程序代写 SIMPLEST CASES

SIMPLEST CASES
Rate of heat conducted through solid of thermal conductivity, k, assuming steady state, 1-d conduction.
Rate of heat transferred from surface, TS, to adjacent fluid, T∞, through convection.
Radiation emitted from a surface, Ts, with emissivity, ε

2 STEADY STATE 1-D CONDUCTION
Incropera p 73-110, Cengel p75-98, Welty et al. p 217-222 THIS WEEK:
Where does the steady-state 1-d conduction equation come from?
Under what conditions can and can’t it be used?
What happens if:
• Thermal conductivity is not uniform?
• Conduction and convection occur in series?
• Conduction through a sphere or cylinder rather than slab?
Learning objectives, tasks, readings, Learn ChemE videos
1. Appreciate that analytical equations in Heat Transfer can all be derived from the general heat equation:
a. Learn ChemE videos “Heat Equation Derivation”
b. Apply simplifying assumptions and understand how they
help you to solve problems.
c. Note – derivations not required in this unit
2. Use the concept of “resistances in series” to determine:
a. Overall heat transfer coefficient, U, for composite walls b. Heat transfer rate through composite walls
c. Both a and b in plane, cylindrical, and spherical walls
By the end of the week you should be able to:
 Calculate q, R’s, T’s from composite wall problems in either plane, cylindrical or spherical co-ordinate systems.

DERIVATION OF GENERAL HEAT EQUATION
Energy balance over a control volume
IN – OUT + GENERATION = ACCUMULATION

Rearrange:
T T T T 2T 2T 2T cpt vx xvy yvz zkx2 y2 z2 q”
Equation H-1

HOW TO SOLVE?
Specify initial condition and boundary conditions, e.g.
 Isothermal surface
 Constant heat flux to surface
 Insulated surface
 Convective surface condition OR, Simplify!
Common simplifying assumptions:
No internal generation (no reactions):
Steady state:
No convection:
Equation H-2
Equation H-3
Equation H-4

DERIVATION: CONDUCTION THROUGH A SOLID WALL
Incropera p. 74-79, Cengel p. 135-162, Welty et al. p 224-233
Determine heat transfer through the wall under the same conditions as last week, but now consider a wall made of plasterboard (2 cm), pine (4 cm) and brick (14 cm):
Assumptions:
 _____________________________________________  _____________________________________________  _____________________________________________  _____________________________________________

Rate of heat conducted through plasterboard:
(1) Equation H-5
Rate of heat conducted through pine:
(2) Equation H-6
Rate of heat conducted through brick:
(3) Equation H-7
Solve simultaneously
Equation H-8

DERIVATION: CONDUCTION THROUGH A SOLID WALL – THE WALL CAN ALSO BE MODELLED AS THREE RESISTANCES IN SERIES:
Composite walls have overall heat transfer coefficient, U:
q UT T 
Rate=_________________________
EquationH-9
Equation H-10
Cf Equation H-8
Equation H-11

1.1 SUMMARY: RESISTANCES IN SERIES AND OVERALL HEAT TRANSFER COEFFICIENTS
CASE 1 – AREA CONSTANT ACROSS ALL LAYERS
q UT T 
io URRR
total  i total
qhTT A i i 1
qk2 TT A x2 2 3
qhTT A o 4 o
q  k1 T T  q  k3 T T  A x 1 2 A x 3 4
hi k1 k2 k3 ho
Rtotal  Rconvi  Rcond1  Rcond 2  Rcond 3  Rconvo  1  x  x  x  1

 Calculate Rtotal, including units to avoid errors.  Comment – which resistance dominates?
– Consider how this is likely to affect temperature profile.
U1  Calculate Rtotal
q UT T 
 A i o or
[U]=Wm-2K-1.
qUAT T 
i o for final answer.

1.2 CYLINDRICAL CO-ORDINATES
EXAMPLE: 1-D CONDUCTION THROUGH A CYLINDRICAL SHELL
Area for heat transfer at radius r1
Assumptions:
 _____________________________________________  _____________________________________________  _____________________________________________
q  k dT A dr
 q  kA dT dr

NOW CONSIDER RESISTANCES IN SERIES IN CYLINDRICAL COORDINATES.
AREA IS NOT CONSTANT ACROSS THE DIFFERENT LAYERS, HENCE USE UA RATHER THAN U
q UT T 

Since area varies across each layer, it must be included in resistance:
total  i total
qhATT, A2rl iii1i1
UAR R  R
q2klT T 223
ln  r3 r2
q  2k lT T  112
q h A T T , oo3o
A 2rl o3
Rtotal  Rconvi  Rcond1  Rcond 2  Rconvo 1 r2r r3r 1
ln  ln  12
r2 r ln 
h A 2k l 2k l h A ii12oo
 Calculate Rtotal, including units to avoid errors.  Comment – which resistance dominates?
– Consider how this is likely to affect temperature profile.
 Calculate Rtotal [UA]=WK-1.
T T  q
qUAT T 
 i o or i o UA for final answer.

1.3 SUMMARY – 1-D, STEADY STATE CONDUCTION
HEAT CONDUCTED ACROSS A THICKNESS ∆X (ERROR! REFERENCE SOURCE NOT FOUND.)
qAkT T  12
1-D CONDUCTION (POLAR CO-ORDINATES)
Heat conducted through a hollow cylinder
12 q  2kl T  T 
Equation H-12
Equation H-13
Resistances added in series as before but area varies with radius so
r  ln 2 r 
Heat conducted through a hollow sphere
12 q4k T T 
qUAT T  in
1 1 r r 
Where 1  R …….R
Equation H-14 Equation H-15
Equation H-16
1 n1 For example, for a cylinder
ln 2 r  R1
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