Page 275 - Schaum's Outline of Theory and Problems of Applied Physics

260 HEAT TRANSFER [CHAP. 22

According to the formula for Q/t, when the ratio k/d is the same for the two materials, their insulating abilities
will also be the same. Hence

k brick 0.6
d brick = d pine = (5cm) = 23 cm
k pine 0.13
SOLVED PROBLEM 22.2
The thermal conductivity of ice is 5.2 ×10 −4 kcal/(m·s· C). At what rate is heat lost by the water in a
◦
6-m by 10-m outdoor swimming pool covered by a layer of ice 1 cm thick if the water is at a temperature
of 0 C and the surrounding air is at a temperature of −10 C?
◦
◦
2
◦
Here A = (6 m)(10 m) = 60 m , d = 0.01 m, and T = 10 C. Hence
2
◦
Q k A T kcal (60 m )(10 C)
= = 5.2 × 10 −4 = 31.2 kcal/s
t d m·s· C 0.01 m
◦
SOLVED PROBLEM 22.3
The handle of a freezer door 5 in. thick is attached by two brass bolts 1 in. in diameter that pass through
4
the entire door and are secured on the inside by nuts. The interior of the freezer is maintained at 0 F, and
◦
2
◦
◦
the room temperature is 65 F; the thermal conductivity of brass is 730 Btu/(ft ·h· F/in.). Find the heat
lost per hour through the bolts.
The total cross-sectional area of the two bolts is
2
0.125 in.

2
A = (2)(πr ) = (2)(π) = 6.82 × 10 −4 ft 2
12 in./ft
and their length is d = 5 in. Since T = 65 F,
◦
2
Q k A T Btu (6.82 × 10 −4 ft )(65 F)
◦
= = 730 = 6.47 Btu/h
2
t d ft ·h· F/in. 5in.
◦
THERMALRESISTANCE
The thermal resistance R of a layer of a material of thickness d and of thermal conductivity k is given by
d
R =
k
thickness
Thermal resistance =
thermal conductivity
The greater the value of R, the greater the resistance to the flow of heat. In British units, thermal resistance is given
2 ◦
in ft · F/(Btu/h), but it is common for R-values to be stated without units. Thus an R-value of 1.5 ft · F/(Btu/h)
2 ◦
would normally be given as just “R–1.5.” In terms of R, the rate of heat conduction is
Q A T
=
t R
An advantage of using thermal resistance is that the R-value for a sandwich consisting of two or more layers
is just the sum of the R-values of the separate layers:

R = R 1 + R 2 + R 3 + ···