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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

The average thermal conductivity of the walls (including windows) and roof of a house in Figure P11.46 is 4.8 · 10−4 kW/m · °C, and their average thickness is 21.0 cm. The house is heated with natural gas, with a heat of combustion (energy released per cubic meter of gas burned) of 9 300 kcal/m3. How many cubic meters of gas must be burned each day to maintain an inside temperature of 25.0°C if the outside temperature is 0.0°C? Disregard surface air layers, radiation, and energy loss by heat through the ground.

images

Figure P11.46

To determine
The amount of gas must be burned each day to maintain a temperature.

Explanation

Section1:

To determine: The total surface area of the house.

Answer: The total surface area of the house is 304m2 .

Explanation:

Given info: The length of the side wall is 10.0 m, length of the end wall is 8.00 m and the height of the side wall is 5.00 m, the roof inclination angle is 37.0° .

Formula to calculate the area of the sidewalls is,

Aside=2(lsh) (1)

  • Aside is the area of the sidewalls,
  • ls is the length of the sidewall,
  • h is the height of the sidewall,

Formula to calculate the area of the end walls is,

Aendwalls=2(leh) (2)

  • Aendwalls is the area of the end walls,
  • le is the length of the end wall,
  • h is the height of the sidewall,

Formula to calculate the area of the gables is,

Agables=2[12(lehg)] (3)

  • Agables is the area of the gables,
  • le is the length of the end wall,
  • hg is the altitude of the gables,

Formula to calculate the area of the roof is,

Aroof=2(lswr) (4)

  • Aroof is the area of the roof,
  • wr is the width of the roof,

The total area of the house is,

A=Aside+Aendwalls+Agables+Aroof (5)

Use (I),(II),(III),(IV) in (V) to rewrite A.

A=2(lsh)+2(leh)+2[12(lehg)]+2(lswr) (6)

From the figure

hg=(le2)tan37.0° (7)

wr=(le/2)cos37.0° (8)

Use (7) and (8) in (4) to rewrite A.

A=2(lsh)+2(leh)+2[12(le(le2)tan37.0°)]+2(ls(le/2)cos37.0°)

Substitute 10.0 m for ls , 8.00 m for le , and 5.00 m for h to find A.

A={2[(10.0m)(5.00m)]+2[(8.00m)(5.00m)]+2[12((8.00m)((8.00m)2)tan37.0°)]+2((10.0m)((8

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