# 4 A Group 8

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Information about 4 A Group 8

Published on January 6, 2009

Author: 4ChEAB08

Source: slideshare.net

## Description

Problem Set #2
4ChE A
Group #8

Assignment No. 2 Group #8 Rachelle Manalo Ronna Anne Oyco Jonas Jane Aguila Albert Jim Gonzales Joeward Manaig

PROBLEM NO. 1 A spherical furnace has an inside radius of 1m, 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°C and the outside surface is at 80°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 1m, 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°C and the outside surface is at 80°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.

PROBLEM NO. 1 - Given r 1 = 1m r 2 =1.2 m k= 0.5 W/mK T 1 =1100°C T 2 =80°C Δx= 1.2- 1 = 0.2

r 1 = 1m

r 2 =1.2 m

k= 0.5 W/mK

T 1 =1100°C

T 2 =80°C

Δx= 1.2- 1 = 0.2

PROBLEM NO. 1 - Solution a.) For a spherical section: A m = 4 π r 1 r 2 = 4 π x1x1.2 m = 15.0796m 2 q= T 1 - T 2 Δx kA m = [ (1100°C + 273) – (80°C + 273) ] K 0.2m 0.5 W x 15.0796m 2 m-K = 38453.09408 J x 3600s x 24hrs. s 1 hr = 3322347329 J ≈ 3.32 x 10 9 J

PROBLEM NO. 1 - Solution b.) At r= 1.1m, T=? q = 38453.09408 J/s A m 4 π (1)(1.1) m 2 = 2781.818182 J s-m 2 q = k T 1 - T 2 A m Δx 2781.818182 J = 0.5 W x [ (1100+273)-(T+273)] K s-m 2 m-K 0.1m T(°C) = 543.6363636 ≈ 543.64 °C

PROBLEM NO. 1 - Answer a.) q = 3.32 x 109 J b.) heat flux = 543.64 °C

a.) q = 3.32 x 109 J

b.) heat flux = 543.64 °C

PROBLEM NO. 16 Application of Unsteady State Conduction 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.

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.

PROBLEM NO. 16 - Given x = 1 m T 0 = 100 ° C T a = 250 ° C H = 40 W/m 2- K α = 0.000025 m 2 /s k = 20 W/m-K M = 5 Slices = 4 Δ x = 0.25 m T = 4000 seconds 1 m Insulated Face SLAB 0.25 m Required: table for the temperature profile of the slab

x = 1 m

T 0 = 100 ° C

T a = 250 ° C

H = 40 W/m 2- K

α = 0.000025 m 2 /s

k = 20 W/m-K

M = 5

Slices = 4

Δ x = 0.25 m

T = 4000 seconds

PROBLEM NO. 16 - Solution N = h Δ x k N = (40)(0.25) 20 N = 0.5 # of time intervals = ? M = Δ x 2 k Δ t 5 = (0.25) 2 m 2 (0.000025 m 2 /s)( Δ t) Δ t = 500 seconds # of time intervals = 4000 s 500 s # of time intervals = 8

N = h Δ x

k

N = (40)(0.25)

20

N = 0.5

PROBLEM NO. 16 - Solution General Working Equation: [ t T n-1 + (M-2) t T n + t T n+1 ] t+ Δ t T n = M Convective Resistance at the Boundary: t+ Δ t T 1 = 1 / M [2N t T a + [M – (2N + 2)] T 1 + 2 t T 2 ] Insulated Boundary Condition t+ Δ t T f = 1 / M [(M – 2) t T f ] + 2 t T f-1 ]

General Working Equation:

[ t T n-1 + (M-2) t T n + t T n+1 ]

t+ Δ t T n =

M

Convective Resistance at the Boundary:

t+ Δ t T 1 = 1 / M [2N t T a + [M – (2N + 2)] T 1 + 2 t T 2 ]

Insulated Boundary Condition

t+ Δ t T f = 1 / M [(M – 2) t T f ] + 2 t T f-1 ]

PROBLEM NO. 16- Solution General Working Equation: [1 t T n-1 + 3 t T n + 1 t T n+1 ] t+ Δ t T n = 5 Convective Resistance at the Boundary: t+ Δ t T 1 = 1 / 5 [2(0.5) t T a + [5 – (2(0.5) + 2)] T 1 + 2 t T 2 ] Insulated Boundary Condition t+ Δ t T f = 1 / 5 [(5 – 2) t T f ] + 2 t T f-1 ] 1 / 5 – 3 / 5 – 1 / 5 1 / 5 – 2 / 5 – 2 / 5 3 / 5 – 2 / 5

General Working Equation:

[1 t T n-1 + 3 t T n + 1 t T n+1 ]

t+ Δ t T n =

5

Convective Resistance at the Boundary:

t+ Δ t T 1 = 1 / 5 [2(0.5) t T a + [5 – (2(0.5) + 2)] T 1 + 2 t T 2 ]

Insulated Boundary Condition

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

PROBLEM NO. 16 - Answer 152.1723 153.6640 160.4991 175.2437 197.6959 4000 151.4057 152.6833 158.8643 173.2194 196.0203 3500 150.7924 151.8146 157.1806 170.9653 194.0855 3000 150.3533 151.0852 155.4640 168.4258 191.7880 2500 150.0960 150.5248 153.7555 165.5284 188.9414 2000 150.0000 150.1600 152.1440 162.1856 185.1680 1500 150.0000 150.0000 150.8000 158.3200 179.6000 1000 150.0000 150.0000 150.0000 154.0000 170.0000 500 150 150 150 150 150 0 f 4 3 2 1 time (s)

PROBLEM NO. 4.1-1 Insulation in a Cold Room Calculate the heat loss per of surface for a temporary insulating wall of a food cold storage room where the outside temperature is 299.9 K and the inside temperature of 276.5 K. The wall is composed of 25.4 mm of corkboard having a k of 0.0433 W/m∙K.

Calculate the heat loss per of surface for a temporary insulating wall of a food cold storage room where the outside temperature is 299.9 K and the inside temperature of 276.5 K. The wall is composed of 25.4 mm of corkboard having a k of 0.0433 W/m∙K.

PROBLEM NO. 16 – Given and Required Given: A= 1 k = 0.0433 W/m∙K T 1 = 299.9 K T 2 = 276. 5 K Wall Thickness = 25.4 mm Required: heat loss per m 2 ( q / a )

Given:

A= 1 k = 0.0433 W/m∙K

T 1 = 299.9 K

T 2 = 276. 5 K

Wall Thickness = 25.4 mm

Required: heat loss per m 2 ( q / a )

PROBLEM NO. 16 – Solution

PROBLEM NO. 16 – Units

PROBLEM 5.3-6 Unsteady State Conduction in a Brick Wall A flat brick wall 1.0 ft thick is the lining on one side of a furnace. If the wall is at a 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 and the physical properties of the brick are k = 0.65 btu/h-ft- ° F and α = 0.02 ft 2 /h.

A flat brick wall 1.0 ft thick is the lining on one side of a furnace. If the wall is at a 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 and the physical properties of the brick are k = 0.65 btu/h-ft- ° F and α = 0.02 ft 2 /h.

PROBLEM 5.3-6 - Given x = 1.0 ft x 1 = 0.5 ft T o = 100 ° F T 1 = 1100 ° F T = 500 ° F h = 2.6 btu/h-ft 2 - ° F k = 0.65 btu/h-ft- ° F α = 0.02 ft 2 /h Required: time for the furnace wall at a point 0.5 ft from the surface reach 500°F

x = 1.0 ft

x 1 = 0.5 ft

T o = 100 ° F

T 1 = 1100 ° F

T = 500 ° F

h = 2.6 btu/h-ft 2 - ° F

k = 0.65 btu/h-ft- ° F

α = 0.02 ft 2 /h

PROBLEM 5.3-6 - Solution Y = ? T 1 - T T 1 - T 0 1100 ° F - 500 ° F 1100 ° F - 100 ° F Y = Y = Y = 0.6

Y = ?

T 1 - T

T 1 - T 0

1100 ° F - 500 ° F

1100 ° F - 100 ° F

PROBLEM 5.3-6 - Solution k hx 1 0.65 btu/h-ft- ° F (0.5 ft)(2.6 btu/h-ft 2 - ° F) m = 0.5 m = m =

k

hx 1

0.65 btu/h-ft- ° F

(0.5 ft)(2.6 btu/h-ft 2 - ° F)

m = 0.5

PROBLEM 5.3-6 - Solution α t x 1 2 ( 0.02 ft 2 /h)(t) (0.5) 2 ft 2 t = 7.5 hours X = X = 0.6 (from Heisler Chart) 0.6 =

α t

x 1 2

( 0.02 ft 2 /h)(t)

(0.5) 2 ft 2

t = 7.5 hours

PROBLEM 5.3-6 - Answer The time for the furnace wall at a point 0.5 ft from the surface to reach 500°F is approximately 7.5 hours .

The time for the furnace wall at a point 0.5 ft from the surface to reach 500°F is approximately 7.5 hours .

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