• Author / Uploaded
• Abdul Hafiz

• Categories
• Transferencia de calor
• Aislamiento térmico
• Calor
• Ingeniería de Edificación
• Fenómenos de transporte

Citation preview

Heat Transfer Tutorials Q1. A furnace wall consists of 125 mm wide refractory brick & 125 mm wide insulating fire brick separated by an air gap. The outside wall is covered with 12 mm thickness of plaster. The inner surface of the wall is at 1100oC and the room temperature is 25oC. The heat transfer coefficient from the outside wall surface to the air in the room is 17 W/m2K, and the resistance to heat flow of the air gap is 0.16 K/W. The thermal conductivities of the refractory brick, insulating brick and plaster are 1.6, 0.3 and 0.14 W/m K, respectively. Calculate: a) the rate of heat loss per unit area (1m2) of the wall surface; b) the temperature at each interface throughout the wall; c) the temperature at the outside surface of the wall.

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T0 = 1100oC

Q=

T5 = 25oC

kA (T o−T 1) x

I=

V RE

Q=

(Δ T ) R Th

1.6 x 1 (1100−T 1 )....(1) 0.125 (T 1 −T 2) (T 1 −T 2) Q= = ...(2) 0.16 (R) 0.3 x 1 Q= (T −T 3 )...(3) 0.125 2 0.14 x 1 Q= (T 3 −T 4)...(4) 0.012 Q=

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Q=17 x 1 x(T 4−25)...(5)

Get the equations in the following format & solve Δ T =Const x Q

Q – 1344 W, t1=995, t2=780, t3=220, t4=104 1.6 x 1 (1100−T 1 )....(1) 0.125 (T 1 −T 2) (T 1 −T 2) Q= = ...(2) 0.16 (R) 0.3 x 1 Q= (T −T 3 )...(3) 0.125 2 0.14 x 1 Q= (T 3 −T 4)...(4) 0.012 Q=17 x 1−(T 4−25)...(5) Q=

0.0781 Q = 1100 – T1 … (6) 0.16 Q = T1 – T2 … (7) 0.4167 Q = T2 – T3 … (8) 0.0857 Q = T3 – T4 … (9) 0.0588 Q = T4 – 25 … (10) (6) + (7) + (8) + (9) + (10), 0.7993 Q = 1100 – 25 or Q = 1344.93 W Substituting, (6) T1 = 994.96oC (7) T2 = 779.77oC (8) T3 = 219.33oC (9) T4 = 104.08oC

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Q2. A steel pipe of 100 mm bore and 7mm wall thickness, carrying steam at 260oC, is insulated with 40mm insulation layer. It is further covered by felt of 60mm thickness. The atmospheric temperature is 15oC. The heat transfer coefficients (h) for the inside and outside surfaces are 550 & 15 W/m2K respectively. Take thermal conductivity for the three substances as follows: Steel – 50 W/mK; Insulation layer – 0.09 W/mK; and Felt – 0.07 W/mK Calculate a) the rate of heat loss per unit length (1m) of pipe, b) the temperature of the outside surface Note: Based on the electrical analogy, take the resistances as follows: ln (r 2 /r 1) 2π K where, r2 > r1, K is the conductivity of material in W/mK Resistance of a layer in a pipe R=

Resistance of the boundary layer (inside or outside) 4 Of 5

R=

1 h x (2 π r x 1)

where, (2πr x 1) is the exposed area of pipe (1 m length) , h is the conductivity of material in W/m2K r1 = 0.05 m r2 = 0.057m r3 = 0.097m r4 = 0.157m ln(r 2 /r1 ) R= 2πK 1 R= h x (2 π r x 1) Q=

(Δ T ) R Th

(Δ T )=R Th x Q

t0 = Steam temperature (260oC); t1 = Steel temperature; t2 = Steel-insulation layer interface temperature; t3 = Insulation_layer-felt interface temperature; t4 = Felt-air interface temperature; t5 = Outside air temperature (15oC) to – t1 = RInner Boundary Layer x Q t1 – t2 = RSteel x Q t2 – t3 = RInsulation Layer x Q t3 – t4 = RFelt x Q t4 – t5= ROuter Boundary Layer x Q 5 Of 5

to – t5 = ( RInner Boundary Layer+ RSteel + RInsulation Layer + Rfelt + ROuter Boundary Layer).Q 260 – 15 = (………….) . Q

Q˙ = ??

(Ans – 116 W)

t4 = ??

(Ans – 22.8oC)

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