Liquid ammonia is heated as it flows at a mean velocity of 2

  • Thread starter Tiberious
  • Start date
  • #1
73
3

Homework Statement



I've completed this problem and received the below feedback but have gone to a state of complete loss on what to do to amend. Can anyone provide a prompt ?

Heat transfer coefficient is correct. But heat transfer rate is wrong - the question says Heat transfer per UNIT AREA, so why would you multiply your heat flux by AREA??

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:
ρ Density=600 kg m^(-3)
C_p Specific heat capacity=4.86 kJ kg^(-1) K^(-1)
μ Dynamic Viscosity (at 27°C)=1.19∙10^(-4 ) kg m^(-1) s^(-1)
μ Dynamic Viscosity= 2.05∙10^(-4 ) kg m^(-1) s^(-1)
k Thermal conductivity= 5.11∙10^(-4 ) kW m^(-1) s^(-1)

The Attempt at a Solution



Heat transfer correlations:
N_u=1.86 〖R_e〗^(1/3) 〖P_r〗^(1/3) 〖(d/l)^(1/3) (μ/μ_w )〗^0.14for laminar flow

N_u=0.023 〖R_e〗^0.8 〖P_r〗^0.33 for turbulent flow.

Determining the Reynolds number,

Re=(ρ∙u∙L)/μ=(ρ∙u∙D)/μ=(600∙2∙0.075)/(2.05∙〖10〗^(-4) )=439024.4

Determining the Prandtl number,

Pr=(C_p∙μ)/k=(4.86∙〖10〗^3∙2.05∙〖10〗^(-4))/(5.11∙〖10〗^(-4)∙〖10〗^3 )=1.95

Re > 2600, so flow is turbulent


Nu=0.023∙〖Re〗^0.8∙〖Pr〗^0.33=0.023∙〖439024.4〗^0.8∙〖1.95〗^0.33=936.3

h=Nu∙k/d

=936.3∙(5.11∙〖10〗^(-4)∙〖10〗^3)/0.075

=6379.5 W/(m^2∙K)

Φ=h∙A∙(T_s-T_f )

=h∙π∙D∙L∙(T_s-T_f )

=6379.5∙π∙0.075∙1.2∙(27-(-23))

Answer:

=90188.W

=90.2 kW
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,558
6,442
It is not clear to me which parts are your attempt and which the feedback you received.
As you say, the answer should be in terms of power per unit area, not Watts or kW.
 
  • #3
73
3
Apologies, I should have made this more clear. This was the feedback. "Heat transfer coefficient is correct. But heat transfer rate is wrong - the question says Heat transfer per UNIT AREA, so why would you multiply your heat flux by AREA??"

So, Φ=h∙A∙(T_s-T_f ) is this the element that I need to adjust ?
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,558
6,442
So, Φ=h∙A∙(T_s-T_f ) is this the element that I need to adjust ?
Where are you getting that equation from? How is φ defined in that context?
There can be confusion in some disciplines (e.g. electromagnetism) between flux density and total flux, but in heat transfer I believe "heat flux" is synonymous with "heat flux density", so the factor A should not be there.
E.g. https://en.m.wikipedia.org/wiki/Heat_flux
 
  • #5
73
3
Would this be the equation that I would need to apply then?

Φ=h∙A (T_s-T_f ) which would equate too 6379.5 * (27-(-23))
 
  • #6
73
3
Sorry remove the A

Φ=h∙(T_s-T_f )
 
  • #7
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,558
6,442
Sorry remove the A

Φ=h∙(T_s-T_f )
Yes. But you have not explained why you had the A in there. Where did that come from?
 
  • #8
20,869
4,546
Yes. I agree with the criticism. They are asking for the local heat flux to the fluid at 1.2 m. Even if you wanted to total rate of heat flow, you couldn't get it by multiplying by A, since the temperature driving force and heat flux are changing along the tube.
 
  • #9
73
3
This seems to be some form of mis understanding. I've been following some examples provided and most of them are having area as a factor in their equation. Honestly, I'm lost with this one.

The below equation is given repeatedly.

The heat transfer rate is Φ = hA (Ts – Tf)

= hπdL(Ts – Tf)

With respect to the OP the final answer is coming out as 318975W.



So would I be correct with the below.

Determining the Reynolds number,

Re=(ρ∙u∙L)/μ=(ρ∙u∙D)/μ=(600∙2∙0.075)/(2.05∙〖10〗^(-4) )=439024.4

Determining the Prandtl number,

Pr=(C_p∙μ)/k=(4.86∙〖10〗^3∙2.05∙〖10〗^(-4))/(5.11∙〖10〗^(-4)∙〖10〗^3 )=1.95

Re > 2600, so flow is turbulent


Nu=0.023∙〖Re〗^0.8∙〖Pr〗^0.33=0.023∙〖439024.4〗^0.8∙〖1.95〗^0.33=936.3

h=Nu∙k/d

=936.3∙(5.11∙〖10〗^(-4)∙〖10〗^3)/0.075

=6379.5 W/(m^2∙K)

Φ=h∙(T_s-T_f ) *

=h∙π (T_s-T_f )

=6379.5∙(27-(-23))

Answer:

=318.975 kW
 
  • #10
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,558
6,442
The heat transfer rate is Φ = hA (Ts – Tf)
I just scanned through half a dozen hits on the net, and nobody uses φ for heat transfer rate. They all use ##\dot Q## or ##\frac{dQ}{dt}## or ##\frac{\Delta Q}{\Delta t}##. The letter φ should be reserved for flux, and heat flux is defined as heat transfer rate per unit area:
##\phi=h(T_s-T_f)##, or ##\frac{k\Delta\theta}x## etc. So ##\dot Q=\phi A##.
The question asks for heat flux, which will have units of W/m2, not heat transfer rate, which is what you originally calculated.
(And as Chester points out, you are not given a specified area, only a point along the pipe, so you have no basis for calculating a heat transfer rate.)

So would I be correct with the below
....
Answer:
=318.975 kW
Yes, that looks reasonable, but somewhat overprecise.
 
  • #11
73
3
Certainly confused on this one, all the examples are using Area. Although they do stipulate calculating the 'Natural convection heat transfer coefficient h'.

Either way, overprecise? would you go with 318975 W (I assume W is correct opposed to W/m2)
 
  • #12
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,558
6,442
all the examples are using Area
Then I would guess those examples calculate heat transfer rate, not heat flux.
overprecise?
The temperatures you are given are only to two significant figures. Technically you should answer nothing more exact than that, so 3.2x102kW, but you should not be marked down for answering 319kW.
 
  • Like
Likes Tiberious
  • #13
20,869
4,546
Certainly confused on this one, all the examples are using Area. Although they do stipulate calculating the 'Natural convection heat transfer coefficient h'.

Either way, overprecise? would you go with 318975 W (I assume W is correct opposed to W/m2)
Do you know the difference between the average heat transfer coefficient and the local heat transfer coefficient?
 
  • #14
73
3
As far as what I can tell, the average heat transfer coefficient is equal to / average temperature and the average heat transfer of the surface. Whereas, the local heat flux is at a given point. Anywhere is the correct ball park ?
 

Related Threads on Liquid ammonia is heated as it flows at a mean velocity of 2

  • Last Post
Replies
7
Views
6K
  • Last Post
Replies
6
Views
3K
Replies
3
Views
590
  • Last Post
Replies
0
Views
1K
Replies
4
Views
13K
  • Last Post
Replies
3
Views
764
Replies
0
Views
936
Replies
1
Views
2K
Top