One Dimensional Steady Heat Conduction problems - PowerPoint PPT Presentation

Loading...

PPT – One Dimensional Steady Heat Conduction problems PowerPoint presentation | free to download - id: 68f1c0-MWMzY



Loading


The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
About This Presentation
Title:

One Dimensional Steady Heat Conduction problems

Description:

One Dimensional Steady Heat Conduction problems P M V Subbarao Associate Professor Mechanical Engineering Department IIT Delhi Simple ideas for complex Problems – PowerPoint PPT presentation

Number of Views:77
Avg rating:3.0/5.0
Slides: 24
Provided by: abc95
Learn more at: http://web.iitd.ac.in
Category:

less

Write a Comment
User Comments (0)
Transcript and Presenter's Notes

Title: One Dimensional Steady Heat Conduction problems


1
One Dimensional Steady Heat Conduction problems
  • P M V Subbarao
  • Associate Professor
  • Mechanical Engineering Department
  • IIT Delhi

Simple ideas for complex Problems
2
Electrical Circuit Theory of Heat Transfer
  • Thermal Resistance
  • A resistance can be defined as the ratio of a
    driving potential to a corresponding transfer
    rate.

Analogy Electrical resistance is to conduction
of electricity as thermal resistance is to
conduction of heat. The analog of Q is current,
and the analog of the temperature difference, T1
- T2, is voltage difference. From this
perspective the slab is a pure resistance to heat
transfer and we can define
3
(No Transcript)
4
(No Transcript)
5
The composite Wall
  • The concept of a thermal resistance circuit
    allows ready analysis of problems such as a
    composite slab (composite planar heat transfer
    surface).
  • In the composite slab, the heat flux is constant
    with x.
  • The resistances are in series and sum to R R1
    R2.
  • If TL is the temperature at the left, and TR is
    the temperature at the right, the heat transfer
    rate is given by

6
Wall Surfaces with Convection
Boundary conditions
7
Heat transfer for a wall with dissimilar materials
  • For this situation, the total heat flux Q is
    made up of the heat flux in the two parallel
    paths
  • Q Q1 Q2
  • with the total resistance given by

8
Composite Walls
  • The overall thermal resistance is given by

9
Desert Housing Composite Walls
10
One-dimensional Steady Conduction in Radial
Systems
Homogeneous and constant property material
11
At any radial location the surface are for heat
conduction in a solid cylinder is
At any radial location the surface are for heat
conduction in a solid sphere is
The GDE for cylinder
12
The GDE for sphere
General Solution for Cylinder
General Solution for Sphere
13
Boundary Conditions
  • No solution exists when r 0.
  • Totally solid cylinder or Sphere have no physical
    relevance!
  • Dirichlet Boundary Conditions The boundary
    conditions in any heat transfer simulation are
    expressed in terms of the temperature at the
    boundary.
  • Neumann Boundary Conditions The boundary
    conditions in any heat transfer simulation are
    expressed in terms of the temperature gradient at
    the boundary.
  • Mixed Boundary Conditions A mixed boundary
    condition gives information about both the values
    of a temperature and the values of its derivative
    on the boundary of the domain.
  • Mixed boundary conditions are a combination of
    Dirichlet boundary conditions and Neumann
    boundary conditions.

14
Mean Critical Thickness of Insulation
Heat loss from a pipe
h,T?
  • If A, is increased, Q will increase.
  • When insulation is added to a pipe, the outside
    surface area of the pipe will increase.
  • This would indicate an increased rate of heat
    transfer

ri
Ts
ro
  • The insulation material has a low thermal
    conductivity, it reduces the conductive heat
    transfer lowers the temperature difference
    between the outer surface temperature of the
    insulation and the surrounding bulk fluid
    temperature.
  • This contradiction indicates that there must be a
    critical thickness of insulation.
  • The thickness of insulation must be greater than
    the critical thickness, so that the rate of heat
    loss is reduced as desired.

15
Electrical analogy
As the outside radius, ro, increases, then in
the denominator, the first term increases but the
second term decreases. Thus, there must be a
critical radius, rc , that will allow maximum
rate of heat transfer, Q The critical radius, rc,
can be obtained by differentiating and setting
the resulting equation equal to zero.
16
(No Transcript)
17
Ti,Tb, k, L, ro, ri are constant terms, therefore
When outside radius becomes equal to critical
radius, or ro rc, we get,
18
Safety of Insulation
  • Pipes that are readily accessible by workers are
    subject to safety constraints. 
  • The recommended safe "touch" temperature range is
    from 54.4 0C to 65.5 0C.  
  • Insulation calculations should aim to keep the
    outside temperature of the insulation around 60
    0C. 
  • An additional tool employed to help meet this
    goal is aluminum covering wrapped around the
    outside of the insulation.  
  • Aluminum's thermal conductivity of 209 W/m K does
    not offer much resistance to heat transfer, but
    it does act as another resistance while also
    holding the insulation in place. 
  • Typical thickness of aluminum used for this
    purpose ranges from 0.2 mm to 0.4 mm. 
  • The addition of aluminum adds another resistance
    term, when calculating the total heat loss

19
Structure of Hot Fluid Piping
20
  • However, when considering safety, engineers need
    a quick way to calculate the surface temperature
    that will come into contact with the workers. 
  • This can be done with equations or the use of
    charts. 
  • We start by looking at diagram

21
At steady state, the heat transfer rate will be
the same for each layer
22
Solving the three expressions for the temperature
difference yields
Each term in the denominator of above Equation
is referred to as the Thermal resistance" of
each layer. 
23
Design Procedure
  • Use the economic thickness of your insulation as
    a basis for your calculation.
  • After all, if the most affordable layer of
    insulation is safe, that's the one you'd want to
    use. 
  • Since the heat loss is constant for each layer,
    calculate Q from the bare pipe.
  • Then solve T4 (surface temperature).  
  • If the economic thickness results in too high a
    surface temperature, repeat the calculation by
    increasing the insulation thickness by 12 mm each
    time until a safe touch temperature is reached.
  • Using heat balance equations is certainly a valid
    means of estimating surface temperatures, but it
    may not always be the fastest. 
  • Charts are available that utilize a
    characteristic called "equivalent thickness" to
    simplify the heat balance equations. 
  • This correlation also uses the surface resistance
    of the outer covering of the pipe. 
About PowerShow.com