What is the area through which the heat flows at r = 5.92 cm?

Click For Summary
SUMMARY

The discussion focuses on calculating the temperature of pipe metal at a radius of 5.92 cm, given a fluid temperature of 350 K and ambient air temperature of 306 K. The heat transfer equation used is dQ/dt = kA(dT/dx), where A is the area through which heat flows. The user initially calculated the temperature at r=5.92 cm to be 321.46 K but identified a potential error in applying the initial conditions related to the area of heat flow. The correct approach requires considering the length of the pipe in the area calculation.

PREREQUISITES
  • Understanding of heat transfer principles, specifically conduction.
  • Familiarity with the heat equation and its components.
  • Knowledge of calculus, particularly differentiation and integration.
  • Experience with boundary value problems in thermal analysis.
NEXT STEPS
  • Review the derivation of the heat equation in cylindrical coordinates.
  • Learn about boundary conditions in heat transfer problems.
  • Study the impact of pipe length on heat transfer area calculations.
  • Explore numerical methods for solving differential equations in thermal systems.
USEFUL FOR

Students and professionals in mechanical engineering, thermal engineering, and physics, particularly those working on heat transfer problems in cylindrical geometries.

dinospamoni
Messages
90
Reaction score
0

Homework Statement



The fluid inside the pipe shown has a temperature of
350 K, but the temperature of the air in the room is only 306 K.
Therefore, heat flows at a constant rate from the fluid, through
the pipe walls, and into the room. The inner pipe radius is 4
cm, and the outer radius is 8 cm. The heat equation is:

dQ/dt = kA(dT/dx)

where x is the direction of heat flow, A is the area through
which the heat flows (i.e., perpendicular to x), and k is the
conductivity of the material through which the heat is flowing.
Determine the temperature of the pipe metal at r = 5.92 cm.

Homework Equations



T(r<4)=350
T(r>8) = 306

The Attempt at a Solution



I've tried this a bunch of times, but can't see to get it. I have done:

Q'=kA(dT/dr)
where Q' is a constant

A=pi*r^2

dT=Q'/(k*pi*r^2) *dr

T=(Q'/k*pi)(-1/r)+c1

I let Q'/k = c2

so

T=c1-c2/r*pi

After imposing the initial conditions:

350 = c1 - c2/4pi

306 = c1 - c2/8pi

from this
c1 = 262
c2= 1105.84

and got the temperature at r=5.92 to be 321.46 K, but this wasn't right.

Any ideas?

I think I went wrong with the initial conditions somewhere
 
Physics news on Phys.org
dinospamoni said:
A is the area through
which the heat flows (i.e., perpendicular to x),
...

A=pi*r^2

There's a mistake here, since the area through which the heat flows must depend on the length of the pipe.
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
22
Views
4K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
11
Views
2K
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K