problem set 2 4b3

Problem Set on ConductionGat Granada Jamie hendrix Angel SalcedoKay VillaflorJan Rannel Alejandro

Group 34ChEB

Problem No. 3

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

0.23m

315°C

38°C

Problem No. 3• Compute the heat loss per square meter of surface for a furnace wall 23

cm. thick. The inner and outer surface temperature are 315°C and 38°C respectively. The variation of the thermal conductivity in W/mK, with temperature in °C is given by the following relation: k= 0.006T-1.4x10-6 T2.

0.23m

315°C

38°C

Solution:k = 0.006T + 1.4 x 10^-6km = 0.006(T2 +T1)/2 – 1.4 x 10^-6 (T2^2 + T2T1 + T1^2)T2 = 588.15KT1 = 311.15 KAm = 1m^2km = 2.4059 Wm/Kq/(Am) = kmdT/dx = 2.4059 x (588.15-311.15) /(.23-0)q/(Am) = 2897.54 w/m^2

Problem No. 11•An insulated steam pipe having an outside

diameter of 0.0245 m is to be covered w/ two layers of insulation each having a thickness of 0.0245 m. 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 transfer be reduced when the better insulating material is next to the pipe then when it is the outer layer?

Problem No. 11• An insulated steam pipe having an outside diameter of 0.0245 m is to be covered w/

two layers of insulation each having a thickness of 0.0245 m. 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 transfer be reduced when the better insulating material is next to the pipe then when it is the outer layer?

0. 245

0. 2450. 245

Do = 0.0735Di = 0.0245

Do = 0.1225Di = 0.0735

Do = 0.0735Di = 0.0245

Do = 0.1225Di = 0.0735

Assume k1 = 1W/mK k2 = 4W/mK

q = ∆T/∆R

R1 = ∆x / KmAm = 0.0245/(1)(0.1401L) = 0.1748/L

Am = ∏L (Do – Di) [ln(Do/Di)]

R2 = ∆x / KmAm = 0.0245/(4)(0.3014L) = 0.0203/L

q = ∆T / [(0.1748/L) + (0.0203/L)]q = ∆T / [(0.1951/L)]

Geankoplis: 4.3-2, 5.3-3

5.3-3

Cooling of 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 455W/m2-K. Calculate the time in hours (hr) for the temperature to reach 388.8K at a depth of 25.4mm. The average physical properties are α=0.340m2/hr and k=208W/m-K.

•Given: To = 505.4 K T1 = 388.8 K T = 388.8 Kα = .340 m^2/hh = 455 W/m^2KK = 208 W/mK

Required: time in hours for T = 388.8K at a depth of 25.4mm

T1 = 388.8K

T0 = 505.4K

25.4 mmT = 388.8K

Solution•Y = (T1 – T)/(T1-To) = (338.8 –

388.8)/(338.8-505.4) = .301-Y = .70Using the graph in fig 5.3-3slope= (h*sqrt(alpha*time))/k – eqn 1x-axis=x/(2*sqrt(alpha*t)) – eqn 2Get the approximate slope.Get the equivalent value of x-axis.Get the time x-axis using eqn 2Get the time of the slope using eqn 1Check if the computed x=axis time and

slope time are the same.

•time from slope in sec (eqn 1)= (slope(208)/455)^2/9.444 x 10 ^-5

•time from x-axis in sec (eqn 2) = (.0254/2(x-axis))^2/9.4444x 10 ^-5

•1st assumption= slope=0.05 computed t=10.85, x-axis=0.05, computed t=683.43

•2nd: slope=0.07, t=22.14; x-axis= 0.25 t=27.34

•The 3rd evaluation will approach a higher error. That is why the final time should be 22.14 sec = 6.22 x 10 ^-3 hours

4.3-2

Insulation of a Furnace.A wall of a furnace 0.224 m thick is constructed

of material having a thermal conductivity of 1.30 W/m-K. The wall will be insulated on the outside with material having an average k of 0.346 W/m-K, so the heat loss from the furnace will be equal to or less than 1830 W/m2, The inner surface temperature is 1588 K and the outer 299 K. Calculate the thickness of insulation required.

Ans. 0.179 m

•Given: Furnace:▫Thickness: .244m▫K = 1.30 W/mk

Insulation:Thickness=

required? K = .346 W/mK q/A = 1830 W/m^2

1588K

299K

•Solution:T bet, Temperature between furnace wall and insulation:

q/A = detla T ( km )/ thicknessdelta T = (q/A) (thickness)/km = 1830 W/m^2 x .244m /1.30

= 343.48KT bet = 1588-343.48 = 1244.52 K(Thickness = (delta T) km/q/A)insulator

Thickness = (1244.52-299)K .346 W/mK / 1830W/m^2 = 0.179 m

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