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Chapter 2
Heat Transfer (ME 531)
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KNOWN: Steady-state, one-dimensional heat conduction through an axisymmetric shape.
FIND: Sketch temperature distribution and explain shape of curve.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Constant properties, (3) No internal heat generation.
ANALYSIS: Performing an energy balance on the object according to Eq. 1, EEin−=out 0 , it
follows that
EE qin−=out x
and that qqxxx≠ b g. That is, the heat rate within the object is everywhere constant. From Fourier’s
law,
qkA
dT xxdx =− ,
and since qx and k are both constants, it follows that
A
dT dx x =Constant.
That is, the product of the cross-sectional area normal to the heat rate and temperature gradient
remains a constant and independent of distance x. It follows that since Ax increases with x, then dT/dx must decrease with increasing x. Hence, the temperature distribution appears as shown above.
COMMENTS: (1) Be sure to recognize that dT/dx is the slope of the temperature distribution. (2)
What would the distribution be when T 2 > T 1? (3) How does the heat flux, q′′x, vary with distance?
KNOWN: Hot water pipe covered with thick layer of insulation.
FIND: Sketch temperature distribution and give brief explanation to justify shape.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (radial) conduction, (3) No internal heat generation, (4) Insulation has uniform properties independent of temperature and position.
ANALYSIS: Fourier’s law, Eq. 2, for this one-dimensional (cylindrical) radial system has the form
rr ()
dT dT qkA k 2 r dr dr
=− =− πA
where Ar= 2 πr and AA is the axial length of the pipe-insulation system. Recognize that for steady-
state conditions with no internal heat generation, an energy balance on the system requires
EE sin==out ince EEg st= 0. Hence
qr = Constant.
That is, qr is independent of radius (r). Since the thermal conductivity is also constant, it follows that
dT r Constant. dr
⎡⎤ ⎢⎥= ⎣⎦
This relation requires that the product of the radial temperature gradient, dT/dr, and the radius, r, remains constant throughout the insulation. For our situation, the temperature distribution must appear as shown in the sketch.
COMMENTS: (1) Note that, while qr is a constant and independent of r, q′r′ is not a constant. How
does qr′′rbg vary with r? (2) Recognize that the radial temperature gradient, dT/dr, decreases with
increasing radius.
KNOWN: Symmetric shape with prescribed variation in cross-sectional area, temperature distribution and heat rate.
FIND: Expression for the thermal conductivity, k.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x-direction, (3) No internal heat generation.
ANALYSIS: Applying the energy balance, Eq. 1, to the system, it follows that, since
EEin= out,
qx=≠Constant f x .()
Using Fourier’s law, Eq. 2, with appropriate expressions for Ax and T, yields
() ( )
xx 23
dT qk A dx dK 6000W=-k 1-x m 300 1 2x-x. dx m
=−
⋅⋅ −⎡ ⎤ ⎢⎣ ⎥⎦
Solving for k and recognizing its units are W/m⋅K,
() ( ) ()( )
2 2
-6000 20 k=. 1-x 300 2 3x 1x 23x
= ⎡⎤−− −+ ⎢⎥⎣⎦
<
COMMENTS: (1) At x = 0, k = 10W/m⋅K and k → ∞ as x → 1. (2) Recognize that the 1-D assumption is an approximation which becomes more inappropriate as the area change with x, and hence two-dimensional effects, become more pronounced.
KNOWN: End-face temperatures and temperature dependence of k for a truncated cone.
FIND: Variation with axial distance along the cone of q q k, and dT / dx,,′′
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in x (negligible temperature gradients in the r direction), (2) Steady-state conditions, (3) Adiabatic sides, (4) No internal heat generation.
ANALYSIS: For the prescribed conditions, it follows from conservation of energy, Eq. 1, that for
a differential control volume, EE oin==out r qqxx+dx. Hence
qx is independent of x.
Since A(x) increases with increasing x, it follows that qq/xx′′ = Ax( ) decreases with increasing x.
Since T decreases with increasing x, k increases with increasing x. Hence, from Fourier’s law, Eq. 2,
qk′′=−
dT x dx ,
it follows that | dT/dx | decreases with increasing x.
COMMENT: How is the analysis changed if a has a negative value?
rr
KNOWN: Irradiation and absorptivity of aluminum, glass and aerogel.
FIND: Ability of the protective barrier to withstand the irradiation in terms of the temperature gradients that develop in response to the irradiation.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Constant properties, (c) Negligible emission and convection from the exposed surface.
PROPERTIES: Table A, pure aluminum (300 K): kal = 238 W/m⋅K. Table A, glass (300 K): kgl = 1 W/m⋅K.
ANALYSIS: From Eqs. 1 and 2.
sabs x=
T -k = q = G = αG x
∂ ′′ ∂
or
x=
T αG = - xk
∂ ∂
The temperature gradients at x = 0 for the three materials are: <
Material
∂∂T/ xx=0 (K/m)
aluminum 8 x 10 3 glass 6 x 10 6 aerogel 1 x 10 9
COMMENT: It is unlikely that the aerogel barrier can sustain the thermal stresses associated with the large temperature gradient. Low thermal conductivity solids are prone to large temperature gradients, and are often brittle.
x
G = 10 x 10 6 W/m 2
al gl
a
α = 0.
α= α=
x
G = 10 x 10 6 W/m 2
al gl
a
α = 0.
α= α=
KNOWN: One-dimensional system with prescribed thermal conductivity and thickness.
FIND: Unknowns for various temperature conditions and sketch distribution.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No internal heat generation, (4) Constant properties.
ANALYSIS: The rate equation and temperature gradient for this system are
x 21
dT dT T T q k and. dx dx L
− ′′ =− = (1,2)
Using Eqs. (1) and (2), the unknown quantities for each case can be determined.
(a)
####### dT ()20 50 K
280 K/m dx 0
−− ==−
x 2
WK q 50 280 14 kW/m. mK m
′′ =− × − = ⋅
⎡⎤ ⎢⎥⎣⎦
(b)
dT () 10 ()30 K
80 K/m dx 0
−−−x 2
WK q 50 80 4 kW/m. mK m
′′ =− × =− ⋅
⎡⎤ ⎢⎥⎣⎦
(c) x 2
WK q 50 160 8 kW/m mK m
′′ =− × =− ⋅
⎡⎤ ⎢⎥⎣⎦
21
dT K TL T0 60 0C 7. dx m
=⋅ + = × +
⎡⎤ ⎢⎥⎣⎦
D
T110C. 2 = D
(d) x 2
WK q 50 80 4 kW/m mK m
′′ =− × − = ⋅
⎡⎤ ⎢⎥⎣⎦
12
dT K TTL 40C0 dx m
=−⋅ = − −⎡⎢ ⎤⎥ ⎣ ⎦
D
T60C. 1 = D
(e) x 2
WK q 50 200 10 kW/m mK m
′′ =− × =− ⋅
⎡⎤ ⎢⎥⎣⎦
12
dT K T T L 30 C 0 200 20 C. dx m
=−⋅ = − =−
⎡⎤ ⎢⎥⎣⎦
DD
< < < < < <
KNOWN: Temperature distribution in solid cylinder and convection coefficient at cylinder surface.
FIND: Expressions for heat rate at cylinder surface and fluid temperature.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Steady-state conditions, (3) Constant properties.
ANALYSIS: The heat rate from Fourier’s law for the radial (cylindrical) system has the form
qkA
dT rrdr =−.
Substituting for the temperature distribution, T(r) = a + br
2 ,
qr=−k 2 rL 2br = -4 kbLr .()ππ 2
At the outer surface ( r = ro), the conduction heat rate is
qr=ro =− 4 πkbLro 2. <
From a surface energy balance at r = ro,
qq hr=ro==conv () 2 πrLo T⎡⎣ (rTo)−∞⎤⎦,
Substituting for qr=ro and solving for T∞,
()o o
2kbr T = Tr ∞ h +
T = a+br
kbr o h
2 o ∞ +
2
oo
2k T = a+br r. h
∞
⎡⎤ ⎢⎥+ ⎣⎦
<
KNOWN: Two-dimensional body with specified thermal conductivity and two isothermal surfaces of prescribed temperatures; one surface, A, has a prescribed temperature gradient.
F IND: Temperature gradients, ∂T/∂x and ∂T/∂y, at the surface B.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) No heat generation, (4) Constant properties.
ANALYSIS: At the surface A, the temperature gradient in the x-direction must be zero. That is,
(∂T/∂x)A = 0. This follows from the requirement that the heat flux vector must be normal to an isothermal surface. The heat rate at the surface A is given by Fourier’s law written as
y,A A A
TWK q k w 10 2m 30 600W/m. ymKm
∂ ∂
⎤ ′ =− ⋅ ⎥ =− × × =− ⎦ ⋅
O n the surface B, it follows that
()∂∂T/ yB= 0 <
in order to satisfy the requirement that the heat flux vector be normal to the isothermal surface B. Using the conservation of energy requirement, Eq. 1, on the body, find
q′y,A−q′x, B=0. or q′x, B=q′y,A
N ote that,
x,B B B
T qkw x
∂ ∂
′ =− ⋅ ⎤ ⎥⎦
a nd hence
()
y,A ()
B B
q 600 W/m T/ x 60 K/m. k w 10 W/m K 1m
∂∂
− ′ −− == = ⋅⋅×
<
COMMENTS: Note that, in using the conservation requirement, qq′in=+ ′y,A and qq′out=+ ′x,B.
KNOWN: A rod of constant thermal conductivity k and variable cross-sectional area Ax(x) = Aoeax w here Ao and a are constants.
FIND: (a) Expression for the conduction heat rate, qx(x); use this expression to determine the temperature distribution, T(x); and sketch of the temperature distribution, (b) Considering the presence
of volumetric heat generation rate, qqexpax=−o ( ), obtain an expression for qx(x) when the left
face, x = 0, is well insulated.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the rod, (2) Constant properties, (3) Steady- state conditions.
ANALYSIS: Perform an energy balance on the control volume, A(x)⋅dx,
EE E in−+=out g 0
= 0
qqxxd−++ xqAxd⋅⋅()x
The conduction heat rate terms can be expressed as a Taylor series and substituting expressions for
nd A(x),
q
a
()xo( )o()
d qqexpaxAexpax dx
−+− ⋅ = 0 (1)
x ()
dT qkAx dx
=− ⋅ (2)
(a) With no internal generation, qo = 0, and from Eq. (1) find
()x
d q dx
−= <
i ndicating that the heat rate is constant with x. By combining Eqs. (1) and (2)
() () 1
ddT dT kAx 0 or Ax C dx dx dx
−−⋅⎛⎞= ⋅= ⎜⎟ ⎝⎠
(3) <
Continued...
PROBLEM 2 (Cont.)
That is, the product of the cross-sectional area and the temperature gradient is a constant, independent of x. Hence, with T(0) > T(L), the temperature distribution is exponential, and as shown in the sketch above. Separating variables and integrating Eq. (3), the general form for the temperature distribution can be determined,
o1()
dT Aexpax C dx
⋅=
dT C A=−1o− 1 exp()ax dx
Tx()=−CAaexp ax C1o ( )− + 2 <
We could use the two temperature boundary conditions, To = T(0) and TL = T(L), to evaluate C 1 and C 2
and, hence, obtain the temperature distribution in terms of To and TL.
( b) With the internal generation, from Eq. (1),
()xoo xoo
d qqA0 or qqA dx
−+==x <
T hat is, the heat rate increases linearly with x.
COMMENTS: In part (b), you could determine the temperature distribution using Fourier’s law and knowledge of the heat rate dependence upon the x-coordinate. Give it a try!
KNOWN: Dimensions of and temperature difference applied across thin gold film.
FIND: (a) Energy conducted along the film, (b) Plot the thermal conductivity along and across the thin dimension of the film, for film thicknesses 30 ≤ L ≤ 140 nm.
SCHEMATIC:
x
y
L = 60 nm
a = 1 μm
b = 250 nm
T 1
T 2
x
y
L = 60 nm
a = 1 μm
b = 250 nm
T 1
T 2
ASSUMPTIONS: (1) One-dimensional conduction in the x- and y-directions, (2) Steady-state conditions, (3) Constant properties, (4) Thermal conductivity not affected by nanoscale effects associated with 250 nm dimension.
PROPERTIES: Table A, gold (bulk, 300 K): k = 317 W/m⋅K.
ANALYSIS: a) From Eq. 2,
xx 12
dT T - T q = -kA = k Lb[ ] dx a
(1)
From Eq. 2,
k = k [1 - 2xmλ fp / (3 L)]π (2)
Combining Eqs. (1) and (2), and using the value of λmfp= 31 nm from Table 2 yields
12 xmfp
T - T q = k[1 - 2λ / (3πL)]Lb[ ] a - -9 - -9 -
W 2×31×10 m 20°C = 317 × [1 - ] × 60 × 10 m × 250 × 10 m × mK⋅ 3×π×60×10 m 1 × 10 m
= 85 × 10 -6 W = 85 μW <
(b) The spanwise thermal conductivity may be found from Eq. 2,
k = k[1 - ymλ fp / (3L)] (3)
Continued...
PROBLEM 2 (Cont.)
The plot is shown below.
COMMENT: Nanoscale effects become less significant as the thickness of the film is increased.
KNOWN: Electrical heater sandwiched between two identical cylindrical (30 mm dia. × 60 mm
length) samples whose opposite ends contact plates maintained at To.
FIND: (a) Thermal conductivity of SS316 samples for the prescribed conditions (A) and their average temperature, (b) Thermal conductivity of Armco iron sample for the prescribed conditions (B), (c) Comment on advantages of experimental arrangement, lateral heat losses, and conditions for which
∆T 1 ≠ ∆T 2.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional heat transfer in samples, (2) Steady-state conditions, (3) Negligible contact resistance between materials.
PROPERTIES: Table A , Stainless steel 316 (T=400 K : k) ss=15 W/m K;⋅ Armco iron
()T=380 K : kiron=⋅67 W/m K.
ANALYSIS: (a) For Case A recognize that half the heater power will pass through each of the samples which are presumed identical. Apply Fourier’s law to a sample
q=kA
T c x
∆ ∆
()
()
c 2
qx 0 100V 0 0 m k= 15 W/m K. AT π 0 m / 4 25 C
∆ ×× = ∆ × D
=⋅ <
The total temperature drop across the length of the sample is ∆T 1 (L/∆x) = 25°C (60 mm/15 mm) =
100 °C. Hence, the heater temperature is Th = 177°C. Thus the average temperature of the sample is
T= T()oh+=T / 2 127 C=400 K <
We compare the calculated value of k with the tabulated value (see above) at 400 K and note the good agreement.
(b) For Case B, we assume that the thermal conductivity of the SS316 sample is the same as that found in Part (a). The heat rate through the Armco iron sample is
Continued .....
PROBLEM 2 (CONT.)
()
()
2 iron heater ss
iron
0 m 15 C q q q 100V 0 15 W/m K 4 0 m q 60 10 W=49 W
π =−=×− ⋅× ×
=−
D
where
qkATxss=∆∆ss c 2 /. 2
Applying Fourier’s law to the iron sample,
()
iron 2 iron 2 c
qx 49 W 0 m k 70 W/m K. AT π 0 m / 4 15 C
∆ × == = ∆ × D
⋅ <
The total drop across the iron sample is 15°C(60/15) = 60°C; the heater temperature is (77 + 60)°C = 137 °C. Hence the average temperature of the iron sample is
T= 137 + 77 C/2=107 C=380 K.()
D D
<
We compare the computed value of k with the tabulated value (see above) at 380 K and note the good agreement.
(c) The principal advantage of having two identical samples is the assurance that all the electrical power dissipated in the heater will appear as equivalent heat flows through the samples. With only one sample, heat can flow from the backside of the heater even though insulated.
Heat leakage out the lateral surfaces of the cylindrically shaped samples will become significant when the sample thermal conductivity is comparable to that of the insulating material. Hence, the method is suitable for metallics, but must be used with caution on nonmetallic materials.
For any combination of materials in the upper and lower position, we expect ∆T 1 = ∆T 2. However, if the insulation were improperly applied along the lateral surfaces, it is possible that heat leakage will
occur, causing ∆T 1 ≠ ∆T 2.
Chapter 2
Course: Heat Transfer (ME 531)
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Students also viewed
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