# Group 6-4ChEA

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Published on January 5, 2009

Author: 4ChEAB08

Source: slideshare.net

Unit Operations Assignment # 2 Problems on Conduction Group 6 – 4ChEA Dalawang Bayan, Rose Ann Dayao, Jeseca Gatdula, John Robert Lamayan, Abegail Macaraeg, Gladies

Unit Operations

Assignment # 2

Problems on Conduction

Group 6 – 4ChEA

Dalawang Bayan, Rose Ann

Dayao, Jeseca

Gatdula, John Robert

Lamayan, Abegail

6. A flat slab of rubber, 2.5 cm thick initially at 20 ºC is to be placed between two heated steel plates maintained at 140 ºC. The heating is to be discontinued when the temperature at the mid plane of the slab is 132 ºC. The rubber has a thermal conductivity of 0.16 W/mK and thermal diffusivity is 8.671x10 -6 m 2 /s. Thermal resistance from metal to rubber may be neglected. Calculate: a.) The length of the heating period, sec b.) The temperature of the rubber 0.65cm from the metal c.) The time required for the rubber to reach 132 ºC at the plane specified at b. GIVEN: x 1 = (0.025/2) = 0.0125m k = 0.16 W/mK T 0 = 20 ºC α = 8.671x10 -6 m 2 /s T 1 = 140 ºC T = 132 ºC PROBLEM SET IN CONDUCTION

6. A flat slab of rubber, 2.5 cm thick initially at 20 ºC is to be placed between two heated steel plates maintained at 140 ºC. The heating is to be discontinued when the temperature at the mid plane of the slab is 132 ºC. The rubber has a thermal conductivity of 0.16 W/mK and thermal diffusivity is 8.671x10 -6 m 2 /s. Thermal resistance from metal to rubber may be neglected. Calculate:

a.) The length of the heating period, sec

b.) The temperature of the rubber 0.65cm from the metal

c.) The time required for the rubber to reach 132 ºC at the plane specified at b.

GIVEN:

x 1 = (0.025/2) = 0.0125m k = 0.16 W/mK

T 0 = 20 ºC α = 8.671x10 -6 m 2 /s

T 1 = 140 ºC

T = 132 ºC

SOLUTION: a.) assume: m=0 n = x = 0 = 0 x 1 0.0125m Y = T 1 – T = 140 - 132 = 0.067 T 1 – T 0 140 – 20 from chart: X = 0.6 X = αt (x 1 ) 2 0.6 = (8.671x10 -6 m 2 /s)(t) (0.0125) 2 t = 10.81 s

a.) assume: m=0

n = x = 0 = 0

x 1 0.0125m

Y = T 1 – T = 140 - 132 = 0.067

T 1 – T 0 140 – 20

from chart: X = 0.6

X = αt

(x 1 ) 2

0.6 = (8.671x10 -6 m 2 /s)(t)

(0.0125) 2

b.) assume: m=0 n = x = 6.5x10 -3 = 0.52 x 1 0.0125m X = αt = (8.671x10 -6 m 2 /s)(10.81 s) = 0.6 (x 1 ) 2 (0.0125 m) 2 from chart: Y = 0.185 Y = T 1 – T T 1 – T 0 0.185 = 140 – T 140 – 20 T = 117.8 ºC T = 390.8 K T = 117.8 ºC T = 390.8 K

b.) assume: m=0

n = x = 6.5x10 -3 = 0.52

x 1 0.0125m

X = αt = (8.671x10 -6 m 2 /s)(10.81 s) = 0.6

(x 1 ) 2 (0.0125 m) 2

from chart: Y = 0.185

Y = T 1 – T

T 1 – T 0

0.185 = 140 – T

140 – 20

T = 117.8 ºC

T = 390.8 K

T = 117.8 ºC

T = 390.8 K

c.) n = x = 6.5x10 -3 = 0.52 x 1 0.0125m Y = 140 – 132 = 0.067 140 – 20 From chart: X = 0.15 X = αt (x 1 ) 2 0.15 = (8.671x10 -6 )(t) (0.0125) 2 t = 2.7 s

c.)

n = x = 6.5x10 -3 = 0.52

x 1 0.0125m

Y = 140 – 132 = 0.067

140 – 20

From chart: X = 0.15

X = αt

(x 1 ) 2

0.15 = (8.671x10 -6 )(t)

(0.0125) 2

14. (US) A cylindrical steel shaft 10 cm. in diameter and 2.4 m long is heat treated to give it desired physical properties. It is heated to a uniform temperature of 600C and then plunged into an oil bath which maintains the surface temperature at 150C. Calculate the radial temperature profile at 2 min and at 3 min after immersion. GIVEN: x 1 = (0.1/2) = 0.05m k = 38 W/mK T 0 = 600 ºC α = 0.0381 m 2 /s T 1 = 150 ºC h = 125 W/ m 2 -K t = 2 min and 3 min REQUIRED: radial temperature profile at 2 min and 3 min

14. (US) A cylindrical steel shaft 10 cm. in diameter and 2.4 m long is heat treated to give it desired physical properties. It is heated to a uniform temperature of 600C and then plunged into an oil bath which maintains the surface temperature at 150C. Calculate the radial temperature profile at 2 min and at 3 min after immersion.

GIVEN:

x 1 = (0.1/2) = 0.05m k = 38 W/mK

T 0 = 600 ºC α = 0.0381 m 2 /s

T 1 = 150 ºC h = 125 W/ m 2 -K

t = 2 min and 3 min

REQUIRED: radial temperature profile at 2 min and 3 min

SOLUTION: a. at 2 min after immersion n= 0 = 0 0.05 m= 38 = 6.08 (125)(0.05) X= 0.0381 (2/60) = 0.508 0.05 2 From Figure 5.3-7 Y = 0.82 = T 1 – T = 150 – T = 519 ºC T 1 – T 0 150 - 600

SOLUTION:

a. at 2 min after immersion

n= 0 = 0

0.05

m= 38 = 6.08

(125)(0.05)

X= 0.0381 (2/60) = 0.508

0.05 2

From Figure 5.3-7

Y = 0.82 = T 1 – T = 150 – T = 519 ºC

T 1 – T 0 150 - 600

SOLUTION: a. at 3 min after immersion n= 0 = 0 0.05 m= 38 = 6.08 (125)(0.05) X= 0.0381 (3/60) = 0.762 0.05 2 From Figure 5.3-7 Y = 0.77 = T 1 – T = 150 – T = 496.5 ºC T 1 – T 0 150 - 600

SOLUTION:

a. at 3 min after immersion

n= 0 = 0

0.05

m= 38 = 6.08

(125)(0.05)

X= 0.0381 (3/60) = 0.762

0.05 2

From Figure 5.3-7

Y = 0.77 = T 1 – T = 150 – T = 496.5 ºC

T 1 – T 0 150 - 600

4.2-2.) Heat Removal of a Cooling Coil. A cooling coil of 1.0ft of 304 stainless-steel tubing having an inside diameter of 0.25 in. and an outside diameter of 0.40 in. is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40 ºF and is 80 ºF on the outside. The thermal conductivity of 304 stainless steel is a function of temperature: k = 7.75 + 7.78x10 -3 T, where k is in BTU/hr.ft.ºF and T is in ºF. Calculate the heat removal in BTU/s and watts. GIVEN: 304 stainlesss-steel tubing D i = 0.25 in k = 7.75 + 7.78x10 -3 T D 0 = 0.40 in α = 4.65x10 -4 m 2 /h T 1 = 40 ºF T 2 = 80 ºF

4.2-2.) Heat Removal of a Cooling Coil. A cooling coil of 1.0ft of 304 stainless-steel tubing having an inside diameter of 0.25 in. and an outside diameter of 0.40 in. is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40 ºF and is 80 ºF on the outside. The thermal conductivity of 304 stainless steel is a function of temperature: k = 7.75 + 7.78x10 -3 T, where k is in BTU/hr.ft.ºF and T is in ºF. Calculate the heat removal in BTU/s and watts.

GIVEN: 304 stainlesss-steel tubing

D i = 0.25 in k = 7.75 + 7.78x10 -3 T

D 0 = 0.40 in α = 4.65x10 -4 m 2 /h

T 1 = 40 ºF T 2 = 80 ºF

A = πL [(D 0 – D i ) / ln (D 0 /D i )] = π (1.0 ft) [( 0.033 – 0.021)/ ln(0.033/0.021)] = 0.083 ft 2 K m = a + bT ave = 7.75 + 7.78x10 -3 [(40+20)/2] = 8.2168 BTU/hr.ft. ºF q = Σ T = (80-40) ºF R T (6.25x10 -3 ft) (8.2168 BTU/hr.ft.ºF)(0.083 ft 2 ) q = 4,364.76 BTU x 1hr = hr 3600 s q = 1.212 BTU x 252 cal x 4.184 J = s 1BTU 1 cal 1.212 BTU/s 1,277.89 Watt

A = πL [(D 0 – D i ) / ln (D 0 /D i )]

= π (1.0 ft) [( 0.033 – 0.021)/ ln(0.033/0.021)]

= 0.083 ft 2

K m = a + bT ave

= 7.75 + 7.78x10 -3 [(40+20)/2]

= 8.2168 BTU/hr.ft. ºF

q = Σ T = (80-40) ºF

R T (6.25x10 -3 ft)

(8.2168 BTU/hr.ft.ºF)(0.083 ft 2 )

q = 4,364.76 BTU x 1hr =

hr 3600 s

q = 1.212 BTU x 252 cal x 4.184 J =

s 1BTU 1 cal

PROBLEMS FROM GEANKOPLIS 5.3-9.) Temperature of Oranges on Trees During Freezing Weather. In orange-growing areas, the freezing of the oranges on the trees during cold nights is of serious economic concern. If the oranges are initially at a temperature of 21.1 ºC, calculate the center temperature of the orange if exposed to air at -3.9 ºC for 6 hr. The oranges are 102mm in diameter and the convective coefficient is estimated as 11.4 W/m 2 .K. The thermal conductivity k is 0.431 W/m.K and α is 4.65x10 -4 m 2 /h. Neglect any latent heat effects. GIVEN: x 1 = (0.102/2) = 0.051m k = 0.431 W/mK T 0 = 21.1 ºC α = 4.65x10 -4 m 2 /h T 1 = -3.9 ºC h = 11.4 W/m 2 .K t = 6 hr

5.3-9.) Temperature of Oranges on Trees During Freezing Weather. In orange-growing areas, the freezing of the oranges on the trees during cold nights is of serious economic concern. If the oranges are initially at a temperature of 21.1 ºC, calculate the center temperature of the orange if exposed to air at -3.9 ºC for 6 hr. The oranges are 102mm in diameter and the convective coefficient is estimated as 11.4 W/m 2 .K. The thermal conductivity k is 0.431 W/m.K and α is 4.65x10 -4 m 2 /h. Neglect any latent heat effects.

GIVEN:

x 1 = (0.102/2) = 0.051m k = 0.431 W/mK

T 0 = 21.1 ºC α = 4.65x10 -4 m 2 /h

T 1 = -3.9 ºC h = 11.4 W/m 2 .K

t = 6 hr

m = k = 0.431 W/mK = 0.74 hx 1 (11.4 W/m 2 K)(0.051) X = αt = (4.65x10 -4 m 2 /hr)(6hr) = 1.07 (x 1 ) 2 (0.051m) 2 From chart: Y = 0.05 Y = T 1 – T T 1 – T 0 0.05 = -3.9 ºC – T -3.9 ºC – 21.1 ºC T = -2.65 ºC

m = k = 0.431 W/mK = 0.74

hx 1 (11.4 W/m 2 K)(0.051)

X = αt = (4.65x10 -4 m 2 /hr)(6hr) = 1.07

(x 1 ) 2 (0.051m) 2

From chart: Y = 0.05

Y = T 1 – T

T 1 – T 0

0.05 = -3.9 ºC – T

-3.9 ºC – 21.1 ºC