What is the Steady State Temperature of a Brick Wall in a Fire?

Click For Summary
SUMMARY

The steady state temperature of the surface of a solid brick wall exposed to a fire at 1000°C on one side and an ambient temperature of 20°C on the warehouse side can be calculated using the heat transfer equation. Given the wall's thickness of 200 mm and thermal conductivity of 0.72 W/m°C, along with a heat transfer coefficient of 12 W/m²°C for the air in the warehouse, the temperature on the warehouse side can be determined. The relevant equations include the heat flow equation dQ/dt = kA(Δ∅)/x and the second order differential equation d²T/dx² = ρc dT/kdt.

PREREQUISITES
  • Understanding of heat transfer principles
  • Familiarity with differential equations
  • Knowledge of thermal conductivity and heat transfer coefficients
  • Basic concepts of thermal resistance in solid materials
NEXT STEPS
  • Study the derivation and application of the heat transfer equation dQ/dt = kA(Δ∅)/x
  • Learn about the thermal resistance concept in solid walls
  • Explore the use of electrical analogies in thermal problems
  • Investigate numerical methods for solving second order differential equations
USEFUL FOR

Students in engineering and physics, professionals in thermal management, and anyone involved in fire safety and building design will benefit from this discussion.

marganda
Messages
2
Reaction score
0

Homework Statement


A fire in a room rapidly raises the temperature of the surface of the wall to a steady value of 1000°C. on the other side of the wall is a large warehouse, whose ambient air temperature is 20°C. If the wall is solid brick, 200 mm thick, with thermal conductivity of 0.72 W/m°C, what is the steady state temperature of the surface of the wall ( on the warehouse side) in °C. You may assume the heat transfer coeffiecient of the air in the warehouse is 12W/m^2°C


Homework Equations



d^2T/dx^2 =ρc dT/ kdt

The Attempt at a Solution



this is a second order differential equation.

Please help me. thanks in advance!
 
Physics news on Phys.org
The equation for heat flow is
dQ/dt = kA(Δ∅)/x
Where dQ/dt = rate of heat flow in Watts (or Watts/m^2)
k = thermal conductivity W/m
A = cross sectional area (1m^2)
(Δ∅)/x = (difference in temp)/thickness
Hope this gets you started
 
technician said:
The equation for heat flow is
dQ/dt = kA(Δ∅)/x
Where dQ/dt = rate of heat flow in Watts (or Watts/m^2)
k = thermal conductivity W/m
A = cross sectional area (1m^2)
(Δ∅)/x = (difference in temp)/thickness
Hope this gets you started

thank you technician :) i use different approarch in solving this problem. I used electrical analogy instead of solving the differentail equation.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K