hw5_solutions

hw5_solutions - ES330 Assignment 5 Solutions Chapter 5 Due...

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ES330 Assignment 5 Solutions Chapter 5 Due Date: Tuesday October 5, 2010 I Problem 1 Consider a river flowing toward a lake at an average velocity of 3 m/s and a volume flow rate of 250 m 3 /s . If 20% of the river water is diverted through a turbine, what is the minimum height drop required to generate 150 MW of power? Solution In order to find the required height, we can start with the following equation for the power: P = ˙ m e mech = ρ ˙ turb " V 2 2 + gz # (1) Solving for z (the height of the drop) we get: z = ± 1 g ² P 0 . 2( ˙ total ) ρ - V 2 2 ! (2) z = ± 1 9 . 81 m/s 2 ² 150 , 000 kW 0 . 2(250 m 3 /s )(1000 kg/m 3 ) - (3 m/s ) 2 2 1 kW · s/kg 1000 m 2 /s 2 ! 1000 m 2 /s 2 1 kW · s/kg ! (3) z = 305 . 4 m 1
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II Problem 2 Water is pumped from a lake to a storage tank 40 m above at a rate of 55 L/s while consuming 23 . 4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine ( a ) the pressure difference between the inlet and exit of the pump and ( b ) the overall efficiency of the pump-motor unit. Solution (a) Find the pressure difference between the inlet and exit of the pump: To find the pressure difference, we need to find the rate of change of mechanical energy of the water supplied to the pump. This can be expressed as: Δ ˙ E mech,fluid = ˙ m e mech,out - ˙ e mech,in ) = ˙ m P 2 - P 1 ρ + V 2 2 - V 2 1 2 + g ( z 2 - z 1 ) ! (4) We will call the surface of the lake point 1 and the surface of the storage tank point 2. Let’s assume that the surface of the water is at z = 0 and kinetic energy is negligible. Also, the pressure at both points is atmospheric and cancels. Therefore, the above equation becomes: Δ ˙ E mech,fluid = ˙ mgz 2 = ρ ˙ gz 2 (5) Δ ˙ E mech,fluid = (1000 kg/m 3 ) ± 55 L s ± 1 m 3 1000 L ²² (9 . 81 m/s 2 )(40 m ) ± 1 kW 1000 kg * m 2 /s 3 ² Δ ˙ E mech,fluid = 21 . 58 kW Now looking at the pump, the change in the mechanical energy of the water is also the mechanical energy supplied to the pump. Also, in the pump, the kinetic energy and elevation change are negligible. Therefore: Δ ˙ E mech,fluid = Δ ˙ E pump = ˙ m ( P 2 - P 1 ρ ) = ρ ˙ ( Δ P ρ ) (6) Solving for Δ P : Δ P = Δ ˙ E pump ˙ = 21 . 58 kW ± 1000( m 2 kg ) /s 3 1 kW ² 55 L s ± 1 m 3 1000 L ² ³ 1 kPa 1000 kg/ ( ms 2 ) ´ (7) Δ P = 392 . 4 kPa 2
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(b) Find overall efficiency of pump motor unit: To find the overall efficiency, we can use the following equation: η pump - motor = Δ ˙ E mech,fluid ˙ W elec,in = 21 . 58 kW 23 . 4 kW = 0 . 922 (8) η pump - motor = 92 . 2% 3
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hw5_solutions - ES330 Assignment 5 Solutions Chapter 5 Due...

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