Explanation Given: The maximum temperature is 80 ° F . The outside temperature is 50 ° F . Time constant of the garage is 2 hr . Calculation: Consider the heat produced is at a constant rate H 0 and no heating and cooling that is, U ( t ) = 0 Consider the outside temperature M ( t ) varies as s sine wave over a 24 hrs period with its minimum at t = 0 and maximum at t = 12 therefore, M ( t ) = M 0 − B cos ω t Here, B is positive constant, and M 0 is the average outside temperature. And, ω = 2 π 24 = π 12 rad / hr The temperature distribution is given by, T ( t ) = e − K t { ∫ e K t [ K ( M 0 − B cos ω t ) + H 0 ] d t } = e − K t { ( K M 0 + H 0 ) ∫ e K t d t − K B ∫ e K t cos ω t d t } = e − K t { ( K M 0 + H 0 ) e K t K − K B e K t K 2 + ω 2 ( K cos ω t + ω sin ω t ) + C } = ( M 0 + H 0 K ) − K B K 2 + ω 2 ( K cos ω t + ω sin ω t ) + C e − K t Simplify the above equation. T ( t ) = ( M 0 + H 0 K ) − K B K 2 + ω 2 ( K cos ω t + ω sin ω t ) + C e − K t = ( M 0 + H 0 K ) − K B K 2 + ω 2 ( K K 2 + ω 2 cos ω t + ω K 2 + ω 2 sin ω t ) + C e − K t Consider, sin ϕ = − K K 2 + ω 2 and cos ϕ = ω K 2 + ω 2 Hence, T ( t ) = ( M 0 + H 0 K ) − K B K 2 + ω 2 ( K K 2 + ω 2 cos ω t + ω K 2 + ω 2 sin ω t ) + C e − K t = ( M 0 + H 0 K ) − K B K 2 + ω 2 ( cos ϕ sin ω t − sin ϕ cos ω t + ) + C e − K t = ( M 0 + H 0 K ) − K B K 2 + ω 2 sin ( ω t − ϕ ) + C e − K t = B 0 − B 1 + ( ω K ) 2 sin ( ω t − ϕ ) + C e − K t …… ( 1 ) Here, B 0 = M 0 + H 0 K Given that the outside temperature varies as s sine wave with a minimum of 50 ° F at 2 : 00 A

7th Edition

ISBN: 9780321977175

Author: Nagle

Publisher: PEARSON CO

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Question

Chapter 3.3, Problem 8E

To determine

The time, at which the building reaches its lowest temperature and its highest temperature, assuming the exponential term has died off.

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Chapter 3.3, Problem 7E

Chapter 3.3, Problem 9E

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