Heat transfer and Fluid Flow Struggling Problems

In summary: You need to consider the density of the gasoline as well as the flow rate, and the area of the siphon.
  • #1
williamcarter
153
4

Homework Statement


Hello, I am struggling with this heat transfer and fluid flow problem.I would really appreciate if someone could have a look over what I have done on it.
I am not sure if what I have done is correct ,I would really appreciate if you could tell me if it is correct or wrong.And also I would really appreciate if you could correct what I did wrong.
Thank you in advance.

Question 2) image: http://imgur.com/aJCzNyB
Question 3) image http://imgur.com/wFl93SA
Table image: http://i.stack.imgur.com/aiO1J.jpg

Homework Equations


Q2.Data:
a)[/B]
layer of snow=x=20*10^-2 m
thermal conductivity(lambda)=0.3W/m*K
Tair(inside)=18 Degrees Celsius
Tinterface=0 Degrees Celsius
Roof area=9m^2
Polycarbonate sheet x'=25*10^-3 m
latent heat of ice=333.5 kj/kg

b)
A=100cm^2
Pin=100KPa
Pout=60KPa
u=15m/s

Q3 Data.
a)

L=30m
D=6*10^(-2)m
composite material x=2*10^(-2)m
lambda(thermal conduct)=0.4W/m*k
Ti (inside temp of insulation)=130 degrees celsius
Tair=10 degrees celsius
h=20W/m^2*k(convective heat transfer coefficient)

b)
D=10^-3m
Specific gravity=0.8=>Petrol ro(density)=800kg/m^3
Pvap=60KPa abs=>Pvap=101.3-60=41.3KPa gauge

The Attempt at a Solution


Q2 Solved
i)[/B]Q=delta T/Rtotal
delta T=(18-0)
Rtotal=1/UA=(1/hi+xsnow/lambda snow)*1/A=(1/25+0.2/03)*1/9=0.07851
=>Q=(18-0)/0.07851
=>Q=229.24W

ii)Q=? if ice melts at 1.8 kg/h
m=1.8kg/h=1.8kg/3600s=5*10^-4 kg/s
Q=m*hmelting (phase change)
Q=5*10^-4*333.5
Q=0.16675 Kw

iii)h=6W/m^2*k (convective heat transfer coefficient)
Toutside air=?
We know Qconv=hA*delta T or q=h*Delta T
Struggle on this one.
I stopped here , I am confused how to solve this one.

b)U bend pipe, force to keep in place
A=ct=100cm^2=0.01m^2
Pin=100Kpa
Pout=60Kpa
u=15m/s

F=?to hold pipe in place
Fm(momentum force)=m*u=ro*q*u=ro*A*u^2
Fmx=m*u=m*(u2-u1) if done on fluid
Fmx=m*u=m*(u1-u2) if done by fluid
Fmx=ro*A*u^2=10^3*0.01*15^2
Fmx=2250N(in X direction)
Fmy= - 2250N n(in Y direction)

Fpx=P1A1-P2A2
FPx=10^5Pa*10^-2 m^2
FPx=10^3 N(in X direction)
Fpy=-10^3 N(in Y direction)

Fx=Fmx+FPx=3250N
Fy=Fmy+Fpy=-3250N

Magnitude of force =sqrt(Fx^2+Fy^2)=sqrt(3250^2+3250^2)=4956N
Force to keep pipe in place=reaction force equal in magnitude but opposite in direction
=> Force to keep pipe in place Fr= -4956 N

Q3.solved
a)i)
q(heat flux)=?
q=U*delta T
1/U=1/h+x/lambda
=>1/U=1/20+2*10^-2/0.4=>1/U=0.1=>U=10
q=U*delta T=10*(130-10)
q=1200 W/m^2*k

ii)if x'=2*x
=>q'=U' *delta T
1/U'=1/h+2x/lambda=1/20+4*10^-2/0.4=>U=6.6
=>q=800W/m^2

iii)increasing insulation layer is also increasing safety level by reducing q(heat flux)

b)
Vol=2L Petrol=2*10^-3 m^3
D=10^-3m
=>A=7.85*10^-5 m^2
Specific gravity=0.8
=>Petrol ro(density)=800kg/m^3
Pvap=60KPa abs
=>Pvap=101.3-60=41.3KPa gauge
We apply Bernoulli equation P1/ro*g+h1+u1^2/2*g=P2/ro*g+h2+u2^2/2*g
Taking points 1 and 2 as below
http://imgur.com/cDR2v51
Point 1 on the bottom of the tank
Point 2 discharged in atmosphere from siphon

time t=Volume/Volumetric flowrate
for point 1:
u1=0(we can neglect velocity due to difference in Area between point1 and 2)
h1=1m
P1=41.3*10^3Pa

for point2:
u2=?we will find it
P2=0(due to atmosphere)
h2=0(datum point)

Plug all those values in main Bernoulli equation
=> u2^2=(P1/ro*g+h1)*2g
u2=sqrt( (P1/ro*g +h1)*2g
u2=sqrt( (41.3*10^3/800*9.81+1)*2*9.81
=>u2=11.08 m/s
we know q(volumetric flowrate)= u*A
=>q=u2*A2(area of siphon)
q=11.08 m/s*7.85*10^-5 m^2
q=8.70*10^-4 m^3/s

but time t=V/q
t=2*10^-3 m^3/8.70*10^-4 m^3/s
=>time=2.29 seconds

ii) find maximum allowable h in order to maintain flow.
I struggle here.
 
Last edited:
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  • #2
williamcarter said:

Homework Statement


Hello, I am struggling with this heat transfer and fluid flow problem.I would really appreciate if someone could have a look over what I have done on it.
I am not sure if what I have done is correct ,I would really appreciate if you could tell me if it is correct or wrong.And also I would really appreciate if you could correct what I did wrong.
Thank you in advance.

Question 2) image: http://imgur.com/aJCzNyB
Question 3) image http://imgur.com/wFl93SA
Table image: http://i.stack.imgur.com/aiO1J.jpg

Homework Equations


Q2.Data:
a)[/B]
layer of snow=x=20*10^-2 m
thermal conductivity(lambda)=0.3W/m*K
Tair(inside)=18 Degrees Celsius
Tinterface=0 Degrees Celsius
Roof area=9m^2
Polycarbonate sheet x'=25*10^-3 m
latent heat of ice=333.5 kj/kg

b)
A=100cm^2
Pin=100KPa
Pout=60KPa
u=15m/s

Q3 Data.
a)

L=30m
D=6*10^(-2)m
composite material x=2*10^(-2)m
lambda(thermal conduct)=0.4W/m*k
Ti (inside temp of insulation)=130 degrees celsius
Tair=10 degrees celsius
h=20W/m^2*k(convective heat transfer coefficient)

b)
D=10^-3m
Specific gravity=0.8=>Petrol ro(density)=800kg/m^3
Pvap=60KPa abs=>Pvap=101.3-60=41.3KPa gauge

The Attempt at a Solution


Q2 Solved
i)[/B]Q=delta T/Rtotal
delta T=(18-0)
Rtotal=1/UA=(1/hi+xsnow/lambda snow)*1/A=(1/25+0.2/03)*1/9=0.07851
=>Q=(18-0)/0.07851
=>Q=229.24W

ii)Q=? if ice melts at 1.8 kg/h
m=1.8kg/h=1.8kg/3600s=5*10^-4 kg/s
Q=m*hmelting (phase change)
Q=5*10^-4*333.5
Q=0.16675 Kw

iii)h=6W/m^2*k (convective heat transfer coefficient)
Toutside air=?
We know Qconv=hA*delta T or q=h*Delta T
Struggle on this one.
I stopped here , I am confused how to solve this one.
You showed that the heat flow through the roof layer is 229 W and that 167 W of this is used to bring about the phase change at the interface between the snow and the roof. So you are left with 62 W for the rate of heat flow from 0 C at the snow interface with the roof to the outside air. What outside air temperature is required to drive this heat flow?
b)U bend pipe, force to keep in place
A=ct=100cm^2=0.01m^2
Pin=100Kpa
Pout=60Kpa
u=15m/s

F=?to hold pipe in place
Fm(momentum force)=m*u=ro*q*u=ro*A*u^2
Fmx=m*u=m*(u2-u1) if done on fluid
Fmx=m*u=m*(u1-u2) if done by fluid
Fmx=ro*A*u^2=10^3*0.01*15^2
Fmx=2250N(in X direction)
Fmy= - 2250N n(in Y direction)

Fpx=P1A1-P2A2
FPx=10^5Pa*10^-2 m^2
FPx=10^3 N(in X direction)
Fpy=-10^3 N(in Y direction)

Fx=Fmx+FPx=3250N
Fy=Fmy+Fpy=-3250N

Magnitude of force =sqrt(Fx^2+Fy^2)=sqrt(3250^2+3250^2)=4956N
Force to keep pipe in place=reaction force equal in magnitude but opposite in direction
=> Force to keep pipe in place Fr= -4956 N
You kinda had the right idea here, but not quite. If you regard the total pipe bend as your control volume, there is no component of force in the y direction. For the x direction, if F is the force exerted by the bend on the fluid in the negative x direction, then $$P_{in}A-P_{out}A-F=-2\rho A v^2$$ where the right hand side is the rate of change in fluid momentum in the positive x direction.
Q3.solved
a)i)
q(heat flux)=?
q=U*delta T
1/U=1/h+x/lambda
=>1/U=1/20+2*10^-2/0.4=>1/U=0.1=>U=10
q=U*delta T=10*(130-10)
q=1200 W/m^2*k

ii)if x'=2*x
=>q'=U' *delta T
1/U'=1/h+2x/lambda=1/20+4*10^-2/0.4=>U=6.6
=>q=800W/m^2

iii)increasing insulation layer is also increasing safety level by reducing q(heat flux)
This problem is not done correctly because you haven't taken into account the cylindrical geometry. Within the insulation, the heat conduction equation should be $$q=-k\frac{dT}{dr}$$, and the heat balance equation should be $$\frac{d(qr)}{dr}=0$$ or equivalently$$qr=const$$ These equations need to be solved simultaneously to get the temperature profile and the heat flux. Then, this solution has to be properly merged with the convective heat transfer.

In part iii, you should be comparing the ratio of the conductive resistance to the convective resistance.
b)
Vol=2L Petrol=2*10^-3 m^3
D=10^-3m
=>A=7.85*10^-5 m^2
Specific gravity=0.8
=>Petrol ro(density)=800kg/m^3
Pvap=60KPa abs
=>Pvap=101.3-60=41.3KPa gauge
We apply Bernoulli equation P1/ro*g+h1+u1^2/2*g=P2/ro*g+h2+u2^2/2*g
Taking points 1 and 2 as below
http://imgur.com/cDR2v51
Point 1 on the bottom of the tank
Point 2 discharged in atmosphere from siphon

time t=Volume/Volumetric flowrate
for point 1:
u1=0(we can neglect velocity due to difference in Area between point1 and 2)
h1=1m
P1=41.3*10^3Pa

for point2:
u2=?we will find it
P2=0(due to atmosphere)
h2=0(datum point)

Plug all those values in main Bernoulli equation
=> u2^2=(P1/ro*g+h1)*2g
u2=sqrt( (P1/ro*g +h1)*2g
u2=sqrt( (41.3*10^3/800*9.81+1)*2*9.81
=>u2=11.08 m/s
we know q(volumetric flowrate)= u*A
=>q=u2*A2(area of siphon)
q=11.08 m/s*7.85*10^-5 m^2
q=8.70*10^-4 m^3/s

but time t=V/q
t=2*10^-3 m^3/8.70*10^-4 m^3/s
=>time=2.29 seconds

ii) find maximum allowable h in order to maintain flow.
I struggle here.
Bernoulli's equation is not applied correctly here. The vapor pressure of Petrol has nothing to do with the pressure at point 1. The pressure at point 1 is determined from the hydrostatic equation within the tank. Or you could apply the Bernoulli equation to the combination of a point at the very top of the tank (where the pressure is atmospheric) and point 2, where the pressure is again atmospheric. Either way, you will get the same answer.

To answer part ii, you need to consider the absolute pressure at location h (at the top of the siphon) and compare this with the vapor pressure of the Petrol. If the pressure at this location is calculated to be less than the vapor pressure of the petrol, you will get cavitation (and the siphon will not work).
 
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  • #3
Please, in the future, submit problems is separate threads. You have actually submitted 4 problems within this thread. This is not acceptable.
 
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  • #4
for Question2) iii)
I did delta Q=Q1-Q2
where Q1=229.2 W from above subpoints of exercise
Q2=166.75 W rate of heat transfer to melt ice
=> delta Q=62.49 W
Q=h*A*delta T(convection) => delta T=Q/h*A =>(Tsnow-Toutside)=Q/hA
=>To=Tsnow-Q/hA
To=0-62.49/6*9
To(outside air temp)= - 1.15 degrees Celsius

for Question2 b) U-bend pipe
F=force to keep pipe in place=Fm(momentum force)+Fp(pressure force)
Fm=ro*A*u^2=10^3*10^-2*15^2=2250N
Fp=P1A1-P2A2=(10^5Pa*10^-2)-(6*10^4Pa*10^-2)
Fp=400N

=>F=Fm+Fp
F=2250+400
F=2650 N( force to keep pipe in place)

For question 3)
i) I tried using Laplace equation and by integrating and putting boundary conditions I got:
Temp profile T(x)=(T2-T1)*x/L+T1
=>T(x)=(10-130)*2*10^-2/30 +130
T(x)=129.92

q(x)=-lambda/L *(T2-T1)
q(x)= -0.4/30 *(10-130)
q(x)=1.6 W/m^2 (heat flux)
Now I do not know how to connect this to heat flux from convection q=h*delta T

For question 3b)
I do not know how to apply hydrostatic pressure
for point 1(let's say is on the bottom of the tank) P1=ro*g*h1=800*9.81*1m=7848Pa ,if at point 2 h2=0(datum)

However for point 3 let's say somewhere in fluid above, at 4m P3=ro*g*h3=800*9.81*4m
P3=31392 Pa, which is incorrect because the deeper we go , the pressure should be bigger but in this case is opposite.

I applied Bernoulli as above but I took the point 1 as at the top of the fluid of the tank.
So for point 1: u1=0,P1=0, h1=4m;
For point 2:u2=? P1=0; h2=0m(datum)
=> u2=sqrt(2*g*h1)
u2=sqrt(2*9.81*4)
u2=8.86 m/s
Area of siphon=pi*D^2/4=pi*(10^-2)^2/4
Asiphon=7.85*10^-5 m^2
q=u2*Asiphon=8.86m/s*7.85*10^-5 m^2
q=6.95*10^-4 m^3/s
t(time)=V/q=2*10^-3 m^3/ 6.95*10^-4 m^3/s
t=2.87 seconds

For question 3 last part with maximum height
I will calculate the hydrostatic Pressure in point 3 which is at top of siphon
P=ro*g*h for Gauge Pressure
P=Patm+ro*g*h to make Absolute Pressure
=>P3=Patm + ro*g*h3=800 kg/m^3 *9.81 m/s^2 *4m
P3=31392 Pa gauge
P3=69933Pa=69.9 Kpa absolute > 60KPa vap absolute
We apply again Bernoulli eq:
for point 3: P3=31392 Pa gauge ; u3=0;h3=?
for point 2: P2=0 ; u2=8.86 m/s (from above) h2=0;
h3=u2^2/2*g -P3/ro*g
h3=8.86^2/2*9.81 - 31392 /800*9.81
h3=9.98*10^ -4 meters
 
Last edited:
  • #5
williamcarter said:
for Question2) iii)
I did delta Q=Q1-Q2
where Q1=229.2 W from above subpoints of exercise
Q2=166.75 W rate of heat transfer to melt ice
=> delta Q=62.49 W
Q=h*A*delta T(convection) => delta T=Q/h*A =>(Tsnow-Toutside)=Q/hA
=>To=Tsnow-Q/hA
To=0-62.49/6*9
To(outside air temp)= - 1.15 degrees Celsius
You omitted the (conductive) temperature difference across the layer of snow.
for Question2 b) U-bend pipe
F=force to keep pipe in place=Fm(momentum force)+Fp(pressure force)
Fm=ro*A*u^2=10^3*10^-2*15^2=2250N
Fp=P1A1-P2A2=(10^5Pa*10^-2)-(6*10^4Pa*10^-2)
Fp=400N

=>F=Fm+Fp
F=2250+400
F=2650 N( force to keep pipe in place)
I didn't check this result. Is it consistent with the equation I gave for the force exerted by the bend on the fluid in my previous post?
For question 3)
i) I tried using Laplace equation and by integrating and putting boundary conditions I got:
Temp profile T(x)=(T2-T1)*x/L+T1
=>T(x)=(10-130)*2*10^-2/30 +130
T(x)=129.92
You need to use the Laplace equation in cylindrical coordinates. Using the equation in cartesian coordinates is not correct.
q(x)=-lambda/L *(T2-T1)
q(x)= -0.4/30 *(10-130)
q(x)=1.6 W/m^2 (heat flux)
Now I do not know how to connect this to heat flux from convection q=h*delta T
The two resistances are in series. $$(\Delta T)_A=\frac{Q}{R_A}$$$$(\Delta T)_B=\frac{Q}{R_B}$$
$$(\Delta T)_{total}=(\Delta T)_A+(\Delta T)_B=\frac{Q}{R_A}+\frac{Q}{R_B}$$
Then solve for Q.
For question 3b)
I do not know how to apply hydrostatic pressure
for point 1(let's say is on the bottom of the tank) P1=ro*g*h1=800*9.81*1m=7848Pa ,if at point 2 h2=0(datum)

However for point 3 let's say somewhere in fluid above, at 4m P3=ro*g*h3=800*9.81*4m
P3=31392 Pa, which is incorrect because the deeper we go , the pressure should be bigger but in this case is opposite.
I don't understand what you did here.
I applied Bernoulli as above but I took the point 1 as at the top of the fluid of the tank.
So for point 1: u1=0,P1=0, h1=4m;
For point 2:u2=? P1=0; h2=0m(datum)
=> u2=sqrt(2*g*h1)
u2=sqrt(2*9.81*4)
u2=8.86 m/s
Area of siphon=pi*D^2/4=pi*(10^-2)^2/4
Asiphon=7.85*10^-5 m^2
q=u2*Asiphon=8.86m/s*7.85*10^-5 m^2
q=6.95*10^-4 m^3/s
t(time)=V/q=2*10^-3 m^3/ 6.95*10^-4 m^3/s
t=2.87 seconds
This is correct.
For question 3 last part with maximum height
I will calculate the hydrostatic Pressure in point 3 which is at top of siphon
P=ro*g*h for Gauge Pressure
This equation should have a minus sign.
P=Patm+ro*g*h to make Absolute Pressure
This equation should also have a minus sign:
P=Patm-ro*g*h=100 kPa-ro*g*h
In order to avoid cavitation, this pressure must be greater than the equilibrium vapor pressure of the liquid.

EDIT: Actually, after further consideration, I realize that I left out a term in the equation for the pressure at the top of the bend. The correct application of the Bernoulli equation should be:
$$P+\rho g h+\rho \frac{v^2}{2}=P_{atm}$$
Therefore, $$P=P_{atm}-\rho g h-\rho \frac{v^2}{2}$$
 
Last edited:
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  • #6
Chestermiller said:
You omitted the (conductive) temperature difference across the layer of snow.
What exactly do you mean that I omitted conductive temp difference, and where exactly am I supposed to add that?
 
  • #7
williamcarter said:
What exactly do you mean that I omitted conductive temp difference, and where exactly am I supposed to add that?
There is a resistance to heat transfer through the layer of snow.
 
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  • #8
Chestermiller said:
There is a resistance to heat transfer through the layer of snow.
Thank you for quick reply, but where am I supposed to plug in that temp difference of conduction?
 
  • #9
williamcarter said:
Thank you for quick reply, but where am I supposed to plug in that temp difference of conduction?
$$\frac{62.49}{9}=0.3\frac{(0-T)}{0.2}$$where T is the temperature at the outer surface of the snow.
 
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  • #10
Chestermiller said:
This problem is not done correctly because you haven't taken into account the cylindrical geometry. Within the insulation, the heat conduction equation should be
q=−kdTdrq=−kdTdr​
q=-k\frac{dT}{dr}, and the heat balance equation should be
d(qr)dr=0d(qr)dr=0​
\frac{d(qr)}{dr}=0 or equivalently
qr=constqr=const​
qr=const These equations need to be solved simultaneously to get the temperature profile and the heat flux. Then, this solution has to be properly merged with the convective heat transfer.

In part iii, you should be comparing the ratio of the conductive resistance to the convective resistance.

I know the first one is from Fourier for conduction
I would really appreciate it if you could give me some more hints, I am stuck on this one.
I knew that d/dr*(r*dT/dr)=0
 
  • #11
If Q is the total rate of heat flow through the insulation (in W), for a cylindrical surface of length L and arbitrary radial location r within the insulation, what is the heat flux at this radial location (in terms of Q, L, and r, but not T)?
 
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  • #12
Chestermiller said:
If Q is the total rate of heat flow through the insulation (in W), for a cylindrical surface of length L and arbitrary radial location r within the insulation, what is the heat flux at this radial location (in terms of Q, L, and r, but not T)?
Q=q*A
Q=q*2pi*r*L
=>q=Q/2*pi*r*L
 
  • #13
williamcarter said:
Q=q*A
Q=q*2pi*r*L
=>q=Q/2*pi*r*L
Good. Now, in terms of the thermal conductivity k and the temperature gradient dT/dr, what is the heat flux q?
 
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  • #14
Chestermiller said:
Good. Now, in terms of the thermal conductivity k and the temperature gradient dT/dr, what is the heat flux q?
Thank you for your answer.

q= -k*dT/dr from Fourier's law of conduction
 
  • #15
You need to learn how to use LaTex. Look under INFO drop down menu and click on Help/How to, then click on LaTex Primer.
 
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  • #16
williamcarter said:
Thank you for your answer.

q= -k*dT/dr from Fourier's law of conduction
Good. Now set them equal. What do you get?
 
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  • #17
Chestermiller said:
Good. Now set them equal. What do you get?
##Q=-k* \frac {dT} {dr} *2\pi*r*L##
 
  • #18
Good. Now solve the ordinary linear differential equation for T as a function of r, subject to the boundary condition that ##T=T_i## at the inner boundary of the insulation, ##r=r_i##. (I feel like we've gone through all of this before).
 
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  • #19
Chestermiller said:
Good. Now solve the ordinary linear differential equation for T as a function of r, subject to the boundary condition that ##T=T_i## at the inner boundary of the insulation, ##r=r_i##. (I feel like we've gone through all of this before).
##T(r)=-\frac Q {k*2*\pi*L}* ln|r|+c1##
 
  • #20
williamcarter said:
##T(r)=-\frac Q {k*2*\pi*L}* ln|r|+c1##
What happened to substituting the boundary condition?
 
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  • #21
Chestermiller said:
What happened to substituting the boundary condition?
I didn't totally understood that part, what exactly to do with that.
 
  • #22
Determine the constant of integration c1 (algebraically) using the boundary condition, and substitute that back into the equation.
 
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  • #23
Chestermiller said:
Determine the constant of integration c1 (algebraically) using the boundary condition, and substitute that back into the equation.
##c1=T(r)+\frac {Q} {2*\pi*k*L}##
Like this ?and now plug in values for r=ri and T=Ti
where ##Ti=130 degrees Celsius## and ri=2*10-2
and ##L=30m## and ##lambda=0.4 W/m*K##
 
  • #24
williamcarter said:
##c1=T(r)+\frac {Q} {2*\pi*k*L}##
Like this ?and now plug in values for r=ri and T=Ti
where ##Ti=130 degrees Celsius## and ri=2*10-2
and ##L=30m## and ##lambda=0.4 W/m*K##
Algebraically, please.
 
  • #25
Chestermiller said:
Algebraically, please.
I did not quite understood what exactly needs to be done here algebrically with the boundary conditions , could you please lend me a hand? I don't know what I am doing wrong on it.
Thank you in advance.
 
  • #26
$$T(r)=-\frac {Q}{2k\pi L}\ln{r}+c1$$
$$c_1=T(r_i)+\frac {Q}{2k\pi L}\ln{r_i}$$where ##r_i## is the radius at the inner surface of the insulation, and ##T(r_i)## is the temperature at that location.
$$T(r)=T(r_i)-\frac {Q}{2k\pi L}\ln(r/r_i)$$
From this equation, what is the temperature ##T(r_0)## at the outer surface of the insulation ##r_0##?
 

Related to Heat transfer and Fluid Flow Struggling Problems

1. What is heat transfer and fluid flow?

Heat transfer and fluid flow are two interconnected processes that involve the movement of energy and matter. Heat transfer refers to the movement of thermal energy from one object to another, while fluid flow refers to the movement of a fluid (liquid or gas) from one location to another. These processes are crucial in understanding the behavior of systems such as engines, power plants, and weather patterns.

2. What are some common applications of heat transfer and fluid flow?

Heat transfer and fluid flow have a wide range of applications in various industries. Some common examples include heating and cooling systems, refrigerators, air conditioning, chemical reactors, and aerodynamics in transportation vehicles. These processes are also important in many natural phenomena such as ocean currents, weather patterns, and the Earth's climate.

3. What are some factors that affect heat transfer and fluid flow?

Several factors can influence the rate and direction of heat transfer and fluid flow. These include temperature, pressure, density, viscosity, and the geometry of the system. The properties of the fluid, such as its thermal conductivity and specific heat capacity, also play a significant role in determining the rate of heat transfer and fluid flow.

4. How do engineers and scientists study and analyze heat transfer and fluid flow?

Engineers and scientists use various mathematical models and computational tools to study and analyze heat transfer and fluid flow. These include fluid dynamics equations, heat transfer equations, and computer simulations. Experimental methods, such as measuring temperature and velocity, are also commonly used to gather data and validate theoretical models.

5. What are some common challenges or problems in heat transfer and fluid flow?

One of the most common challenges in heat transfer and fluid flow is understanding and predicting the behavior of complex systems. This can be challenging due to the non-linear nature of fluid flow and the presence of various factors that can affect heat transfer. Other common problems include heat and mass transfer limitations, flow instabilities, and turbulence, which require advanced techniques and models to address.

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