application of steady-state heat transfer
DESCRIPTION
Application of Steady-State Heat Transfer. Steady-state heat transfer. Temperature in a system remains constant with time. Temperature varies with location. T 1. T 2. T 1 > T 2. Conductive heat transfer in a rectangular slab. Example. - PowerPoint PPT PresentationTRANSCRIPT
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Application of Steady-State Heat Transfer
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Steady-state heat transfer
• Temperature in a system remains constant with time.
• Temperature varies with location.
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Conductive heat transfer in a rectangular slab
dx
dTkAq
x
T
TkdTdx
x
x A
qx
11
T
TdTk
x
xdx
A
qX
11
x
TkAq
x
T1 > T2
T1
T2
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ExampleFor the stainless steel plate 1 cm thick is
maintained at 110C, while the other face is at 90 C. Calculate temperature at 0.5 cm from the
110C-temperature face.
Given :
heat flux = 34,000 W/m2
thermal conductivity of stainless steel = 17 W/m C
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Conductive Heat Transfer through a Tubular Pipe
• Consider a long hollow cylinder
l
r
rodr
rl)r(A 2
Tiri
To
T
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Conductive Heat Transfer through a Tubular Pipe
• Consider a long hollow cylinder
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Example
A 2 cm thick steel pipe (k= 43 W/mC) with 6 cm inside diameter
is being used to convey steam from a boiler to process
equipment for a distance of 40 m. The inside pipe surface
temperature is 115C, and the outside pipe surface temperature
is 90C. Under steady state conditions, calculate total heat
loss to the surrounding.
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Heat conduction in multilayered systems
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Composite rectangular wall (in series)
x3
k3
T3
x2
k2
T2
x1
k1
T1
q
q
R3
R2
R1
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= composite thermal resistance
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q
q
R3
R2
R1
x1 k1
T1x2 k2
T2x3 k3
T3
q = A T k / x = A T / (x/k)
T1 + T2 + T3 = T
R = Resistance = x/k = 1/C 1/RT = 1/R1+1/ R2+1/ R3
= (1/(x1 / k 1))+ (1/(x2 / k 2))+ (1/(x3 / k 3))
Composite rectangular wall (in parallel)
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and it is resistance which is additive when several conducting layers lie between the hot and cool regions, because A and Q are the same for all layers. In a multilayer partition, the total conductance is related to the conductance of its layers by:
So, when dealing with a multilayer partition, the following formula is usually used:
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ExampleA cold storage wall (3m X 6m) is constructed of a 15 cm
thick concrete (k = 1.37 W/mC). Insulation must be provided to maintain a heat transfer rate through the wall at or below 500 W. If k of insulation is 0.04 W/mC. The outside surface temperature of the wall is 38C and the
inside wall temperature is 5C.
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Example
How many joules of thermal energy
flow through the wall per second? -------------------------------------------
Heat is like a fluid: whatever flows through the insulation must also flow
through the wood.
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k (insulation) = 0.20 J/(s-m-C)k (wood) = 0.80 J/(s-m-C)
Across insulation: Hins = (0 .2 0 )(4 0 )(2
- 5 T)/0.076 -2631 6 105= .
3. T Across wood: 0 80 40Hwood = ( . )( )(
- T 4 )/0 .0 1 9 - = 1 6 8 4 .2 T 6 736.8
Heat is like afluid: whateverffff f fffffff fff ffffffffff f fff ffff ffff fff
ough thewood:
Hwood = Hins - 1684.2 T 6736.8 = 2631.
- 6 105.3 T 1789.5 T = 9368.4
5 235T = . C
H=Hwood=Hins
1684 2 5 23H= . ( . - 5 67368 2080) . = J/s
- 2631 6 105H= . . 3 (5.235) = 2080 J/s
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1 32 54
B
A
D
C E
G
F
RB
RE
RA
RD
RC
RG
RF
Series and parallel one-dimensional heat transfer through a composite wall and
electrical analog
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Composite cylindrical tube(in series)
r1
r2
r3
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ExampleA stainless steel pipe (k= 17
W/mC) is being used to convey heated oil. The inside surface
temperature is 130C. The pipe is 2 cm thick with an inside diameter
of 8 cm. The pipe is insulated with 0.04 m thick insulation (k=
0.035 W/mC). The outer insulation temperature is 25C. Calculate the temperature of interface between steel and
isulation. Assume steady-state conditions.
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THERMAL CONDUCTIVITY CHANGE WITH TEMPERATURE
k = k0(1+T)
dx
dTkAqx
)(21TT
X
Akq m
x
221
TT km is thermal conductivity at T =
Heat transfer through a slab
))(2
)(( 21
2212
0 TTTTx
Akq
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THERMAL CONDUCTIVITY CHANGE WITH TEMPERATURE
Heat transfer through a cylindrical tube
dr
kAdTqr
dr
dTrLTkqr )2))(1(( 0
dTTLkr
drqr )1(2 0
)))((1(/ln
200
0 TTTTrr
Lkq ii
ior
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Problem1. Find the heat transfer per unit area through the composite
wall. Assume one-dimensional heat flow.
Given:
kA = 150 W/mC
kB = 30 W/mC
kC = 50 W/mC
kD = 70 W/mC
AB = AD
T = 370C
CA
D
Bq
AA = AC = 0.1 m2
T = 66C2.5 cm
7.5 cm
5.0 cm
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Problem
2. One side of a copper block 5 cm thick is maintained at 260C. The other side is covered with a layer of fiber glass 2.5 cm thick. The outside of the fiber glass is
maintained at 38C, and the total heat flow through the copper-fiber-glass combination is 44 kW. What is the
area of the slab?
3. A wall is constructed of 2.0 cm of copper, 3.0 mm of asbestos, and 6.0 cm of fiber glass. Calculate the heat
flow per unit area for an overall temperature difference of 500C.
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Problem
4. A certain material has a thickness of 30 cm and a thermal conductivity of 0.04 W/mC. At a particular
instant in time the temperature distribution with x, the distance from the left face, is T = 150x2 - 30x, where x is in meters. Calculate the heat flow rates at x = 0 and x =
30 cm. Is the solid heating up or cooling down?5. A certain material 2.5 cm thick, with a cross-sectional area of 0.1 m2, has one side maintained at 35C and the other at 95C. The temperature at the center plane of the material is 62C, and the heat flow through the material is 1 kW. Obtain an expression for the thermal conductivity
of the material as a function of temperature.