ChE206ChE206Problem Set on Problem Set on

ConductionConductionGroup 2Group 2

Boniquit, Edward JohnBoniquit, Edward JohnDoliente, Lorenzo MiguelDoliente, Lorenzo MiguelMonsanto, Randy DavidMonsanto, Randy David

Soriano, Ron DaleSoriano, Ron Dale4ChE-B4ChE-B

Problem 2A furnace wall is to consist in series of 18 cm of kaolin firebrick, 15 cm of kaolin insulating brick, and sufficient fireclay brick to reduce the heat loss to 350 W/m2 when the face temperatures are 815 and 38 ºC, respectively. If an effective air gap of 3 mm (assume k for air as 0.17 W/mK) can be incorporated between the fireclay and insulating brick when erecting the wall impairing its structural support, what thickness of fireclay brick will be required?

Given:

1 2

Km

W0.17k 3 ⋅

=C815T1 °= C38T2 °=

T′

Req’d: x

4

3

T ′′

T ′′′1

2

4

3

Kaolin FirebrickKaolin Insulating BrickAirFireclay Brick

18cm 15cm x

2mm

Problem 2

Km

W0.17k 3 ⋅

=

Given:

1 2

C815T1 °= C38T2 °=

T′

Req’d: x4

3

T ′′

T ′′′1

2

4

3

Kaolin FirebrickKaolin Insulating BrickAirFireclay Brick

Sol’n:Assume: A1 = A2 = A3 = A4 = 1m2

Assume:

C757.5T2

700815T

ave,1

ave,1

°=

+=

C700T °=′

( )CT °

°⋅⋅ Ffthr

BTUk

200

757.5

760

0.050

k

0.113

Km

W0.1950k

Ffthr

BTUKm

W

1.73Ffthr

BTU0.1127k

m,1

m,1

⋅=

°⋅⋅

⋅⋅°⋅⋅

=

18cm 15cm x

2mm

Problem 2

Km

W0.17k 3 ⋅

=

Given:

1 2

C815T1 °= C38T2 °=

T′

Req’d: x4

T ′′

T ′′′1

2

4

3

Kaolin FirebrickKaolin Insulating BrickAirFireclay Brick

Sol’n: 350qq 1 ==

( ) ( )C491.92T

350

10.1950100

18T-815

°=′

=′

18cm 15cm x

2mm3

5%42.30%%ε

100491.92

491.92-700%ε

>=

×=

Assume: C700T °=′

C46.653T2

92.914815T

ave,1

ave,1

°=

+=

Km

W0.1747k

Ffthr

BTUKm

W

1.73Ffthr

BTU0.1010k

m,1

m,1

⋅=

°⋅⋅

⋅⋅°⋅⋅

=

350qq 1 ==

( ) ( )C454.38T

350

10.1747100

18T-815

°=′

=′

5%8.26%%ε

100454.38

454.38-491.92%ε

>=

×=

Assume: C454.38T °=′

C69.634T2

38.544815T

ave,1

ave,1

°=

+=

Km

W0.1711k

Ffthr

BTUKm

W

1.73Ffthr

BTU0.0989k

m,1

m,1

⋅=

°⋅⋅

⋅⋅°⋅⋅

=

350qq 1 ==

( )( )C446.79T

350

10.1711100

18T-815

°=′

=′

5%1.70%%ε

100446.79

446.79-454.38%ε

<=

×=

Problem 2

Km

W0.17k 3 ⋅

=

Given:

1 2

C815T1 °= C38T2 °=

T′

Req’d: x4

T ′′

T ′′′1

2

4

3

Kaolin FirebrickKaolin Insulating BrickAirFireclay Brick

18cm 15cm x

2mm3

Problem 2

Assume: C200T °=′′

C40.323T2

200446.79T

ave,2

ave,2

°=

+=

Km

W0.2078k

Ffthr

BTUKm

W

1.73Ffthr

BTU0.1201k

m,2

m,2

⋅=

°⋅⋅

⋅⋅°⋅⋅

=

350qq 2 ==

( ) ( )C194.14T

350

10.2078100

15T-446.79

°=′′

=′′

5%3.02%%ε

100194.14

194.14-200%ε

<=

×=

Km

W0.17k 3 ⋅

=

Given:

1 2

C815T1 °= C38T2 °=

T′

Req’d: x4

T ′′

T ′′′1

2

4

3

Kaolin FirebrickKaolin Insulating BrickAirFireclay Brick

18cm 15cm x

2mm3

( )CT °

°⋅⋅ Ffthr

BTUk

323.40

500

1150

k

0.15

0.26

Problem 2

C01.114T2

38190.02T

ave,4

ave,4

°=

+=

Km

W0.9030k

Ffthr

BTUKm

W

1.73Ffthr

BTU0.5220k

m,4

m,4

⋅=

°⋅⋅

⋅⋅°⋅⋅

=

( )( )

39.22cmx

0.3922mx

10.9030

x38-190.02

350

==

=

Km

W0.17k 3 ⋅

=

Given:

1 2

C815T1 °= C38T2 °=

T′

Req’d: x4

T ′′

T ′′′1

2

4

3

Kaolin FirebrickKaolin Insulating BrickAirFireclay Brick

18cm 15cm x

2mm3

( )CT °

°⋅⋅ Ffthr

BTUk

114.01

200

600

k

0.58

0.85

( )( )C190.02T

350

10.171000

2T-194.14

°=′′

=′′′

Problem 10An experimental heat transfer apparatus consists of a 5 cm. schedule 80 steel pipe covered with two layers of insulation. The inside layer is 2.5 cm thick and consists of diatomaceous silica, asbestos and a bonding material; the outside layer is 85% magnesia and is 4 cm thick. The following data were obtained during a test run:

length of test section = 3 m. heating medium inside pipe = dowtherm A vapour

temperature of inside of steel pipe = 400°C temperature of outside of magnesia insulation = 52°C

dowtherm condensed in test section = 9 kg/hr temperature of condensate = 400°C

latent heat of condensation of dowtherm of upper conditions = 206 kJ/kgDetermine the mean thermal conductivity of the magnesia insulation. The thermal conductivity of the inner layer is as follows:

Temperature (°C) k (W/mK) 93 0.0885

260 0.1050 426 0.1209

Problem 10

2in1.969in ≈=

Given: Nominal Pipe Size 2.54cm

1in5cm ⋅=

4cmthickness

2.5cmthickness

outer

inner

==

Sch. 80 Steel

orWs

J515q

1kJ

1000J

3600s

1h

h

kJ1854q

kg

kJ206

h

kg9q

oncondensati

oncondensati

oncondensati

=

⋅⋅=

=

oncondensatisteel qqq ==

( ) ( )

C399.91T

4.93

6.03ln

100

4.936.03

3π64

100

0.55T-400

515

°=′

−

′=

Inner Insulation

Outer Insulation

Problem 10Assume: C252T °=′′

C46.312T2

22591.993T

innerave,

innerave,

°=

+=

Km

W0.1102km,1 ⋅

=oncondensatiinner qqq ==

( )( )

C93.227T

515

03.6

53.8ln

100

5.2

30.1102

100

2.5T-399.91

°=′′

=

′′

π

5%1.29%%ε

100227.93

225-227.93%ε

<=

×=

oncondensatiouter qqq ==

( ) ( )

Km

W0.1194k

515

8.53

12.53ln

100

4

3πk

100

452-227.93

m

m

⋅=

=

Problem 4.3-3Heat Loss Through Thermopane Double Window. A double window called thermopane is one in which two layers of glass are separated by a later of dry, stagnant air. In a given window, each of the glass layers is 6.35 mm thick separated by a 6.35-mm space of stagnant air. The thermal conductivity of the glass is 0.869 W/m·K and that of air is 0.026 over the temperature range used. For a temperature drop of 27.8 K over the system, calculate the heat loss for a window 0.914 m x 1.83 m. (Note: This calculation neglects the effect of the convective coefficient on one outside surface of one side of the window, the convective coefficient on the other surface, and convection inside the window.)

Problem 4.3-3

G G

A

G

A

Glass

Air

6.35 mm

6.35 mmKm

W0.869kG ⋅

=

Km

W0.026kA ⋅

=

Window Dimensions:

0.914 m x 1.83 m

27.8KΔT =

Given:

Req’d: q

Sol’n:

( ) ( ) ( )

W

K104.37R

1.830.9140.8691000

6.35

R

3-G

G

×=

=

( ) ( )

W

K1460.0R

1.831000

0.9140.026

1000

6.35

R

A

A

=

=

( )

W

K1547.0R

1460.0104.372R

T

-3T

=

+×=

W70.179q0.1547

27.8q

=

=

Problem 5.2-2Quenching Lead Shot in a Bath. Lead shot having an average diameter of 5.1 mm is at an initial temperature of 204.4°C. To quench the shot, it is added to a quenching oil bath held at 32.2°C and falls to the bottom. The time of fall is 15 s. Assuming an average convection of h = 199 W/m2·K, what will be the temperature of the shot after the fall? For lead, ρ = 11,370 kg/m3 and cP = 0.138 kJ/kg·K.