7:09 a mybb.qu.edu.qa Issue Date: 15 sept 2018 Due Date: 23 sept 2018 (irn class

Question

7:09 a mybb.qu.edu.qa Issue Date: 15 sept 2018 Due Date: 23 sept 2018 (irn class) Total Points: 100 Question 1: (10 Points) Consider a 1.2-m-high and 2-m-wide glass window whose thickness is 6 mm and thermal conductivity is k-0.78 W/m C. Determine the steady rate of heat transfer through this glass window and the temperature of its inner surface for a day during which the room is maintained at 24°C while the temperature of the outdoors is -5 C. Take the convection heat transfer coefficients on the inner and outer surfaces of the win- dow to be h 10 W/m2 C and h2-25 W/m2 - °C, and dis- regard any heat transfer by radiation. Question 2: (20 Points) The wall of a refrigerator is constructed of fiberglass in- sulation (k-0.035 W/m C) sandwiched between two layers of 1-mm-thick sheet metal (k-15.1 W/m·). The refriger- ated space is maintained at 3°C. and the average heat transfer coefficients at the inner and outer surfaces of the wall are 4 W/m2 C and 9 W/m2 °C, respectively. The kitchen tem- perature averages 25 C. It is observed that condensation occurs on the outer surfaces of the refrigerator when the temperature of the outer surface drops to 20 C. Determine the minimum thickness of fiberglass insulation that needs to be used in the wall in order to avoid condensation on the outer surfaces. Question 3: (20 Points) Steam at 450 F is flowing through a steel pipe (k-8.7 Btu/h ft F) whose inner and outer diameters are 3.5 in and 4.0 in., respectively, in an environment at 55 F. The pipe is insulated with 2-in.-thick fiberglass insulation (-0.020 Btu/h ft.F). If the heat transfer coefficients on the inside and the outside of the pipe are 30 and 5 Btulh fi F respectively determine the rate of heat loss from the steam per foot length of the pipe. What is the error involved in neglecting the thermal resistance of the steel pipe in calculations? Question 4: (20 H_

Explanation / Answer

Ans 1

Resistance due to inner convection

R1 = 1/(h1 x A)

= 1/(10 W/m2-°C x 1.2 x 2 m2)

= 0.041666 °C/W

Resistance due to conduction

R2 = t/(k x A)

= (6 mm x 1m/1000mm) / (0.78 W/m-°C x 1.2 x 2 m2)

= 0.0032051 °C/W

Resistance due to outer convection

R3 = 1/(h2 x A)

= 1/(25 W/m2-°C x 1.2 x 2 m2)

= 0.016666 °C/W

Total thermal resistance

R = R1 + R2 + R3

= 0.041666 + 0.0032051 + 0.016666

= 0.0615371 °C/W

Heat transfer

Q = (T1 - T2) /R

= [24 - (-5)] °C / (0.0615371 °C/W)

= 471.26 W

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