Answers of Cn2125

PROBLEM 15.15 (WWWR) A 0.20-m-thick brick wall (k=1.3W/m•K) separated the combustion zone of a furnace from its surroundings at 25℃. For an outside wall surface temperature of 100℃, with a convective heat transfer coefficient of 18 W/m2•K, what will be the inside wall surface temperature at steady-state conditions?

Solution :

q k = Tw = h(Ts − Tsurrounding ) A L
Tw =
=

h(Ts − Tsurrounding )i L k

Tinside =?

0.2m

Ts = 100°c

(18W / m.k )(75k )(20 × 10−2 m) 1.3(W / m.k )

Tsurrounding = 25°c

= 207.7k

Tinside = 100 + Tw = 307.7 C

PROBLEM 15.21 (WWWR)

The cross section of a storm window is shown in the sketch. How much heat will be lost through a window measuring 1.83m by 3.66 m on a cold day when the inside and outside air temperatures are, respectively, 295K and 250K? Convective coefficients on the inside and outside surfaces of a window are 20 W/m2•K and 15 W/m2•K, respectively. What temperature drop will exist across each of the glass panes? What will be the average temperature of the air between the glass panes?

Air space 0.8cm wide

Window glass--0.32 cm thick

Solution:

250k hi

Air space 0.8cm wide

295k h0

Window glass 0.32 cm thick

Ti

To

Ri

RGL

RAS

RGL

Ro

q=

T =? ∑R

Ri =

1 = 7.46 × 10−3 (20)(1.83)(3.66) 0.0032 = 6.125 × 10−4 (0.78)(1.83)(3.66) 0.008 = 0.0488 (0.0245)(1.83)(3.66)

RGL = RAS = R0 =

1 = 9.95 × 10−3 (15)(1.83)(3.66)

∑ R = 0.06744 q= T 45 = = 667W ∑ R 0.06744

PROBLEM 15.22 (WWWR) Compare the heat lost through the storm window described in problem 15.25 with the same conditions existing except that the window is a single pane of glass 0.32 cm thick. Solution:

For single pane of glass only ∑ R = Ri + RGL + RO = 0.018 q= T 45 = = 2500W ∑ R 0.018

PROBLEM 2.2 (ID)
KNOWN: Hot water pipe covered with thick layer of insulation. FIND: Sketch temperature distribution and give brief explanation to justify shape. SCHEMATIC:

ASSUMPTIONS: (1)