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# Chapter 3 : STEADY HEAT CONDUCTION

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## Mohamed Hakim

on 1 May 2014

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#### Transcript of Chapter 3 : STEADY HEAT CONDUCTION

Example 3-8 ( Heat Lost Through an Insulated Steam Pipe)
(page 159)
Steam at T∞1 = 320 °C flow in a cast iron pipe ( k=80W/m.K) whose inner and outer diameters are D1=5cm and D2=5.5cm respectively. The pipe is covered with 3-cm-thick glas wool insulation with k=0.05W/m.K. Heat is lost to the surrounding at T∞2 = 5°C by natural convection and radiation, with a combined heat transfer coefficient of h2=18wm2.k, determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drop across the pipe shell and the insulation.
The heat loss for a given pipe length are determined by multiplying the above quantity by pipe length L
step 7
Chapter 3 :

FROM GROUP 3

Chapter 3 :

FROM GROUP 3

Represented by :-

MOHAMED HAKIM B. AHMAD SHAH DD110054
MOHAMED SYAFIZ B. AMIRUDIN DD110051
MOHD AIKAL SYAZWAN B. ABDUL RAZAK CD100303
MOHD AMIN HAFIZ B. SULAIMAN BD100014
MOHD AZREN B. MOHAMED DD110006
MOHD DZULKIFLY B. IBRAHIM DD110053

SOLUTION :-
A steam pipe covered with glass wool insulation is subjected to convection on its surfaces. The rate of heat transfer per unit length and the temperature drops across the pipe and the insulation are to be determine.
we need to do assumption ;
1. Heat transfer is steady since there is
no indicator of any change with time.
2. Heat transfer is one-dimensional since there is
thermal symmetry about the centerline and no
vibration in the axial direction
3. Thermal conductivities are constans
4.The thermal contact resistance at the interface is
negligible.
Properties
given :-
Analysis
Taking L=1m, the ares of the surface exposed to convection are determine to be
k = 80W/m.K
(FOR CAST iRON)

k = 0.05W/m.K
(FOR GLASS WOOL INSULATION)
A1=2(3.142)r1L
= 2(3.142)(0.025m)(1m)
= 0.157m2
A3=2(3.142)r3L
=2(3.142)(0.0575m)(1m)
= 0.361m2
step 1
Ri=Rconv,1
= 1/(h_1 A_1 )
= 1/((60W/m^2 .K)(0.157m^(2 )))
= 0.106°C
step 2
R1=Rcpipe
= ( ln⁡〖(r_2/ r_2 〗))/(2(3.142)k_1 L)
= ln⁡(2.75/2.5)/(2(3.142)(80W/m.K)(1m))
= 0.0002°C/W
step 3
R2=Rinsulation
= ( ln⁡〖(r_3/ r_2 〗))/(2(3.142)k_2 L)
= ln⁡(5.75/2.75)/(2(3.142)(0.05W/m.K)(1m))
= 2.35°C/W
step 4
Ro=Rconv,2
= 1/(h_2 A_3 )
= 1/(((18 W)/m^2 .K)(0.361m^(2 )))
= 0.154°C/W
step 5
Rtotal = Ri+R1+R2+Ro
= 0.106+0.002+2.35+0.154
= 2.61°C/W
step 6

Q ̇ = T_(∞1- T_∞ 2)/R_total
= ([ 320-5)]°C)/(2.61°C/W)
= 121 w (per m pipe length)

The total resistance is determine
The steady rate of heat loss from the steam becomes

DeltaT_pipe = Q ̇R_pipe
= (121W) (0.0002°C/W)
= 0.02°C
DeltaT_insulation = Q ̇R_insulation
= (121W) (2.35°C/W)
= 284°C
Conclusion
That is, the temperatures between the inner and the outer surface of the pipe differ by 0.02°C, whereas the temperatures between inner and outer surface of the insulation differ by 284°C
thanks you for watching our group video.....
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