Explanation Write the expression for the energy balance equation for the closed system. E ˙ in − E ˙ out = ( d E d t ) system (I) Here, rate of net energy transfer in to the control volume is E ˙ in , rate of net energy transfer exit from the control volume is E ˙ out and rate of change in internal, kinetic and potential energies of system is ( d E / d t ) system . For steady state, the rate of change in internal, kinetic and potential energies of the system is zero. ( d E / d t ) system = 0 Substitute 0 for Δ E ˙ system in Equation (I). E ˙ in − E ˙ out = 0 E ˙ in = E ˙ out W ˙ elect, in = m ˙ air V out 2 2 (II) Here, rate of electric power input to the system is W ˙ elect, in , mass flow rate of the air is m ˙ air and output velocity of the air is V out . Write the formula to calculate mass flow rate of the air ( m ˙ air ) . m ˙ air = ρ V out A (III) Here, density of the air is ρ and area of the duct is A . Conclusion: Substitute 0.075 lbm / ft 3 for ρ , 22 ft / s for V out and 3 ft × 3 ft for A in Equation (III)

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Chapter 2.8, Problem 48P

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The minimum electric power consumption of the motor.

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Chapter 2.8, Problem 47P

Chapter 2.8, Problem 49P

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