ch02
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ch02


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95°C
with insulation thickness around 10 mm. We conclude that by selecting the proper insulation
thickness and controlling the convection coefficient, the reaction could be operated around 95°C such
that the outer surface temperature would not exceed 45°C.
PROBLEM 2.44
KNOWN: One-dimensional system, initially at a uniform temperature Ti, is suddenly
exposed to a uniform heat flux at one boundary, while the other boundary is insulated.
FIND: (a) Proper form of heat equation and boundary and initial conditions, (b) Temperature
distributions for following conditions: initial condition (t \u2264 0), and several times after heater
is energized; will a steady-state condition be reached; (c) Heat flux at x = 0, L/2, L as a
function of time; (d) Expression for uniform temperature, Tf, reached after heater has been
switched off following an elapsed time, te, with the heater on.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) No internal heat generation, (3)
Constant properties.
ANALYSIS: (a) The appropriate form of the heat equation follows from Eq. 2.15. Also, the
appropriate boundary and initial conditions are:
Initial condition: T x,0 Ti\ufffd \ufffd = Uniform temperature
\u2202
\u2202 \u3b1
\u2202
\u2202
2 1T
 x
 T
 t2
= Boundary conditions: x q k T / x)o 0= \u2032\u2032 = \u22120 \u2202 \u2202
x L T / x)L= =\u2202 \u2202 0
(b) The temperature distributions are as follows:
<
No steady-state condition will be reached since \ufffd \ufffd \ufffdE E and Ein st in= is constant.
(c)The heat flux as a function of time for positions x = 0, L/2 and L is as follows:
<
(d) If the heater is energized until t = te and then switched off, the system will eventually
reach a uniform temperature, Tf. Perform an energy balance on the system, Eq. 1.11b, for
an interval of time \u2206t = te,
E E E Q q A dt q A t E Mc T Tin st in in o s
te
o s e st f i= = = \u2032\u2032 = \u2032\u2032 = \u2212\ufffd0 \ufffd \ufffd
It follows that \u2032\u2032 = \u2212 = + \u2032\u2032q A t Mc T T or T T q A t
Mco s e f i f i
o s e\ufffd \ufffd . <
PROBLEM 2.45
KNOWN: Plate of thickness 2L, initially at a uniform temperature of Ti = 200°C, is suddenly
quenched in a liquid bath of T\u221e = 20°C with a convection coefficient of 100 W/m
2
\u22c5K.
FIND: (a) On T-x coordinates, sketch the temperature distributions for the initial condition (t \u2264 0), the
steady-state condition (t \u2192 \u221e), and two intermediate times; (b) On xq t\u2032\u2032 \u2212 coordinates, sketch the
variation with time of the heat flux at x = L, (c) Determine the heat flux at x = L and for t = 0; what is
the temperature gradient for this condition; (d) By performing an energy balance on the plate,
determine the amount of energy per unit surface area of the plate (J/m2) that is transferred to the bath
over the time required to reach steady-state conditions; and (e) Determine the energy transferred to the
bath during the quenching process using the exponential-decay relation for the surface heat flux.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, and (3) No internal heat
generation.
ANALYSIS: (a) The temperature distributions are shown in the sketch below.
(b) The heat flux at the surface x = L, ( )xq L, t ,\u2032\u2032 is initially a maximum value, and decreases with
increasing time as shown in the sketch above.
(c) The heat flux at the surface x = L at time t = 0, ( )xq L, 0 ,\u2032\u2032 is equal to the convection heat flux with
the surface temperature as T(L,0) = Ti.
( ) ( ) ( ) ( )2 2x conv iq L, 0 q t 0 h T T 100 W / m K 200 20 C 18.0 kW / m\u221e\u2032\u2032 \u2032\u2032= = = \u2212 = \u22c5 \u2212 ° = <
From a surface energy balance as shown in the sketch considering the conduction and convection
fluxes at the surface, the temperature gradient can be calculated.
Continued \u2026..
PROBLEM 2.45 (Cont.)
in outE E 0\u2212 =\ufffd \ufffd
( ) ( ) ( )x conv x
x L
Tq L,0 q t 0 0 with q L,0 k
x
=
\u2202 \uf8f6
\u2032\u2032 \u2032\u2032 \u2032\u2032
\u2212 = = = \u2212 \uf8f7\u2202 \uf8f8
( ) 3 2conv
L,0
T q t 0 / k 18 10 W / m / 50 W / m K 360K / m
x
\u2202 \uf8f6
\u2032\u2032= \u2212 = = \u2212 × \u22c5 = \u2212\uf8f7\u2202 \uf8f8 <
(d) The energy transferred from the plate to the bath over the time required to reach steady-state
conditions can be determined from an energy balance on a time interval basis, Eq. 1.11b. For the
initial state, the plate has a uniform temperature Ti; for the final state, the plate is at the temperature of
the bath, T\u221e.
in out st f i inE E E E E with E 0,\u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032\u2212 = \u2206 = \u2212 =
( )[ ]out p iE c 2L T T\u3c1 \u221e\u2032\u2032\u2212 = \u2212
( )[ ]3 6 2outE 2770kg / m 875J / kg K 2 0.010 m 20 200 K 8.73 10 J / m\u2032\u2032 = \u2212 × \u22c5 × \u2212 = + × <
(e) The energy transfer from the plate to the bath during the quenching process can be evaluated from
knowledge of the surface heat flux as a function of time. The area under the curve in the ( )xq L, t\u2032\u2032 vs.
time plot (see schematic above) represents the energy transferred during the quench process.
( ) Btout xt 0 t 0E 2 q L, t dt 2 Ae dt
\u221e \u221e
\u2212
= =
\u2032\u2032 \u2032\u2032= =\u222b \u222b
( )Btout
0
1 1E 2A e 2A 0 1 2A / B
B B
\u221e
\u2212
\uf8ee \uf8f9 \uf8ee \uf8f9
\u2032\u2032 = \u2212 = \u2212 \u2212 =\uf8ef \uf8fa \uf8ef \uf8fa\uf8f0 \uf8fb \uf8f0 \uf8fb
4 2 3 1 6 2
outE 2 1.80 10 W / m / 4.126 10 s 8.73 10 J / m
\u2212 \u2212
\u2032\u2032 = × × × = × <
COMMENTS: (1) Can you identify and explain the important features in the temperature
distributions of part (a)?
(2) The maximum heat flux from the plate occurs at the instant the quench process begins and is equal
to the convection heat flux. At this instant, the gradient in the plate at the surface is a maximum. If
the gradient is too large, excessive thermal stresses could be induced and cracking could occur.
(3) In this thermodynamic analysis, we were able to determine the energy transferred during the
quenching process. We cannot determine the rate at which cooling of the plate occurs without solving
the heat diffusion equation.
PROBLEM 2.46
KNOWN: Plane wall, initially at a uniform temperature, is suddenly exposed to convective heating.
FIND: (a) Differential equation and initial and boundary conditions which may be used to find the
temperature distribution, T(x,t); (b) Sketch T(x,t) for these conditions: initial (t \u2264 0), steady-state, t \u2192
\u221e, and two intermediate times; (c) Sketch heat fluxes as a function of time for surface locations; (d)
Expression for total energy transferred to wall per unit volume (J/m3).
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal heat
generation.
ANALYSIS: (a) For one-dimensional conduction with constant properties, the heat equation has the
form,
\u2202
\u2202 \u3b1
\u2202
\u2202
2 1T
 x
 T
 t2
=
Initial, t T x,0 T uniform
Boundaries: x = 0 T / x) adiabatic
 x = L k T / x) = h T L, t T convection
i
0
L
\u2264 =
=
\u2212 \u2212
\ufffd
\ufffd
\ufffd\ufffd
\ufffd
\ufffd
\ufffd \u221e
0
0
: \ufffd \ufffd
\ufffd \ufffd
\u2202 \u2202
\u2202 \u2202
(b) The temperature distributions are shown on the sketch.
Note that the gradient at x = 0 is always zero, since this boundary is adiabatic. Note also that the
gradient at x = L decreases with time.
(c) The heat flux, \u2032\u2032q x, tx\ufffd \ufffd, as a function of time, is shown on the sketch for the surfaces x = 0 and x
= L.
Continued \u2026..
and the
conditions are:
PROBLEM 2.46 (Cont.)
For the surface at x q tx= \u2032\u2032 =0 0 0, ,\ufffd \ufffd since it is adiabatic. At x = L and t = 0, \u2032\u2032q L,0x\ufffd \ufffd is a
maximum
\u2032\u2032 = \u2212
\u221e
q L,0 h T L,0 Tx\ufffd \ufffd \ufffd \ufffd
where T(L,0) = Ti. The gradient, and hence the flux, decrease with time.
(d) The total energy transferred to the wall may be expressed as
E q A dt
E hA T T L, t dt
in conv s
in s
= \u2032\u2032
= \u2212
\u221e
\u221e
\u221e
\ufffd
\ufffd
0
0 \ufffd \ufffd\ufffd 	
Dividing both sides by AsL, the energy transferred per unit volume is
E
V
h
L
T T L, t dt J / min 3= \u2212
\u221e
\u221e
\ufffd \ufffd \ufffd0
COMMENTS: Note that the heat flux at x = L is into the wall and is hence in the negative x
direction.
PROBLEM 2.47
KNOWN: Plane wall, initially at a uniform temperature Ti, is suddenly exposed to convection with a
fluid at