Unit OperationsUnit OperationsAssignment # 2Assignment # 2

Problems on ConductionProblems on Conduction

Group 6 – 4ChEAGroup 6 – 4ChEADalawang Bayan, Rose AnnDalawang Bayan, Rose AnnDayao, Jeseca Dayao, Jeseca Gatdula, John RobertGatdula, John RobertLamayan, AbegailLamayan, AbegailMacaraeg, GladiesMacaraeg, Gladies

6. A flat slab of rubber, 2.5 cm thick initially at 20 ºC is to be placed between 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 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 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 8.671x10thermal conductivity of 0.16 W/mK and thermal diffusivity is 8.671x10 -6-6 m m22/s. /s. Thermal resistance from metal to rubber may be neglected. Calculate:Thermal resistance from metal to rubber may be neglected. Calculate: a.) The length of the heating period, seca.) The length of the heating period, sec b.) The temperature of the rubber 0.65cm from the metalb.) 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 c.) The time required for the rubber to reach 132 ºC at the plane specified at

b.b.

GIVEN: GIVEN:

xx11 = (0.025/2) = 0.0125m = (0.025/2) = 0.0125m k = 0.16 W/mKk = 0.16 W/mK

TT00= 20 ºC = 20 ºC α = 8.671x10α = 8.671x10-6-6 m m22/s/s

TT11= 140 ºC = 140 ºC

T = 132 ºC T = 132 ºC

PROBLEM SET IN CONDUCTIONPROBLEM SET IN CONDUCTION

SOLUTION:SOLUTION:

a.) assume: m=0a.) assume: m=0

n = x = 0 = 0 n = x = 0 = 0 xx11 0.0125m0.0125m

Y = TY = T11 – T = 140 - 132 = 0.067 – T = 140 - 132 = 0.067

TT11 – T – T00 140 – 20 140 – 20

from chart: X = 0.6from chart: X = 0.6

X = αtX = αt

(x(x11))22

0.6 = (8.671x100.6 = (8.671x10-6-6 m m22/s)(t) /s)(t)

(0.0125)(0.0125)22

t = 10.81 s t = 10.81 s

b.) assume: m=0b.) assume: m=0

n = x = 6.5x10n = x = 6.5x10-3-3 = 0.52 = 0.52

xx11 0.0125m0.0125m

X = αtX = αt = (8.671x10 = (8.671x10-6-6 m m22/s)(10.81 s) = 0.6/s)(10.81 s) = 0.6

(x(x11))22 (0.0125 m)(0.0125 m)22

from chart: Y = 0.185from chart: Y = 0.185

Y = TY = T11 – T – T

TT11 – T – T00

0.185 = 140 – T 0.185 = 140 – T 140 – 20140 – 20

T = 117.8 ºC T = 117.8 ºC T = 390.8 KT = 390.8 K

T = 117.8 ºC T = 117.8 ºC T = 390.8 KT = 390.8 K

c.)c.)

n = x = 6.5x10n = x = 6.5x10-3-3 = 0.52 = 0.52

xx11 0.0125m0.0125m

Y = 140 – 132 = 0.067Y = 140 – 132 = 0.067

140 – 20140 – 20

From chart: X = 0.15From chart: X = 0.15

X = αtX = αt

(x(x11))22

0.15 = (8.671x100.15 = (8.671x10-6-6)(t))(t)

(0.0125)(0.0125)22

t = 2.7 st = 2.7 s

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

GIVEN: GIVEN: xx11 = (0.1/2) = 0.05m = (0.1/2) = 0.05m k = 38 W/mKk = 38 W/mK

TT00= 600 ºC = 600 ºC α = 0.0381 mα = 0.0381 m22/s/s

TT11= 150 ºC = 150 ºC h = 125 W/ mh = 125 W/ m22-K-K

t = 2 min and 3 mint = 2 min and 3 min

REQUIRED: radial temperature profile at 2 min and 3 minREQUIRED: radial temperature profile at 2 min and 3 min

SOLUTION:SOLUTION:

a. at 2 min after immersiona. at 2 min after immersion

n= 0 = 0n= 0 = 0

0.05

m= 38 = 6.08

(125)(0.05)(125)(0.05)

X= 0.0381 (2/60) = 0.508X= 0.0381 (2/60) = 0.508

0.050.0522

From Figure 5.3-7From Figure 5.3-7

Y = 0.82 = TY = 0.82 = T11 – T = 150 – T = 519 ºC – T = 150 – T = 519 ºC

TT11 – T – T0 0 150 - 600150 - 600

SOLUTION:SOLUTION:

a. at 3 min after immersiona. at 3 min after immersion

n= 0 = 0n= 0 = 0 0.05

m= 38 = 6.08 (125)(0.05)(125)(0.05)

X= 0.0381 (3/60) = 0.762X= 0.0381 (3/60) = 0.762 0.050.0522

From Figure 5.3-7From Figure 5.3-7

Y = 0.77 = TY = 0.77 = T11 – T = 150 – T = 496.5 ºC – T = 150 – T = 496.5 ºC

TT11 – T – T0 0 150 - 600150 - 600

4.2-2.) 4.2-2.) Heat Removal of a Cooling Coil. Heat Removal of a Cooling Coil. A cooling coil of A cooling coil of 1.0ft of 304 stainless-steel tubing having an inside diameter 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 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 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 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: conductivity of 304 stainless steel is a function of temperature: k = 7.75 + 7.78x10k = 7.75 + 7.78x10-3-3 T, where k is in BTU/hr.ft.ºF and T is in T, where k is in BTU/hr.ft.ºF and T is in ºF. Calculate the heat removal in BTU/s and watts.ºF. Calculate the heat removal in BTU/s and watts.

GIVEN: 304 stainlesss-steel tubingGIVEN: 304 stainlesss-steel tubing

DDii = 0.25 in = 0.25 in k = 7.75 + 7.78x10k = 7.75 + 7.78x10-3-3 T T

DD0 0 = 0.40 in = 0.40 in α = 4.65x10α = 4.65x10-4-4 m m22/h/h

TT1 1 = 40 ºF = 40 ºF TT22 = 80 ºF = 80 ºF

A = πL [(DA = πL [(D00 – D – Dii) / ln (D) / ln (D00/D/Dii)])]

= π (1.0 ft) [( 0.033 – 0.021)/ ln(0.033/0.021)]= π (1.0 ft) [( 0.033 – 0.021)/ ln(0.033/0.021)] = 0.083 ft= 0.083 ft22

KKmm = a + bT = a + bTaveave

= 7.75 + 7.78x10= 7.75 + 7.78x10-3-3 [(40+20)/2] [(40+20)/2] = 8.2168 BTU/hr.ft. ºF = 8.2168 BTU/hr.ft. ºF

q = Σ T = (80-40) ºF q = Σ T = (80-40) ºF

RRTT (6.25x10(6.25x10-3-3 ft) ft)

(8.2168 BTU/hr.ft.ºF)(0.083 ft(8.2168 BTU/hr.ft.ºF)(0.083 ft22) )

q = 4,364.76 BTU x 1hr =q = 4,364.76 BTU x 1hr = hr 3600 shr 3600 s

q = 1.212 BTU x 252 cal x 4.184 J =q = 1.212 BTU x 252 cal x 4.184 J = s 1BTU 1 cals 1BTU 1 cal

1.212 BTU/s1.212 BTU/s

1,277.89 Watt1,277.89 Watt

PROBLEMS FROM GEANKOPLISPROBLEMS FROM GEANKOPLIS

5.3-9.) 5.3-9.) Temperature of Oranges on Trees During Freezing Weather.Temperature of Oranges on Trees During Freezing Weather. In In orange-growing areas, the freezing of the oranges on the trees during cold 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 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 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 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/mthe convective coefficient is estimated as 11.4 W/m22.K. The thermal .K. The thermal conductivity k is 0.431 W/m.K and α is 4.65x10conductivity k is 0.431 W/m.K and α is 4.65x10-4-4 m m22/h. Neglect any latent /h. Neglect any latent heat effects.heat effects.

GIVEN: GIVEN:

xx11 = (0.102/2) = 0.051m = (0.102/2) = 0.051m k = 0.431 W/mKk = 0.431 W/mK

TT00= 21.1 ºC = 21.1 ºC α = 4.65x10α = 4.65x10-4-4 m m22/h/h

TT11= -3.9 ºC = -3.9 ºC h = 11.4 W/mh = 11.4 W/m22.K.K

t = 6 hrt = 6 hr

m = k = 0.431 W/mK = 0.74m = k = 0.431 W/mK = 0.74

hxhx11 (11.4 W/m(11.4 W/m22K)(0.051)K)(0.051)

X = αt = (4.65x10X = αt = (4.65x10-4-4 m m22/hr)(6hr) = 1.07 /hr)(6hr) = 1.07

(x(x11))22 (0.051m) (0.051m)22

From chart: Y = 0.05From chart: Y = 0.05

Y = TY = T11 – T – T

TT11 – T – T00

0.05 = -3.9 ºC – T0.05 = -3.9 ºC – T

-3.9 ºC – 21.1 ºC -3.9 ºC – 21.1 ºC

T = -2.65 ºCT = -2.65 ºC