1. DE GUZMAN, John Wilbert
HIZON, Donn Angelo M.
PEDRIGAL, Ian Sygfryd
REYES, Mervick Ann B.
TINDUGAN, Farrah Kaye Z,
4 ChEB Group 8

2. #8.
A spherical furnace has an inside radius of 1 m, and an
outside radius of 1.2 m. The thermal conductivity of
the wall is 0.5 W/mK. The inside furnace
temperature is 1100oC and the outside surface is at
80oC.
 a. Calculate the total heat loss for 24hrs operation.
 b. What is the heat flux and temperature at a radius of
1.1 m?

3. Given: ri=1m ro=1.2m
k=0.5 W/mK

4. a.
For spherical section:
Am = 4 π ri ro
= 4 π (1) (1.2)
= 15.079 m2
Δ x = ro – ri
= 1.2 – 1
= 0.2 m

5. q = 38451.534 W

6. b.
= 2781.7873 W/m2
T’ = 543.70 oC

7. #16.
 (US) A large slab 1m thick is initially at a uniform
temperature of 150˚C. Suddenly its front face is exposed
to a fluid maintained at 250 ˚C, but its rear face
remained insulated. The fluid has a convective
coefficient of 40W/m2•K. Assume the solid has a thermal
diffusivity of 0.000025 m2/s and a thermal conductivity
of 20W/m•K.
 Using a numerical finite difference method with M=5 and
4 slices, construct a table for the temperature profile of
the slab up to a time of 4000 sec.

8. Given:
Δx= 0.25 m
Δx
1 2 3 4 f
insulation
q
Ta= 250˚C
α= 0.000025 m2/s
κ= 20 W/ m•K
h= 40 W/ m2•K

9. Required:
 Temp profile of time, t=4000s
Solution:

10. 1 2 3 4 f
0 s
500 s
1000 s
1500 s
2000 s
2500 s
3000 s
3500 s
4000 s

11. WORKING EQUATIONS:
a. n=1 (CONVECTIVE)
t+ΔtT1= (1/5) [(2)(0.5)tTa + {5-[(2)(0.5)+2]}tT1 +2tT2]
t+ΔtT1= 50 + (0.4)tT1 +(0.4)tT2
b. n= 2 to 4
t+ΔtTn= (1/5) [tTn-1 + (5-2)tTn +tTn+1]
t+ΔtTn= (0.2)tTn-1 + (0.6)tTn +(0.2)tTn+1
c. n=f (INSULATION)
t+ΔtTf= (1/5) [(5-2)tTf + tTf-1]
t+ΔtTf= (0.6)tTf +(0.4) tTf-1

12. 1 2 3 4 f
0 s 150 150 150 150 150
500 s 170 150 150 150 150
1000 s 178 154 150 150 150
1500 s 182.8 158 150.8 150 150
2000 s 186.32 161.52 152.08 150.16 150
2500 s 189.136 164.592 153.584 150.512 150.064
3000 s 191.4912 167.2992 155.1712 151.0368 150.2432
3500 s 193.5162 169.712 156.7699 151.705 150.5606
4000 s 195.2913 171.8844 158.3453 152.4891 151.0184
ANSWERS:

13. GEANKOPLIS:
Problem 4.1-1
Insulation in a Cold Room. Calculate the heat loss per m2 of
surface area for a temporary insulating wall of a food cold
storage room where the outside temperature is 299.9 K and
the inside temperature is 276.5 K. The wall is composed of 25.4
mm of corkboard having a k of 0.0433 W/m•K
Given:
Δx
Δx= 0.0254 m
T1= 299.9 K
T2= 276.5 K
Basis: A=1 m2
Req’d: q
q
q= 39.89 W/m2

14. Problem 5.3-6
A flat brick wall 1.0 ft thick is the lining on one side of a
furnace. If the wall is at uniform temperature of 100°F and
one side is suddenly exposed to a gas at 1100°F, calculate
the time for the furnace wall at a point 0.5 ft from the
surface to reach 500°F. The rear side of the wall is
insulated. The convection coefficient is 2.6 btu/h-ft2
-°F.
The physical properties of the brick are k=0.65 btu/h-ft-°F
and α=0.02 ft2/h.

15. brick wall
To=100F
q
required: time for the wall at a point 0.5ft from the surface to reach 500F
0.5ft
1.0 ft
furnace
T1= 1100F

16. Given:
 To= 100°F
 T1= 1100°F
 h= 2.6 btu/h-ft2
-°F
 k=0.65 btu/h-ft-°F
 α=0.02 ft2/h
 x1= 0.5 ft

17. Solution:
m= k/(hx1) =0.65/(2.6 x 05) = 0.5
Y= (Ti-T)/(T1-To) = (1100-500)/(1100-100) = 0.6
Plotting these on Fig.5.3-6(Heisler),
x≈ 0.4 x= αt/(x1)2
t = 0.4(x1)2/ α = 0.4(0.5)2/0.2 = 0.05h = 3mins