Energy Management in Industry(sECOND SEMESTER)Palembang, March 25th, 2014

DOUBLE MASTER DEGREEENERGY/ENVIRONMENTAL TECHNOLOGY AND MANAGEMENTMAGISTER OF CHEMICAL ENGINEERING2014

CRITICAL RADIUS OF INSULATIONAdding more insulation to a wall or to the attic always decreases heat transfer. The thicker the insulation, the lower the heat transfer rate. This is expected, since the heat transfer area, A, is constant, and adding insulation always increases the thermal resistance of the insu-lation without increasing the convection resistance.Adding insulation to a cylindrical pipe or a spherical shell, however, is a different matter.The additional insulation increases the conduction resistance of the insulation layer but decreases the convection resistance of the surface due to the increase in the outer surface area for convection.

The heat transfer from the pipe may increase or decrease, depending on which effect dominates.The critical radius of insulation, rcr, cyl for a cylindrical body can be calculated by using the following equation: rcr, cyl = where:ki = the thermal conductivity of the insulationh=the external convection heat transfer coefficient .

The rate of heat transfer from the cylinder increases with the addition of insulation for r2 < rcr, cyl , reaches a maximum when r2 = rcr, cyl , and starts to decrese for r2 > rcr, cyl . Thus, insulating the pipe may actually increase the rate of heat transfer instead of decreasing it when r2 < rcr, cyl .

Worked Example3. A pipeline system is used for distributing steam in a factory. The total length of the pipeline is 80 (m) with the thickness of 3 mm and the inside diameter of 74 (mm). Thermal conductivities of the pipe are 52 (W/m.C) at 200 C and 50 (W/m.C) at 300 C. The pipe is covered with an insulation having thermal conductivity of 0.2184 (W/m.C). The temperature at inner surface of the pipe is 260 C while the room temperature in which the pipeline located is 86 F.

The convection heat transfer coefficient between insulation and room air is 5.20 (W/m2.C) and the convection heat transfer coefficient between steam and inside surface of the pipe is 10.5 (W/m2.C). With consideration of thermal conductivity of the pipe at inside surface temperature, determine:a).Heat flow rate if the insulation thickness is equal to critical insulation thickness.

b). Heat flow rate of the naked pipe if the convection heat transfer coefficient between outside surface of the pipe and the room air is the same as convection heat transfer coefficient between outside surface of the insulation and the room air.c). The temperature at outer surface of the pipe at point a).d).Insulation thickness needed to decrease heat loss as much as 20 % from the heat loss without insulation.

Heat energy released by steam in 1.5 hours at point a). and point d).

Solution:Pipe length, L = 80 (m)Pipe thickness, t = 3 (mm) = 0.003 (m)Room temperature, Tr = 86F = (86-32)/1.8 = 30 CTemperature at inside surface of pipe, Tp,i = 260 CConvection heat transfer coefficient between the insulation and the room air, ho = 5.2 (W/m2.C)Convection heat transfer coefficient between the steam and inside surface of pipe, hi =10.5 (W/m2.C)

The inside diameter of the pipe, dp,i = 74 (mm)The inside radius of the pipe, rp,i = 37 (mm) = 0.037 (m)The outside radius of the pipe, rp,o = 40 (mm) = 0.040 (m)

At 200 C: thermal conductivity of the pipe, kp =52 (W/m.C)At 300 C: thermal conductivity of the pipe, kp =50 (W/m.C)

By interpolation, thermal conductivity of the pipe at 260 C:

Critical radius of insulation,where :ki= thermal conductivity of insulation = 0.2184 (W/m.C)ho = 5.2 (W/m2.C)

Therefore:

a). Heat flow rate if the insulation thickness is equal to critical insulation thickness:

a). Qc = 24054.85 (W)

b). Heat flow rate of the naked pipe (Qnaked):

b). Qnaked = 24,027.15 (W)

c).Determination of the temperature at outer surface of the pipe at point a).

Tp,o = 259.9 Cc). The temperature at outer surface of the pipe at point a). is, Tp,o = 259.9 C

d).Determination of insulation thickness needed to decrease heat loss as much as 20 % from the heat loss without insulation.

Q0.8naked = 0.8 x Qnaked = 0.8 x 24027.15 = 19221.72 (W)

For: rx = 0.080 X = 1.218147181 rx = 0.098 X = 1.324659453

rx = 0.0960 (m)Insulation thickness, ti = 0.0960 0.040 = 0.056 (m)d). The insulation thickness needed, ti = 56.0 (mm)

e). Determination of Heat energy released by steam in 1.5 hours at point a). and point d).

Heat energy released by steam in 1.5 hours at point a).:Qa) in 1.5 h = 24054.85 (J/s) x 1.5 x 3600 (s) = 129896190 (J).

Heat energy released by steam in 1.5 hours at point d).:Qd) in 1.5 h = 19221.72 (J/s) x 1.5 x 3600 (s) = 103797288 (J).

HOMEWORK

Subject: Energy Management in IndustryDelivered: Tuesday, March 25th, 2014Collected: Tuesday, April 1st, 2014Lecturer: Dr. Riman SipahutarProblem:1Steam at Th = 572 F flows in a cast iron pipe (k = 80 W/m.C) whose inner and outer diameters are D1 = 60 mm and D2 = 66 mm, respectively. The pipe is covered with 35-mm thick glass wool insulation with k = 0.05 W/m.C. Heat is lost to the surroundings at Tc = 77 F by natural convection and radiation, with a combined heat transfer coefficient of hc = 16 W/m2.C. Taking the heat transfer coefficient inside the pipe to be hh = 60 W/m2.C. Determine:The rate of heat loss from the steam for the pipe length of 10 m.The temperature drops across the pipe shell and the insulation in degree C.The temperatures at inner and outer surfaces of the pipe in degree C.The temperatures at inner and outer surfaces of the insulation in degree C. The rate of heat loss from the steam for the pipe length of 10 m, without insulation.The percentage of energy saving for the system by using 35-mm thick glass wool insulation with k = 0.05 (W/m.C) compared with the system without insulation.

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