# You are expected to know the Energy, and

You are expected to know the Energy, and Momentum balances Formula sheet No explanation is made on purpose Do not assume that you need to use every formula In this test always assume that Kentrance = 0.5, Kexit =1, Kbend = 0.25 If the flow in the pipe is in fully developed turbulence 1 k /D 2 log10 s f 3.7 h pipe L k s (gal iron ) 0.15mm k s (steel) 0.002mm F cos In direction along contact surface I y cp y yA L V2 f D 2g I triangle bh 3 / 36 Irec tan gle bh 3 / 12 Iellipse a 3b 4 Power At 100% eff =

In this test assume that 3 5 2 air 1.2 kg / m , air 1.46x 10 m / s Qh water 1 x10 6 m / s In a two dimensional velocity field If C = 0 velocity can be expressed in terns of a velocity potential C v y D x v x y v y v x x y v x , v y x

y If D= 0 velocity can be expressed in terns of a stream function v x , vy y x Voller 1 Calculate the average capillary rise of water ( = 9810 N/m3) between two vertical plates spaced 1mm apart. ( = 0.073 N/m) The plate one the left is normal glass with a contact angle of 0 o The plate on the right is surface treated glass with a contact angle of 60 0 Capillary force F, balance weight of liquid W For unit width into page F x1x (cos(0) cos(60)) x1.5 0.1095N W hx1xxw hx9810x 0.001 9.81h N h .1095 / 9.81 0.01116 m or 11 .16mm Voller 2 The figure shows the cross-section of a submerged hinged gate ( weight 100 kN) in water ( = 9810 N/m3). The gate has an area of A = 4 m2 And a geometry such that the centroid axis (parallel to the hinge and into the page) is a slant distance 2 m below the hinge and The line of action of the center of pressure is a slant distance of 2.5 m below the hinge Determine the height of water h required to just open the gate. h 300 Pr essure at centroid depth P ( h 2 x sin(30)) (h 1) 9810 x (h 1) N / m 2 Moment Analysis

Opening due to pressure 9810x(h 1)xAx2.5 9810x(h 1)x4x2.5 Closing due to weight Wcos(30)x2 Moment Balance 9810x(h 1)x4x2.5 Wx 3 (h 1) Wx 3 / 98100 h 100000 x1.732 / 98100 1 0.7655m Voller 3 300 The water ( = 1000 kg/m3) in a jet of area A = 0.05 m2 is deflected by a cone. If an external horizontal force of Fx = -1kN (acting to the left) is applied to the cone Calculate, assuming the cone moves in the horizontal direction, the magnitude of the velocity of the cone relative to the water jet (i.e., the value of Vjet-Vcone ) Momentum balance Q(Vjet Vcone ) Q(Vjet Vcone ) cos(30) 1000 Q (Vjet Vcone ) xA 0.05(Vjet Vcone ) SO 1000x 0.05x ( Vjet Vcone ) 2 (1 cos(30)) 1000 (Vjet Vcone ) 1 /(1 cos(30)) / 0.05 12.22 m / s Voller 4 Water ( = 9810 N/m3) flows from Tank A to Tank B at discharge of Q = 1 m3/s Through a pipe of Length 1000 m Diameter 0.5 m Surface roughness 0.1 mm Before entering Tank B the water is directed through a turbine with an 80% efficiency Assuming that the flow in the pipe is fully developed turbulence and approximating the Cross section as = 0.2 m2 calculate the power drawn from the turbine. A Power At 100% eff =

Qh 110 m B Kentrance = 0.5, Kexit =1, Kbend = 0.25 10 m 1 k /D 2 log10 s f 3.7 h pipe f L L V2 D 2g turbine Energy from A to B 110 10 h L h T h T 100 h L V2 L V2 )f 2g D 2g V Q / A 5m / s f 0.01373 h L 37.54 h L (2 x so h T 62.48 so Power 9810 x1x 62.48x.8 0.49034 MW

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