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4ChEB Group 8

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Information about 4ChEB Group 8

Published on January 7, 2009

Author: 4ChEAB08

Source: slideshare.net

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DE GUZMAN, John Wilbert HIZON, Donn Angelo M. PEDRIGAL, Ian Sygfryd REYES, 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 1100 o C and the outside surface is at 80 o C. a. Calculate the total heat loss for 24hrs operation. b. What is the heat flux and temperature at a radius of 1.1 m?

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 1100 o C and the outside surface is at 80 o C.

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: r i =1m r o =1.2m k=0.5 W/mK

a. For spherical section: Am = 4 π r i r o = 4 π (1) (1.2) = 15.079 m 2 Δ x = r o – r i = 1.2 – 1 = 0.2 m

q = 38451.534 W

b. = 2781.7873 W/m 2 T’ = 543.70 o C

#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/m 2 • K. Assume the solid has a thermal diffusivity of 0.000025 m 2 /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.

(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/m 2 • K. Assume the solid has a thermal diffusivity of 0.000025 m 2 /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: Δ x= 0.25 m Δ x 1 2 3 4 f insulation q T a = 250˚C α = 0.000025 m 2 /s κ= 20 W/ m•K h= 40 W/ m 2 •K

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

Temp profile of time, t=4000s

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: n=1 (CONVECTIVE) t+Δt T 1 = (1/5) [(2)(0.5) t T a + {5-[(2)(0.5)+2]} t T 1 +2 t T 2 ] t+Δt T 1 = 50 + (0.4) t T 1 +(0.4) t T 2   b. n= 2 to 4 t+Δt T n = (1/5) [ t T n-1 + (5-2) t T n + t T n+1 ]   t+Δt T n = (0.2) t T n-1 + (0.6) t T n +(0.2) t T n+1   c. n=f (INSULATION) t+Δt T f = (1/5) [(5-2) t T f + t T f-1 ]   t+Δt T f = (0.6) t T f +(0.4) t T f-1    

n=1 (CONVECTIVE)

t+Δt T 1 = (1/5) [(2)(0.5) t T a + {5-[(2)(0.5)+2]} t T 1 +2 t T 2 ]

t+Δt T 1 = 50 + (0.4) t T 1 +(0.4) t T 2

 

b. n= 2 to 4

t+Δt T n = (1/5) [ t T n-1 + (5-2) t T n + t T n+1 ]

 

t+Δt T n = (0.2) t T n-1 + (0.6) t T n +(0.2) t T n+1

 

c. n=f (INSULATION)

t+Δt T f = (1/5) [(5-2) t T f + t T f-1 ]

 

t+Δt T f = (0.6) t T f +(0.4) t T f-1

 

 

ANSWERS: 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

GEANKOPLIS: Problem 4.1-1 Insulation in a Cold Room. Calculate the heat loss per m 2 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: Δ x Δ x= 0.0254 m T 1 = 299.9 K T 2 = 276.5 K Basis: A=1 m 2 Req’d: q q q= 39.89 W/m 2

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-ft 2 -°F. The physical properties of the brick are k=0.65 btu/h-ft-°F and α=0.02 ft 2 /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°F T1= 1100°F h= 2.6 btu/h-ft 2 -°F k=0.65 btu/h-ft-°F α=0.02 ft 2 /h x 1 = 0.5 ft

To= 100°F

T1= 1100°F

h= 2.6 btu/h-ft 2 -°F

k=0.65 btu/h-ft-°F

α=0.02 ft 2 /h

x 1 = 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/(x 1 ) 2 t = 0.4(x 1 ) 2 / α = 0.4(0.5) 2 /0.2 = 0.05h = 3mins

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