Appendix I Heisler Charts for Transient Heat Conduction Problems.mdi

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C H A P T E R 4 Unsteady-State Conduction 143
Table 4-1 Examples of lumped-capacity systems.
Approximate
value of h,
Physical situation k, W/m · ◦C W/m2 · ◦C
h(V/A)
k
1. 3.0-cm steel cube cooling in room air 40 7.0 8.75× 10−4
2. 5.0-cm glass cylinder cooled by a 50-m/s airstream 0.8 180 2.81
3. Same as situation 2 but a copper cylinder 380 180 0.006
4. 3.0-cm hot copper cube submerged in water 380 10,000 0.132
such that boiling occurs
We may point out that uncertainties in the knowledge of the convection coefficient of
±25 percent are quite common, so that the condition Bi=h(V/A)/k< 0.1 should allow fo
some leeway in application.
Do not dismiss lumped-capacity analysis because of its simplicity. Because of uncer-
tainties in the convection coefficient, it may not be necessary to use more elaborate analysis
techniques.
Steel Ball Cooling in Air EXAMPLE 4-1
A steel ball [c= 0.46 kJ/kg · ◦C, k= 35 W/m · ◦C] 5.0 cm in diameter and initially at a uniform
temperature of 450◦C is suddenly placed in a controlled environment in which the temperature
is maintained at 100◦C. The convection heat-transfer coefficient is 10 W/m2 · ◦C. Calculate the
time required for the ball to attain a temperature of 150◦C.
Solution
We anticipate that the lumped-capacity method will apply because of the low value of h and high
value of k. We can check by using Equation (4-6):
h(V/A)
k
= (10)[(4/3)π(0.025)
3]
4π(0.025)2(35)
= 0.0023 < 0.1
so we may use Equation (4-5). We have
T = 150◦C ρ= 7800 kg/m3 [486 lbm/ft3]
T∞ = 100◦C h= 10 W/m2 · ◦C [1.76Btu/h · ft2 · ◦F]
T0 = 450◦C c= 460 J/kg · ◦C [0.11 Btu/lbm · ◦F]
hA
ρcV
= (10)4π(0.025)
2
(7800)(460)(4π/3)(0.025)3
= 3.344× 10−4 s−1
T − T∞
T0− T∞ = e
−[hA/ρcV ]τ
150− 100
450− 100 = e
−3.344×10−4τ
τ = 5819 s= 1.62 h
4-3 TRANSIENT HEAT FLOW IN A
SEMI-INFINITE SOLID
Consider the semi-infinite solid shown in Figure 4-3 maintained at some initial temperature
Ti. The surface temperature is suddenly lowered and maintained at a temperature T0, and we
144 4-3 Transient Heat Flow in a Semi-Infinite Solid
Figure 4-3 Nomenclature for transient
heat flow in a semi-infinite
solid.
qo = –kA x=0
∂T
∂x
T1
T0
x
seek an expression for the temperature distribution in the solid as a function of time. This
temperature distribution may subsequently be used to calculate heat flow at any x position
in the solid as a function of time. For constant properties, the differential equation for the
temperature distribution T(x, τ) is
∂2T
∂x2
= 1 ∂T
α ∂τ
[4-7]
The boundary and initial conditions are
T(x, 0)= Ti
T(0, τ)= T0 for τ > 0
This is a problem that may be solved by the Laplace-transform technique. The solution is
given in Reference 1 as
T(x, τ)− T0
Ti− T0 = erf
x
2
√
ατ
[4-8]
where the Gauss e
or function is defined as
erf
x
2
√
ατ
= 2√
π
∫ x/2√ατ
e−η2 dη [4-9]
It will be noted that in this definition η is a dummy variable and the integral is a function of
its upper limit. When the definition of the e
or function is inserted in Equation (4-8), the
expression for the temperature distribution becomes
T(x, τ)− T0
Ti− T0 =
2√
π
∫ x/2√ατ
e−η2 dη [4-10]
The heat flow at any x position may be obtained from
qx=−kA∂T
∂x
Performing the partial differentiation of Equation (4-10) gives
∂T
∂x
= (Ti− T0) 2√
π
e−x2/4ατ ∂
∂x
(
x
2
√
ατ
)
[4-11]
= Ti− T0√
πατ
e−x2/4ατ
C H A P T E R 4 Unsteady-State Conduction 145
Figure 4-4 Response of semi-infinite solid to (a) sudden change in surface temperature and
(b) instantaneous surface pulse of Q0/A J/m2.
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
(a)
0.5 1 1.5 2 2.5
x
2 α τ x
2
4ατ
τ (sec)
0.01
0.05
0.1
1.0
5
20
100
0.01 0.1
0.000001
0.00001
0.0001
0.001
0.01
0.1
1
10
(b)
1 10
At the surface (x= 0) the heat flow is
q0= kA(T0− Ti)√
πατ
[4-12]
The surface heat flux is determined by evaluating the temperature gradient at x= 0 from
Equation (4-11). A plot of the temperature distribution for the semi-infinite solid is given
in Figure 4-4. Values of the e
or function are tabulated in Reference 3, and an a
eviated
tabulation is given in Appendix A.
Constant Heat Flux on Semi-Infinite Solid
For the same uniform initial temperature distribution, we could suddenly expose the surface
to a constant surface heat flux q0/A. The initial and boundary conditions on Equation (4-7)
would then become
T(x, 0)= Ti
q0
A
=−k ∂T
∂x
]
x=0
for τ > 0
The solution for this case is
T − Ti= 2q0
√
ατ/π
kA
exp
(−x2
4ατ
)
− q0x
kA
(
1− erf x
2
√
ατ
)
[4-13a]
Energy Pulse at Surface
Equation (4-13a) presents the temperature response that results from a surface heat flux that
emains constant with time. A related boundary condition is that of a short, instantaneous
pulse of...