Co, Cynthia

Eniceo, Raniel

Maramba, Paulo

Reyes, Charmaine

Ramos, Hesed

GROUP # 3

4.3-2

A wall of furnace 0.244m thick is constructed of material having a

thermal conductivity of 1.30W/mK. The wall will be insulated on the outside

material having an average k of 0.346 W/mK, so the heat loss from the

furnace will ve equal or less than 1830 W/m2. the inner surface temperature is

1588K and the outer 299K. Calculate the thickness of insulation required.

0.244m x

BA RR

Tq

BA AA

Akx

AkxT

q

B

B

A

A

BB

BB Ak

xR

AA

AA Ak

xR

B

B

A

A

kx

kxT

A

q

mKWx

mKWm

K

m

W

/346.0/3.1244.0

)2991588(18302

X=0.179m

2

1830

m

W

A

q

T1=1588K

T2=299K

kA=1.3W/mK

kB=0.346W/mK

mxA 244.0

4.3-2

5.3-3.

Cooling a slab of Aluminum

A large piece of Aluminum that can be considered a semi-infinite solid initially has a uniform temperature of 505.4K. The surface is suddenly

exposed to an environment at 338.8K with a surface convection

coefficient of 455 W/m2K. Calculate the time in hours for the

temperature to reach 388.8 K at a depth of 25.4 mm. the average

physical properties are ά = 0.340 m2.h and k=208 W/mK.

X1= 25.4mm=0.0254m

To =505.4 K

T1 = 338.8 K

H= 455 W/m2K

t=? If T=388.88K

ά=0.340 m2/h

K=208 W/mK

X= άt/X12

Y= T1-T/ T1-To

= [(338.8-388.8)/(338.8-505.4)]

= 0.30

M= k/hX1

= 208 / (455 x0.0254)

=18

X= άt/X12

22 = 0.340 t / 0.02542

t=0.0417 hrs.

Problem # 3 Compute the heat loss per square meter of surface for a furnace wall 23cm thick. The inner and outer surface temperature are 315oC and 38oC respectively. The variation of the thermal conductivity in W/mL, with temperature in oC is given by the following relation : k=0.006T-1.4x10-6 T2

k=bT + cT2

Km=bTm + cT2m

Km = bTm + c/3 (T12 + T1T2 + T2

2)

KM = 1.059 – 0.05256

Km=1.0064 W/mK

Rt = 0.23 / (1 X 1.0064)

= 0.2285

Q= (315-38)/0.2285

Q = 1212.25 W

Problem # 11

An insulated steam pipe having an outside diameter of 0.0245m is to be covered with 2 layers of

insulation each having a thickness of 0.0245m. The

average thermal conductivity of one material is approximately four times that of the other. Assuming that the inner and

outer surface temperature of the composite insulation are fixed,

how much will the heat be reduced when the better

insulating material is next to the pipe then when it is the outer

layer?

#11 D1=0.0245mD2=0.0245+0.0245=0.049mD3=0.49+0.0245=0.735m

DL; L=1mA1=0.077m2

A2=0.154m2

A3=0.231m2

B A1T3T

2 B 2111.0

1

2ln

12m

A

AAA

AA

219.0

2

3ln

23m

A

AAA

AB

Case1 Case2

k

4k k

4k

Case1 )(95.3

)19.0(40245.0

)111.0(0245.0 31

31 TTk

kk

TTq

Case2)(43.5

)19.0(0245.0

)111.0(40245.0 31

31 TTk

kk

TTq

ratio between case 1 and case 2 is 0.73. the better insulating material should be placed at the outer most layer.