/K, which I cannot connect with anything I know.

 

[electr]

Re is reynolds number and PR=Cpμ/κ is the Prandtl number...i thought this is a forum to help people not give exams,someone that knows about this can understand what i m doing and don't need to write the equation because i substitude with the values so its the same thing,i see many people don't write them ,if you don't know thank you for your time and hope someone that knows about it can answer if i m wrong ,i posted it here just to be 100 % sure

 

[haruspex]

Re is reynolds number and PR=Cpμ/κ is the Prandtl number...i thought this is a forum to help people not give exams,someone that knows about this can understand what i m doing and don't need to write the equation because i substitude with the values so its the same thing,i see many people don't write them ,if you don't know thank you for your time and hope someone that knows about it can answer if i m wrong ,i posted it here just to be 100 % sure

The "relevant equations" section of the template is there for a reason and should be filled in. It allows a far broader audience to offer assistance. I frequently respond to questions in areas of which I have no prior knowledge yet manage to resolve the issue. Mostly the student is not having a problem with the subject matter but with the application of equations and the handling of the algebra.

Thermal conductivity = 5.11 × 10–4 kW m–1 s–1

This is what threw me. I believe you mean kW m–1 K–1. That makes Pr dimensionless, as it should be.

Pr=2.05 x 10 ^-4 x 4.86 x 10^3 / 5.11 x 10 ^-4 x 10^3 = 1,949

Is that a decimal comma? It's about 1.9, right?

> 2000 so turbulent flow

Nu= 0.023Re^0.8Pr^0.14 for turbulent flow

Nu=0.023 (4,39 x 10^5)^0.8 x (1.949)^0.33

You wrote that ^(1/3) is for is laminar flow. But that only makes a small difference to the Nusselt number.

estimate the local heat transfer flux at l = 1.2 m

I note that it says "at", not along. I.e. it sounds like they are asking for the heat transfer rate per linear metre at the 1.2m mark. That means you would multiply 1m, not 1.2m, but again that is just a small difference. Besides, if they mean that there seems no point in specifying 1.2m as the position, so I could be wrong.
If it does mean along the first 1.2m then isn't there a problem with the calculation? The temperature difference would have been greater closer to the source.

So although I cannot find a significant flaw in your calculation, 90kW still seems huge to me.
If you get no other help, try pinging Chestermiller. He's very good on thermodynamics.

Edit: I calculated the temperature rise expected in the gas based on 90kW/m. I get 7K per m travelled. So I now accept that 90kW might be reasonable.

 

[Chestermiller]

It all looks OK to me, except that the exponent in the Sieder Tate equation should be 0.4, and, the last equation is incorrect. The local heat flux should just be $q = h\Delta T$

 

[electr]

thank you for your replies,chester what you mean by 0.4 ,i didnt understand it sorry,and for the Q only be Q=6378.183 x (27 - (-23)) =31890915 ??

 

[haruspex]

for the Q only be Q=6378.183 x (27 - (-23)) =31890915 ??

There does not seem to be universal agreement on the terminology. I see some authors use heat flux for the total rate of heat flow (over whatever area it happens to be) and heat flux density for the flow rate per unit area. But most seem to agree with Chet that heat flux is per unit area.
I would say that latter is how it is being used in this question, since we do not know the temperature of the gas along the first 1.2m of pipe.

 

[Chestermiller]

thank you for your replies,chester what you mean by 0.4 ,i didnt understand it sorry,and for the Q only be Q=6378.183 x (27 - (-23)) =31890915 ??

Nu= 0.023Re^0.8Pr^0.4 for turbulent flow, not Nu= 0.023Re^0.8Pr^0.14 or Nu= 0.023Re^0.8Pr^0.33

When they ask for the local heat flux in a heat transfer context, they mean the local rate of heat flow per unit area. That is just $q=h\Delta T$, not the overall cumulative rate of heat flow over the entire length of pipe.

 

[electr]

now i see ,sorry my bad, write it wrong... your correct about the Nu= 0.023Re^0.8Pr^0.33 (when i was writing the question saw the laminar value...),so the 90 kw is not the correct one and all i need to do is the q=hΔT for my final answer? and something else chest,i have one more question but some guys already posted it here and i don't know if i have to post it again,because i did tha mistake before,but i would like to post my answer for you to check

 

[electr]

and thank you too haruspex for taking the time to help me

 

[Chestermiller]

now i see ,sorry my bad, write it wrong... your correct about the Nu= 0.023Re^0.8Pr^0.33 (when i was writing the question saw the laminar value...),so the 90 kw is not the correct one and all i need to do is the q=hΔT for my final answer? and something else chest,i have one more question but some guys already posted it here and i don't know if i have to post it again,because i did tha mistake before,but i would like to post my answer for you to check

Bird, Stewart, and Lightfoot, Transport Phenomena, give the following equation for Re > 20000:
$$Nu=0.026 Re^{0.8}Pr^{1/3}\left(\frac{\mu _B}{\mu _0}\right)^{0.14}$$where $\mu_B$ is the bulk viscosity and $\mu_0$ is the wall viscosity.

 

[electr]

so chest is my solution ok?and should i post the other question too?like i said some other guys already posted it here but still want u to check it

 

[electr]

its just my last questions to finish the HNC that's why i m a m a bit impatient to finish

 

[Chestermiller]

so chest is my solution ok?and should i post the other question too?like i said some other guys already posted it here but still want u to check it

Do whoever pleases you.

 

[electr]

ok,i posted it if you have time take a look,and for this question am i correct or should i give it a try again?

 

[Chestermiller]

ok,i posted it if you have time take a look,and for this question am i correct or should i give it a try again?

Where is it posted?

 

[electr]

https://www.physicsforums.com/threads/heat-transfer.913709/#post-5756268

this is the link