Heat Transfer Pipe Problem dilemma

[williamcarter]

Homework Statement

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

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

b)

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)?

 

[benny_91]

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.

 

[williamcarter]

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

 

[benny_91]

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

 

[williamcarter]

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?

 

[benny_91]

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.

 

[williamcarter]

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?

 

[benny_91]

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.

 

[Chestermiller]

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)?

 

[williamcarter]

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

 

[Chestermiller]

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?

 

[williamcarter]

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.

 

[williamcarter]

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?

 

[Chestermiller]

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.

 

[williamcarter]

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.

 

[Chestermiller]

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?

 

[williamcarter]

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

 

[Chestermiller]

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

 

[williamcarter]

Don't forget that minus sign in the heat conduction equation. q = - k dT/dx

Thank you, understood now.

 

[williamcarter]

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.

 

[Chestermiller]

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.

 

[williamcarter]

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!