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hw7_solutions - ES330 Assignment 7 Solutions Chapter 6 Due...

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ES330 Assignment 7 Solutions Chapter 6 Due Date: Thursday October 21, 2010 I Problem 1 A fan with 36 - in - diameter blades moves 1500 cfm (cubic feet per minute) of air at 70 o F at sea level. Determine ( a ) the force required to hold the fan and ( b ) the minimum power input required for the fan. Choose the control volume sufficiently large to contain the fan, and the gage pressure and the air velocity on the inlet side to be zero. Assume air approaches the fan through through a large area with negligible velocity and air exits the fan with a uniform velocity at atmospheric pressure through an imaginary cylinder whose diameter is the fan blade diameter. Solution (a) Force Required to Hold the Fan For steady, one-dimensional flow with no external forces, we can write the force balance as Σ ~ F = Σ out ˙ m ~ V - Σ in ˙ m ~ V (1) For this case, the force needed to hold the fan is in the x-direction ( F fan ), and Equation 1 becomes: F fan = ˙ mV 2 - ˙ mV 1 (2) From the problem statement we know V 1 = 0, and we also know ˙ m = ρV A , therefore: F fan = ˙ mV 2 = ρV 2 2 A (3) In order to find F fan , we need the density of the air at the given conditions. We can use the ideal gas law knowing that, at sea level, the pressure is 1 atm or 14.696 psi . ρ = P RT = 14 . 696 psi 53 . 34 lbf ( ft ) lbm ( o R ) 1 psi 144 lbf/ft 2 (70 + 459 . 67) o R = 0 . 0749 lbm ft 3 (4) 1
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We can also find V 2 because we know ˙ m = ρ = ρV A , therefore: V 2 = A = 1500 ft 3 min ( 1 min 60 s ) π 4 h 36 in 1 ft 12 in i 2 = 3 . 537 ft/s (5) We can now solve for F fan using Equation 13: F fan = 0 . 0749 lbm ft 3 3 . 537 ft s 2 π 4 36 in 1 ft 12 in 2 ! 1 lbf 32 . 2 lbf ( ft ) s 2 ! (6) F fan = 0 . 206 lbf (b) Minimum Power Input We can use the energy equation to solve for the required energy input for the fan: ˙ W net = ˙ m P 2 ρ + V 2 2 2 + gz 2 - ˙ m P 1 ρ + V 2 1 2 + gz 1 (7) We know z 1 = z 2 , V 1 = 0, and P 1 = P 2 = P atm , therefore, the above equation reduces to: ˙ W net = ˙ m V 2 2 2 = ρ V 2 2 2 = (0 . 0749 lbm ft 3 )(1500 ft 3 min 1 min 60 s )(3 . 537 ft s ) 2 2 1 lbf 32 . 2 lbm ( ft ) s 2 ! 1 W 0 . 73756 lbf ( ft ) s ! (8) ˙ W net = 0 . 494 W 2
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II Problem 2 A reducing elbow is used to deflect water (density= ρ ) with a mass flow rate of ˙ m in a horizontal pipe upward by an angle of θ from the flow direction while accelerating it. The elbow discharges water into the atmosphere. The diameter of the elbow is d in at the inlet and d in / 2 at the exit. The elevation difference between the centers of the exit and inlet is z . The mass of the elbow and the water in it is m e . Derive an expression for the anchoring force needed to hold the elbow in place if the flow is inviscid. Solution We can start by writing a force balance around the elbow (see The figure above showing the forces). We also have to include the pressure force at the inlet. We will use gage pressure, since both sides experience atmospheric pressure, and therefore the atmospheric pressure can be neglected.
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