Heat Transfer Pipe Problem dilemma

In summary: Heat is transferred firstly by conduction within the pipe , then by convection from pipe surface to the air.To answer the questions I would say , yes Qtotal passes through tube wall, yes Qtotal passes through boundary layer.
  • #1
williamcarter
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4

Homework Statement


Problem data:[/B]
http://imgur.com/a/hEz16
caseb1.JPG


Homework Equations


Rconduction=ln(rout/rin)/2*pi*Length*thermal conduct
Rconvection=1/2*pi*Length*routside*convective heat transfer outside

=>Qconduct=2*pi*Length*thermal conduct*(Tin-Tout)/ln(rout/rin)
=>Qconv=2*pi*Length*routside*convective heat transfer outside*(Tout-Tair)

The Attempt at a Solution


Solution:
a)[/B]
http://imgur.com/a/cH9hV
QHT2.JPG


b)
caseb.JPG


For a)
I do not understand why we need to find Toutside by equating Qcond=Qconv?
And why we need to calculate Q from the pipe just as Qconvection instead of Q total?
Why we cannot do Q=delta T/Resistance Total
where:
Rtotal=Rconduction+Rconvection,because we do not need temperatures in those 2 R formulas
and delta T=(Tin-Tair) specifficaly 300 and 5 degrees celsius

As for b) They do Rtotal and then Qtotal=delta T/Rtotal
and then they do Qtotal=Qconv and they get T2.
Why we cannot proceed the same for a)?
 
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  • #2
Actually you can use that method (adding up individual resistances) and you will get the same result. I guess the question requires you to find the temperature of the outer surface of the pipe hence employ this method.
 
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  • #3
benny_91 said:
Actually you can use that method (adding up individual resistances) and you will get the same result. I guess the question requires you to find the temperature of the outer surface of the pipe hence employ this method.

Does this mean that Qtotal=Qconvection? We neglect Qconduction ?
They were asking for Q emerging from pipe
If I do like them as Q from pipe =Qconvection=75760 W
if I do like Q=delta T/Rtotal , where Rtotal=Rconduction+Rconvection it gives me Q=75758 W
 
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  • #4
williamcarter said:
Does this mean that Qtotal=Qconvection? We neglect Qconduction ?
They were asking for Q emerging from pipe
You are missing an important point here. Note that the heat that is conducted through the pipe is then convected from the surface of the pipe to the atmosphere. Heat transfer passes through different stages: first conduction and then convection.
As long as there is no internal heat generation in the heat transfer medium and steady state in maintained (the pipe in this case) Qtotal=Qconduction=Qconvection
 
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  • #5
benny_91 said:
You are missing an important point here. Note that the heat that is conducted through the pipe is then convected from the surface of the pipe to the atmosphere. Heat transfer passes through different stages: first conduction and then convection.
As long as there is no internal heat generation in the heat transfer medium and steady state in maintained (the pipe in this case) Qtotal=Qconduction=Qconvection

Yes I agree and I understood it is conduction+convection , it is obvious but I do not understand why at a) they did Q as it was just convection?and at b) they did Qtotal and equated Qtotal=Qconvection to get Temp2?
 
  • #6
williamcarter said:
Yes I agree and I understood it is conduction+convection , it is obvious but I do not understand why at a) they did Q as it was just convection?and at b) they did Qtotal and equated Qtotal=Qconvection to get Temp2?
If by T2 you mean outer surface temperature of the pipe then you can either use Qtotal=Qconvection or Qtotal=Qconduction. You will get the same value of T2.
 
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  • #7
benny_91 said:
If by T2 you mean outer surface temperature of the pipe then you can either use Qtotal=Qconvection or Qtotal=Qconduction. You will get the same value of T2.

Thank you for your quick answer, for a) they did Q from pipe as Qconvection and they got 75760 W .If I do Qtotal=delta T/Rtotal
where delta T=(300-5) and Rtotal =Rcond+Rconv . I should get same answer right?
Why were they searching for Toutside and T2 at surface of pipe?If they were not asking for them?
 
  • #8
williamcarter said:
Thank you for your quick answer, for a) they did Q from pipe as Qconvection and they got 75760 W .If I do Qtotal=delta T/Rtotal
where delta T=(300-5) and Rtotal =Rcond+Rconv . I should get same answer right?
Why were they searching for Toutside and T2 at surface of pipe?If they were not asking for them?
Absolutely. If 300 is the temperature of the inner surface of the pipe and 5 is the temperature of the outer convecting fluid you should get the same answer.
 
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  • #9
williamcarter said:
I do not understand why we need to find Toutside by equating Qcond=Qconv?
And why we need to calculate Q from the pipe just as Qconvection instead of Q total?
Does all the heat flow (Q total) pass through the tube wall?
Does all the heat flow then pass through the convective boundary layer?
So, is Qcond = Qconv = Q total (or not)?
 
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  • #10
Chestermiller said:
Does all the heat flow (Q total) pass through the tube wall?
Does all the heat flow then pass through the convective boundary layer?
So, is Qcond = Qconv = Q total (or not)?
Heat is transferred firstly by conduction within the pipe , then by convection from pipe surface to the air.
To answer the questions I would say , yes Qtotal passes through tube wall, yes Qtotal passes through boundary layer.
There is no insulation at the moment, furthermore no generation,no accumulation so I would say Qtotal=Qconduction=Qconvection
 
  • #11
williamcarter said:
Heat is transferred firstly by conduction within the pipe , then by convection from pipe surface to the air.
To answer the questions I would say , yes Qtotal passes through tube wall, yes Qtotal passes through boundary layer.
There is no insulation at the moment, furthermore no generation,no accumulation so I would say Qtotal=Qconduction=Qconvection
Good. So in part (a), for conduction through the pipe, we have:
$$Q=2\pi kL\frac{(300-T_0)}{\ln{(r_{out}/r_{in})}}$$
and for conduction through the convective boundary layer, we have:
$$Q=2\pi r_{out}hL(T_0-5)$$where ##T_0## is the temperature at the outer surface of the pipe. If we solve for the temperature differences, we obtain:
$$(300-T_0)=\frac{\ln{(r_{out}/r_{in})}}{2\pi kL}Q$$
$$(T_0-5)=\frac{1}{2\pi r_{out}hL}Q$$
So, you have two equations in the two unknowns, ##T_0## and Q. If you add these two equations together, you can eliminate ##T_0## to obtain:
$$(300-5)=\left[\frac{\ln{(r_{out}/r_{in})}}{2\pi kL}+\frac{1}{2\pi r_{out}hL}\right]Q$$
You can then solve this equation for Q and substitute back into either of the other equations to get ##T_0##.

Now, when you add the insulation layer in part (b), you end up with three equations in three unknowns, Q and the two unknown temperatures at the two interfaces. You set up and solve these three equations in basically the same way as part (a). Questions?
 
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  • #12
Chestermiller said:
Good. So in part (a), for conduction through the pipe, we have:
$$Q=2\pi kL\frac{(300-T_0)}{\ln{(r_{out}/r_{in})}}$$
and for conduction through the convective boundary layer, we have:
$$Q=2\pi r_{out}hL(T_0-5)$$where ##T_0## is the temperature at the outer surface of the pipe. If we solve for the temperature differences, we obtain:
$$(300-T_0)=\frac{\ln{(r_{out}/r_{in})}}{2\pi kL}Q$$
$$(T_0-5)=\frac{1}{2\pi r_{out}hL}Q$$
So, you have two equations in the two unknowns, ##T_0## and Q. If you add these two equations together, you can eliminate ##T_0## to obtain:
$$(300-5)=\left[\frac{\ln{(r_{out}/r_{in})}}{2\pi kL}+\frac{1}{2\pi r_{out}hL}\right]Q$$
You can then solve this equation for Q and substitute back into either of the other equations to get ##T_0##.

Now, when you add the insulation layer in part (b), you end up with three equations in three unknowns, Q and the two unknown temperatures at the two interfaces. You set up and solve these three equations in basically the same way as part (a). Questions?
Everything clear, thanks.
 
  • #13
Chestermiller said:
Good. So in part (a), for conduction through the pipe, we have:
$$Q=2\pi kL\frac{(300-T_0)}{\ln{(r_{out}/r_{in})}}$$
and for conduction through the convective boundary layer, we have:
$$Q=2\pi r_{out}hL(T_0-5)$$where ##T_0## is the temperature at the outer surface of the pipe. If we solve for the temperature differences, we obtain:
$$(300-T_0)=\frac{\ln{(r_{out}/r_{in})}}{2\pi kL}Q$$
$$(T_0-5)=\frac{1}{2\pi r_{out}hL}Q$$
So, you have two equations in the two unknowns, ##T_0## and Q. If you add these two equations together, you can eliminate ##T_0## to obtain:
$$(300-5)=\left[\frac{\ln{(r_{out}/r_{in})}}{2\pi kL}+\frac{1}{2\pi r_{out}hL}\right]Q$$
You can then solve this equation for Q and substitute back into either of the other equations to get ##T_0##.

Now, when you add the insulation layer in part (b), you end up with three equations in three unknowns, Q and the two unknown temperatures at the two interfaces. You set up and solve these three equations in basically the same way as part (a). Questions?

I understood everything, but I still have a question.
We know Q=m*cp*ΔT=m*cp*(Tout-Tin)
Why we are doing ΔT=Tin-Tout?
 
  • #14
williamcarter said:
I understood everything, but I still have a question.
We know Q=m*cp*ΔT=m*cp*(Tout-Tin)
Why we are doing ΔT=Tin-Tout?
This question does not relate to the present problem. It relates to a flow problem.
 
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  • #15
Chestermiller said:
This question does not relate to the present problem. It relates to a flow problem.
Thank you for your quick answer.

In general we know Q=m*cp*ΔT=m*cp*(Tout-Tin)
but in this problem, in their answer they are doing ΔT=(Tin-Tout)
I am confused why they did like this in their answer on this exercise.
 
  • #16
williamcarter said:
Thank you for your quick answer.

In general we know Q=m*cp*ΔT=m*cp*(Tout-Tin)
but in this problem, in their answer they are doing ΔT=(Tin-Tout)
I am confused why they did like this in their answer on this exercise.
Do you see any Cp in the exercise?
 
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  • #17
Chestermiller said:
Do you see any Cp in the exercise?
No , I don't.
I cannot understand why they did in this problem ΔT=Tin-Tout=300-5.Why they did like that?
Usually we know ΔT=Tout-Tin=Tfinal-Tinitial not Tinitial-Tfinal
 
  • #18
williamcarter said:
No , I don't.
I cannot understand why they did in this problem ΔT=Tin-Tout=300-5.Why they did like that?
Usually we know ΔT=Tout-Tin=Tfinal-Tinitial not Tinitial-Tfinal
Don't forget that minus sign in the heat conduction equation. q = - k dT/dx
 
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  • #19
Chestermiller said:
Don't forget that minus sign in the heat conduction equation. q = - k dT/dx
Thank you, understood now.
 
  • #20
Chestermiller said:
Don't forget that minus sign in the heat conduction equation. q = - k dT/dx
But also in convection they did To-Tair,not Tair-To.
I always thought we take ΔT=Tout-Tin not Tin -Tout, this is what confused me.
 
  • #21
williamcarter said:
But also in convection they did To-Tair,not Tair-To.
I always thought we take ΔT=Tout-Tin not Tin -Tout, this is what confused me.
Good point. I guess convection is just one of those instances when you have to use your judgment and know that the heat flows from hot to cold. However, in the case of conduction, no judgment is required.
 
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  • #22
Chestermiller said:
Good point. I guess convection is just one of those instances when you have to use your judgment and know that the heat flows from hot to cold. However, in the case of conduction, no judgment is required.
Thank you!
 

1. What is the main cause of heat transfer pipe problems?

The main cause of heat transfer pipe problems is usually due to poor design or incorrect installation. This can result in inadequate insulation, improper sizing, or insufficient flow rate, leading to issues such as heat loss, pipe corrosion, and reduced efficiency.

2. How can heat transfer pipe problems be prevented?

To prevent heat transfer pipe problems, it is important to carefully plan and design the system before installation. This includes choosing the right pipe material, insulation, and ensuring proper sizing and flow rates. Regular maintenance and inspections are also crucial to identify and address any potential issues before they become major problems.

3. What are the common signs of heat transfer pipe problems?

Some common signs of heat transfer pipe problems include uneven heating or cooling of spaces, higher than usual energy bills, water leaks, and unusual noises from the pipes. It is important to address these signs promptly to avoid further damage to the system.

4. How can heat transfer pipe problems be diagnosed?

Diagnosing heat transfer pipe problems often involves conducting a thorough inspection of the system. This may include checking for leaks, measuring flow rates and temperatures, and examining the insulation. In some cases, thermal imaging or other diagnostic tools may be used to identify any issues.

5. What are the most effective solutions for heat transfer pipe problems?

The most effective solution for heat transfer pipe problems will depend on the specific issue at hand. In some cases, simple repairs or adjustments may be sufficient, while in others, a complete replacement of the pipes or insulation may be necessary. It is important to consult with a professional to determine the best course of action for your specific situation.

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