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

In summary: If you had calculated ##\dot Q##, I would have said that was correct. But you are still overlooking the fact that you are not given a specified area, only a point along the pipe. So you have no basis for calculating a heat transfer rate, and no basis for multiplying by A, even if you could calculate it.In summary, the conversation discusses a problem related to heat transfer in a circular pipe with liquid ammonia flowing through it. The feedback states that the heat transfer rate is wrong and should be calculated as heat transfer per unit area. The properties of liquid ammonia are given, along with heat transfer correlations for laminar and turbulent flow. The question asks for the local heat transfer flux at a
  • #1
Tiberious
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 turbulentNu=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
 
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  • #2
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
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
Tiberious said:
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
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
Sorry remove the A

Φ=h∙(T_s-T_f )
 
  • #7
Tiberious said:
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
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
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 turbulentNu=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
Tiberious said:
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.)

Tiberious said:
So would I be correct with the below
...
Answer:
=318.975 kW
Yes, that looks reasonable, but somewhat overprecise.
 
  • #11
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
Tiberious said:
all the examples are using Area
Then I would guess those examples calculate heat transfer rate, not heat flux.
Tiberious said:
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.
 
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  • #13
Tiberious said:
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
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 ?
 

1. What is the purpose of heating liquid ammonia as it flows at a mean velocity of 2?

The purpose of heating liquid ammonia as it flows at a mean velocity of 2 is to increase its temperature and convert it into a gas form. This process is known as vaporization and is important in various industrial and scientific applications.

2. How does the mean velocity affect the heating process of liquid ammonia?

The mean velocity plays a crucial role in the heating process of liquid ammonia. As the velocity increases, the amount of heat transferred to the liquid also increases, resulting in a faster vaporization process.

3. What is the ideal temperature for heating liquid ammonia?

The ideal temperature for heating liquid ammonia depends on the specific application. In general, the temperature should be high enough to vaporize the liquid, but not too high to cause excessive pressure buildup. It is typically between -33.4°C and -78.5°C.

4. What safety precautions should be taken when heating liquid ammonia?

When handling and heating liquid ammonia, it is essential to wear appropriate protective gear, including gloves, goggles, and a lab coat. The heating process should also be done in a well-ventilated area to prevent the buildup of ammonia gas, which can be harmful when inhaled.

5. Can liquid ammonia be heated without flowing at a mean velocity of 2?

Yes, liquid ammonia can be heated without flowing at a mean velocity of 2. However, the heating process may take longer due to the slower transfer of heat to the liquid. The mean velocity of 2 is a commonly used value that helps optimize the heating process for efficient vaporization.

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