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Engineering Fundamentals: An Intro...

5th Edition
Saeed Moaveni
Publisher: Cengage Learning
ISBN: 9781305084766

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BuyFindarrow_forward

Engineering Fundamentals: An Intro...

5th Edition
Saeed Moaveni
Publisher: Cengage Learning
ISBN: 9781305084766
Chapter 11, Problem 39P
Textbook Problem
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Nine old 12 ft2 windows with U = 1.2 Btu h ft 2 ° F were replaced with new windows having U = 0.3 Btu h ft 2 ° F . Calculate the energy savings on a day during a 5 hour period, when T in = 68 ° F , T outside = 10 ° F .

To determine

Find the energy saving for the period of five hours.

Explanation of Solution

Given data:

The U-factor for old windows, Uold=1.2Btuhft2°F.

The U-factor for new windows, Unew=0.3Btuhft2°F.

Area of the windows, A=12ft2.

Inside temperature, Tin=68°F.

Outside temperature, Toutside=10°F.

Formula used:

The formula for the heat loss through the window,

q=UA(TinToutside) (1)

Here,

U is the U-factor of the windows,

A is the surface area of the windows,

Tin is the inside temperature of windows,

Toutside is the outside air temperature,

Calculation:

Substitute 1.2Btuhft2°F for U, 12ft2 for A, 68°F for Tin, and 10°F for Toutside in equation (1) to find heat loss through one old window (qold),

qold=(1.2Btuhft2°F)(12ft2)(68°F10°F)=(1.2Btuhft2°F)(12ft2)(58°F)=(1.2×12×58)Btuhqold=835.2Btuh

The heat loss through one old window for 5 hours is,

qold=(835.2Btuh)(5h)=(835.2×5)Btuqold=4176Btu

Hence the heat loss through 9 old windows will be,

qold=9×4176Btuqold=37584Btu (1)

Substitute 0

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Chapter 11 Solutions

Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)
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