(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 :**

STEADY HEAT CONDUCTION

FROM GROUP 3

STEADY HEAT CONDUCTION

FROM GROUP 3

**Chapter 3 :**

STEADY HEAT CONDUCTION

FROM GROUP 3

STEADY HEAT CONDUCTION

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

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

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