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Information about 4ChEB G2 Unit Operations#2

This presentation has solutions to problem set 2

numbers 5 and 13.

This also contains a discussion on the problems from Geankoplis numbers 4.2-3 and 5.3-5

numbers 5 and 13.

This also contains a discussion on the problems from Geankoplis numbers 4.2-3 and 5.3-5

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Problem 2 A furnace wall is to consist in series of 18 cm of kaolin firebrick, 15 cm of kaolin insulating brick, and sufficient fireclay brick to reduce the heat loss to 350 W/m 2 when the face temperatures are 815 and 38 ºC, respectively. If an effective air gap of 3 mm (assume k for air as 0.17 W/mK) can be incorporated between the fireclay and insulating brick when erecting the wall impairing its structural support, what thickness of fireclay brick will be required? Given: 1 2 Req’d: x 4 3 1 2 4 3 Kaolin Firebrick Kaolin Insulating Brick Air Fireclay Brick

A furnace wall is to consist in series of 18 cm of kaolin firebrick, 15 cm of kaolin insulating brick, and sufficient fireclay brick to reduce the heat loss to 350 W/m 2 when the face temperatures are 815 and 38 ºC, respectively. If an effective air gap of 3 mm (assume k for air as 0.17 W/mK) can be incorporated between the fireclay and insulating brick when erecting the wall impairing its structural support, what thickness of fireclay brick will be required?

Problem 2 Given: 1 2 Req’d: x 4 3 1 2 4 3 Kaolin Firebrick Kaolin Insulating Brick Air Fireclay Brick Sol’n: Assume: A 1 = A 2 = A 3 = A 4 = 1m 2 Assume:

Problem 2 Given: 1 2 Req’d: x 4 1 2 4 3 Kaolin Firebrick Kaolin Insulating Brick Air Fireclay Brick Sol’n: 3 Assume:

Problem 2 Assume: Given: 1 2 Req’d: x 4 1 2 4 3 Kaolin Firebrick Kaolin Insulating Brick Air Fireclay Brick 3

Problem 2 Assume: Given: 1 2 Req’d: x 4 1 2 4 3 Kaolin Firebrick Kaolin Insulating Brick Air Fireclay Brick 3

Problem 2 Given: 1 2 Req’d: x 4 1 2 4 3 Kaolin Firebrick Kaolin Insulating Brick Air Fireclay Brick 3

Problem 10 An experimental heat transfer apparatus consists of a 5 cm. schedule 80 steel pipe covered with two layers of insulation. The inside layer is 2.5 cm thick and consists of diatomaceous silica, asbestos and a bonding material; the outside layer is 85% magnesia and is 4 cm thick. The following data were obtained during a test run: length of test section = 3 m. heating medium inside pipe = dowtherm A vapour temperature of inside of steel pipe = 400°C temperature of outside of magnesia insulation = 52°C dowtherm condensed in test section = 9 kg/hr temperature of condensate = 400°C latent heat of condensation of dowtherm of upper conditions = 206 kJ/kg Determine the mean thermal conductivity of the magnesia insulation. The thermal conductivity of the inner layer is as follows: Temperature (°C) k (W/mK) 93 0.0885 260 0.1050 426 0.1209

An experimental heat transfer apparatus consists of a 5 cm. schedule 80 steel pipe covered with two layers of insulation. The inside layer is 2.5 cm thick and consists of diatomaceous silica, asbestos and a bonding material; the outside layer is 85% magnesia and is 4 cm thick. The following data were obtained during a test run:

length of test section = 3 m.

heating medium inside pipe = dowtherm A vapour

temperature of inside of steel pipe = 400°C

temperature of outside of magnesia insulation = 52°C

dowtherm condensed in test section = 9 kg/hr

temperature of condensate = 400°C

latent heat of condensation of dowtherm of upper conditions = 206 kJ/kg

Determine the mean thermal conductivity of the magnesia insulation. The thermal conductivity of the inner layer is as follows:

Temperature (°C) k (W/mK)

93 0.0885

260 0.1050

426 0.1209

Problem 10 Given: Nominal Pipe Size Sch. 80 Steel Inner Insulation Outer Insulation

Problem 10 Assume:

Problem 4.3-3 Heat Loss Through Thermopane Double Window. A double window called thermopane is one in which two layers of glass are separated by a later of dry, stagnant air. In a given window, each of the glass layers is 6.35 mm thick separated by a 6.35-mm space of stagnant air. The thermal conductivity of the glass is 0.869 W/m·K and that of air is 0.026 over the temperature range used. For a temperature drop of 27.8 K over the system, calculate the heat loss for a window 0.914 m x 1.83 m. ( Note: This calculation neglects the effect of the convective coefficient on one outside surface of one side of the window, the convective coefficient on the other surface, and convection inside the window.)

Heat Loss Through Thermopane Double Window. A double window called thermopane is one in which two layers of glass are separated by a later of dry, stagnant air. In a given window, each of the glass layers is 6.35 mm thick separated by a 6.35-mm space of stagnant air. The thermal conductivity of the glass is 0.869 W/m·K and that of air is 0.026 over the temperature range used. For a temperature drop of 27.8 K over the system, calculate the heat loss for a window 0.914 m x 1.83 m. ( Note: This calculation neglects the effect of the convective coefficient on one outside surface of one side of the window, the convective coefficient on the other surface, and convection inside the window.)

Problem 4.3-3 G G A G A Glass Air 6.35 mm 6.35 mm Window Dimensions: 0.914 m x 1.83 m Given: Req’d: q Sol’n:

Problem 5.2-2 Quenching Lead Shot in a Bath. Lead shot having an average diameter of 5.1 mm is at an initial temperature of 204.4°C. To quench the shot, it is added to a quenching oil bath held at 32.2°C and falls to the bottom. The time of fall is 15 s. Assuming an average convection of h = 199 W/m 2 ·K, what will be the temperature of the shot after the fall? For lead, ρ = 11,370 kg/m 3 and c P = 0.138 kJ/kg·K.

Quenching Lead Shot in a Bath. Lead shot having an average diameter of 5.1 mm is at an initial temperature of 204.4°C. To quench the shot, it is added to a quenching oil bath held at 32.2°C and falls to the bottom. The time of fall is 15 s. Assuming an average convection of h = 199 W/m 2 ·K, what will be the temperature of the shot after the fall? For lead, ρ = 11,370 kg/m 3 and c P = 0.138 kJ/kg·K.

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