As shown in the figure, an air conditioner operating at steady state maintains a dwelling at 70°F on a day when the outside temperature Is 103.5°F. The rate of heat transfer Into the dwelling through the walls and roof is 30,000 Btu/h and the net power Input to the air conditioner compressor is 3.45 hp. maps. ΤΟ Γ Exporter 300 Ratigant loop Outsid Condenser mo Determine the coefficient of performance of the air conditioner. Determine the power Input required, In hp, and the coefficient of performance for a reversible air conditioner providing the same cooling effect while operating between the same cold and hot temperatures.

As shown in the figure, an air conditioner operating at steady state maintains a dwelling at 70°F on a day when the outside temperature Is 103.5°F. The rate of heat transfer Into the dwelling through the walls and roof is 30,000 Btu/h and the net power Input to the air conditioner compressor is 3.45 hp. maps. ΤΟ Γ Exporter 300 Ratigant loop Outsid Condenser mo Determine the coefficient of performance of the air conditioner. Determine the power Input required, In hp, and the coefficient of performance for a reversible air conditioner providing the same cooling effect while operating between the same cold and hot temperatures.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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A force of 600 N is applied on a rigid bar as shown in the figure. The moment of F about origin O in clockwise direction is. Select one: a. 2078.4 N.m b. 732 N.m c. 2732 N.m d. 1000 N.m
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Compare the magnitudes of the equilibrant vectors measured from the experiment with those obtained from the graphical and component methods. Example: A: 200 g 60° above +x axis B: 300 g 45° above -x axis C: 400 g 30°below -x-axis A, A cos a = 1.96 N x cos 60° = 0.98 N B₂B cos b = 2.94 N x cos 135º = -2.08 N C, C cos g = 3.92 N x cos 210°= -3.39 N R₂-A, + B + C₂ = -4.49 N A, A sin a = 1.96 N x sin 60° = 1.70 N By B sin b = 2.94 N x sin 135º = 2.08 N C, C sin g = 3.92 N x sin 210° = -1.96 N Ry= Ay+ By + Cy = 1.82 N Questions: I: (a) A: 200 g along +x axis B: 100 g 45° above -x axis A₂ = A cos a = B₂ =B cos b = R₂-A₂+ B₁₂= A₂ = A sin a = B, = B sin b = R₂ = A + B₂ = R=(R₂. R₂):_ Quadrant R = √ (R₂²+R₂²) = Direction:q=tan [R, /R.] (c) A: 100 g along -y axis B: 200 g along -x axis A = A cos a = B = B cos b = R₂-A₂+ B₂ = A = A sin a = B, B sin b = R=A, +B₂ = R=(R₁, R₂): Quadrant R = √ (R₂²+R₂²) = Direction:q tan¹ [R, /R]
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