DE GUZMAN, John WilbertHIZON, Donn Angelo M.PEDRIGAL, Ian SygfrydREYES, Mervick Ann B.

TINDUGAN, Farrah Kaye Z,

4 ChEB Group 8

#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?

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

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

q = 38451.534 W

b.

= 2781.7873 W/m2

T’ = 543.70 oC

#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.

Given:

Δx= 0.25 m

Δx

1 2 3 4 f

insulation

q

Ta= 250˚C

α= 0.000025 m2/sκ= 20 W/ m•Kh= 40 W/ m2•K

Required:Temp profile of time, t=4000s

Solution:

1 2 3 4 f

0 s

500 s

1000 s

1500 s

2000 s

2500 s

3000 s

3500 s

4000 s

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  

 

1 2 3 4 f

0 s 150 150 150 150 150

500 s 170 150 150 150 1501000 s 178 154 150 150 1501500 s 182.8 158 150.8 150 1502000 s 186.32 161.52 152.08 150.16 1502500 s 189.136 164.592 153.584 150.512 150.0643000 s 191.4912 167.2992 155.1712 151.0368 150.24323500 s 193.5162 169.712 156.7699 151.705 150.56064000 s 195.2913 171.8844 158.3453 152.4891 151.0184

ANSWERS:

GEANKOPLIS:Problem 4.1-1Insulation 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

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.

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

Given:To= 100°FT1= 1100°Fh= 2.6 btu/h-ft2-°Fk=0.65 btu/h-ft-°F α=0.02 ft2/hx1= 0.5 ft

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