Co, Cynthia

EniCEo, RaniEl

MaRaMba, Paulo

REyEs, ChaRMainE

RaMos, hEsEd

# 3GROUP

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 =

Ak

x

Ak

xT

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= 25.4/2 = 12.7mm

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 x 0.0127)

= 36

X= άt/X12

52 = 0.340 t / 0.0127 2

t=0.025 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 = 0.006[(315+38)/2] + -1.4x10-6 [(315+38)2/4]

Km= 1.01539 W/mK

Rt = 0.23

--------

(1m2) (1.01539)

= 0.2265

Q= (315-38)/0.2265

Q = 1222.86 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

Α=πDL; 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 −=

+

−=

Case2-Case1=1.48k(T1-T3)

If the better insulating material is next to the pipe, heat transfer will be reduced by 1.48k(T1-T3)