北京科技大学传热学第2章习题答案

heat flux through the outer surface is determined to be
•
•
•
qS
= QS A
=
QS 2πr2 L
=
0.85 × 300W
2π (0.04m)(6m)
=
169.1W
/
m2
Noting that heat transfer is one-dimensional in the radial r direction and heat flux is in the negative r
Taking the direction normal to the surface of the wall to be the x direction with x=0 at the left surface, the
mathematical formulation of this problem can be expressed as
Properties: The thermal conductivity is given to be k = 14W / m ⋅ DC .
Analysis: (a) Noting that the 85% of the 300W generated by the strip heater is transferred to the pipe, the
Assumptions: 1 Heat conduction is steady and one-dimensional since the surface area of the base plate is large relative to its thickness, and the thermal conditions on both sides of the plate are uniform . 2 Thermal conductivity is constant. 3 There is no heat generation in the plate. 4 Heat loss through the upper part of the iron is negligible.
2-17 A compressed air pipe is subjected to uniform heat flux on the outer surface and convection on the inner surface. The mathematical formulation, the variation of temperature in the pipe, and the surface temperature are to be determined for steady one-dimensional heat transfer.
of the element can be expressed as
( ) ( ) ΔEelement = Et+Δt − Et = mC Tt+Δt − Tt = ρCΔxΔy Tt+Δt − Tt
Substituting,
•
•
•
•
Q x + Q y − Q x+Δx − Q y+Δy
=
ρCΔxΔy Tt+Δt − Tt Δt
element for heat conduction in the x and y directions are Ax = Δy ×1 and Ay = Δx ×1 , respectively,
and taking the limit as Δx, Δy, and Δt → 0 yields
1
∂ 2T + ∂ 2T = 1 ∂T ∂x2 ∂y 2 α ∂t
Properties: The thermal conductivity is given to be k = 20W / m ⋅ DC .
Analysis: (a) Noting that the upper part of the iron is well insulated and thus the entire heat generated in
the body is ρ and the specific heat is C. Noting that heat conduction is two-dimensional and assuming no heat generation ,an energy balance on this element during a small time interval Δt can expressed as
⎧Rate − of − heat
⎫ ⎧Rate − of − heat − conduction⎫ ⎧Rate − of − change − of ⎫
⎪⎨conduction − at − the
⎪ ⎬
−
⎪⎨at
−
the
−
surfaces
−
at
⎪ ⎬
=
⎪⎨the
−
energy
−
content
⎪ ⎬
2
/
(0.006
m⋅D C
−
x)m
x
+
85D
C
= 2500(0.006 − x) + 85
(c) The temperature at x=0 (the inner surface of the plate) is
T (0) = 2500(0.006 − 0) + 85 = 100DC
Note that the inner surface temperature is higher than the exposed surface temperature, as expected.
2-10 The variation of the absorption of solar energy in a solar pond with depth is given. A relation for the total rate of heat generation in a water layer at the top of the pond is to be 0
qD L κ
Substituting C1 and C2 into general solution, we variation of temperature is determined to be
Tx
=
•
− qD κ
x
+ T2
+
•
qD L
κ
=
•
qD
(L
−
κ
x) + T2
( ) =
50,000W / m 20W
Radiatio
determined.
L
Solar
x
pond
Assumptions: Absorption of solar radiation by water is modeled
as heat generation.
Analysis: The total rate of heat generation in a water layer of surface area A and thickness L at the top of
2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
•
x =0:
− κC1
=
•
qD
→
C1
=
−
qD κ
•
x= L:
T (L)
=
C1L + C2
= T2
→ C2
= T2
− C1L
→ C2
= T2
+
Assumptions: 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe.
direction, the mathematical formulation of this problem can be expressed as
⎪⎩surface − at − x − and − y⎪⎭ ⎪⎩x + Δxandy + Δy
⎪⎭ ⎪⎩of − the − element
⎪⎭
or
•
••
•
Qx + Qy − Q x+Δx − Q y+Δy
=
ΔEelement Δt
Noting that the volume of the element is Velement = ΔxΔyΔz = ΔxΔy ×1 , the change in the energy content
the pond id determined by integration to be
∫ ∫ ( ) •
•
G = gdV
V
L•
= gDe −bx
x=0
Adx
=
•
A gD
e −bx −b
•
L = A gD (1 − e−bL )
0
b
2-12 We consider a small rectangular element of length Δx , width Δy , and height Δz = 1 , the density of
2-16 The base plate of a household iron is subjected to specified heat flux on the left surface and to specified temperature on the right surface. The mathematical formulation, the variation of temperature in the plate, and the inner surface temperature are to be determined for steady one-dimensional heat transfer.
d 2T dx 2
=0
And
−κ
dT (0) dx
=
•
qD
=
50,000W
/ m2
T (L) = T2 = 85DC
(b) Integrating the differential equation twice with respect to x yields
dT dx
= C1
T (x) = C1x + C2
Dividing by ΔxΔy gives
•
•
•
•
− 1 Qx+Δz − Qx − 1 Qy+Δy − Q y = ρC Tt+Δt − Tt
Δy Δx
Δx Δy
Δt
Taking the thermal conductivity κ to be constant and noting that the heat transfer surface areas of the
the resistance wires is transferred to the base plate, the heat flux through the inner surface is determined to
be
•
•
qD
=
QD Abase
=
800W 160 ×10−4
m
2
= 50,000W / m2
⎟⎞ ⎠
=
−
∂ ∂x
⎜⎛ ⎝
k
∂T ∂x
⎟⎞ ⎠
=
−k
∂ 2T ∂x 2
•
•
lim 1 Q y+Δy − Q y Δy→0 ΔxΔz Δy
=
1 ∂Qy ΔxΔz ∂y
=
1 ΔxΔz
∂ ∂y
⎜⎜⎝⎛ −
kΔxΔz
∂T ∂y
⎟⎟⎠⎞
=
−
∂ ∂y
⎜⎜⎝⎛ k
∂T ∂y
⎟⎟⎠⎞
=
−k
∂ 2T ∂y 2
Here the property α = k / ρC is the thermal diffusivity of the material.
Since from the definition of the derivative and Fourier’s law of heat conduction,
•
•
lim 1 Q x+Δx − Q x Δx→0 ΔyΔz Δx
=
1 ΔyΔz
∂Qx ∂x
=
1 ΔyΔz
∂ ∂x
⎜⎛ ⎝
−
kΔyΔz
∂T ∂x