Group 7 4ChE A

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Information about Group 7 4ChE A

Published on January 5, 2009

Author: 4ChEAB08

Source: slideshare.net

Unit Operations Assignment on Conduction Jacinto, Stephanie Morenos, Jan Monil Sta. Rosa, Clariza Villanueva, Von Kirby 4ChE A Group 7

Problem no. 7 A plane wall is composed of an 20 cm layer of refractory brick (k = 1.3W/mK) and a 5 cm layer of insulating material with k for the insulating material varying linearly as k = 0.034 + 0.00018 t where t is the temperature in °C. the inside surface temperature of the brick is 1100°C and the outside surface temperature of the insulating material is 38 °C. Calculate the temperature at the boundary of the brick and insulation.

A plane wall is composed of an 20 cm layer of refractory brick (k = 1.3W/mK) and a 5 cm layer of insulating material with k for the insulating material varying linearly as k = 0.034 + 0.00018 t where t is the temperature in °C. the inside surface temperature of the brick is 1100°C and the outside surface temperature of the insulating material is 38 °C. Calculate the temperature at the boundary of the brick and insulation.

Problem no. 7 Given: T 1100 o C 38 o C T 1 = 1100 o C T 2 = 30 o C T= ? K brick = 1.3 W/mK K insulation = 0.034 + 0.00018T Basis: 1 m 2 Steady State: q 1 =q 2 20cm 5cm 1 2 Req’d: T Sol’n:

Given:

Problem no. 7 Interations: Answer: Ti = 827.99ºC 827.99 0.4467 432.99 827.98 827.98 0.4467 433.01 828.03 828.03 0.4467 432.87 827.74 827.74 0.4461 433.74 829.49 829.49 0.45 428.39 818.79 818.79 0.4270 461.61 885.21 885.21 0.6066 269 500 Ti,q (ºC) R(K/W) Tave(ºC) Ti(ºC)

Interations:

Problem no. 15 (US) A large sheet of glass 50 cm thick is initially at 150°C throughout. It is plunged into a stream of running water having a temperature of 15°C. How long will it take to cool the glass to an average temperature of 38°C? For glass: κ = 0.70 W/mK; ρ = 2480 kg/m3, Cp = 0.84 kJ/kgK

(US) A large sheet of glass 50 cm thick is initially at 150°C throughout. It is plunged into a stream of running water having a temperature of 15°C. How long will it take to cool the glass to an average temperature of 38°C? For glass: κ = 0.70 W/mK; ρ = 2480 kg/m3, Cp = 0.84 kJ/kgK

Problem no. 15 Given: 50cm To=150ºC T ave =38ºC Ti=15ºC Cp=0.84KJ/kg·K K=0.70W/m·K ρ = 2480 kg/m3 Req’d: t=? Sol’n:

Given:

Problem no. 15 Using the Average Temperature Table ( Fig. 5.3-13 on page 377) At Y = 0.17, X = 0.63

Using the Average Temperature Table

( Fig. 5.3-13 on page 377)

At Y = 0.17, X = 0.63

Geankoplis 4.1-2 Determination of Thermal Conductivity. In determining the thermal conductivity of an insulating material, the temperatures were measured on both sides of a flat slab of 25 mm of the material and were 318.4 and 303.2 K. the heat flux was measured as 35.1 W/m2. Calculate the thermal conductivity in btu/h.ft.ºF and in W/m.K

Determination of Thermal Conductivity. In determining the thermal conductivity of an insulating material, the temperatures were measured on both sides of a flat slab of 25 mm of the material and were 318.4 and 303.2 K. the heat flux was measured as 35.1 W/m2. Calculate the thermal conductivity in btu/h.ft.ºF and in W/m.K

Geankoplis 4.1-2 Given: Req’d: κ in Sol’n: T 1 =318.4K T 2 = 303.2K 25mm

Given:

Req’d: κ in

Sol’n:

Geankoplis 5.3-7 Cooling a Steel Rod. A long steel rod 0.305 m in diameter is initially at a temperature of 588 K. It is immersed in an oil bath maintained at 311 K. The surface convective coefficient is 125 W/m2.K. Calculate the temperature at the center of the rod after 1 h. The average physical properties of the steel are k=38 W/m.K and α=0.0381 m2.h

Cooling a Steel Rod. A long steel rod 0.305 m in diameter is initially at a temperature of 588 K. It is immersed in an oil bath maintained at 311 K. The surface convective coefficient is 125 W/m2.K. Calculate the temperature at the center of the rod after 1 h. The average physical properties of the steel are k=38 W/m.K and α=0.0381 m2.h

Geankoplis 5.3-7 Given: D= 0.305 m To=588 K T1=311 K h=125 W/m 2 ∙K t=1 h k=38 W/m 2 ∙K α=0.0381 m 2 /h Req’d: T at the center Use Gurney and Lurie Chart for cylinders When X=1.6 4 and m = 1.99 , n=0 Y=0.29 T = 391 K Sol’n:

Given:

D= 0.305 m

To=588 K

T1=311 K

h=125 W/m 2 ∙K

t=1 h

k=38 W/m 2 ∙K

α=0.0381 m 2 /h

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