10.3 Solutions
437
Neglecting the area of the faces, area of the cylinder A = 2πr L. For steady
state, dQ/dt = constant. We can then write (1) as
dr
r
= −
2π Lk
dQ/dt
dT
Integrating
r 2
r 1
dr
r
= ln(r 2 /r 1 ) = −
2π Lk
dQ/dt
(T 2 − T 1 )
or
dQ/dt =
2π Lk
ln(r 2 /r 1 )
(T 1 − T 2 )
10.26 dQ/dt = k A(T 1 − T 2 )/d
= (0.59A) ×
10
0.01
= 590 A J/s
( 1 )
Let x m/s ice be added at the bottom of the layer.
Mass of ice formed per second
M = ρ × A
(2)
The required energy per second
E = ρ × AL
(3)
Equating (1) and (3), ρ × AL = 590 A
∴
x =
590A
ρ AL
=
590
917 × 333 × 10 3 = 1.932×10
−6 m/s = 0.00695 m/h
10.27 −
dθ
dt
= C(θ − θ 0 )
(Newton’s law of cooling)
θ 1 − θ 2
t
= C
θ 1 + θ 2
2
− θ 0
(C = constant)
85 − 75
2
= C
85 + 75
2
− 30
∴
C = 0.1
55 − 45
t
= 0.1
55 + 45
2
− 30
∴
t = 5 min