# shmwk06 - Vin x − Fx = ρ QV)out − ρ QV)in = ρ...

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Unformatted text preview: Vin x − Fx = ( ρ QV )out − ( ρ QV )in = ρ Q (Voutx − Vinx ) x x ( ρ Q )out = ( ρ Q )in x VinVinxVout Vin ( − cosθ ) θ = 30° − cosθ x 1/2 1/2 Fx (π D 2 / 4 ) 1200 N (π / 4 )( 0.04 m )2 Fx = ρ QVin (1 + cosθ ) = Q= = 3 3 Ain ρ (1 + cosθ ) 10 kg/m (1 + cos30° ) 3 = 0.028 m /s Dy − ( F ' ) x = ( ρ Q 'V ' )out − ( ρ Q 'V ' )in = ρ Q (V ' )outx − (V ' )inx ρ Q 2 (1 + cosθ ) x x Q ' V ' F ' 2 ( F ' ) x = ρ Q ' (V ' )in (1 + cosθ ) = ρ (V ' )in Ain (1 + cosθ ) 1/2 1/2 ( F ')x 1800 N V ' )in = = 27.7 m/s = 3 ( 2 2 3 ρ (1 + cosθ ) (π D / 4) 10 kg/m (1 + cos30°)(π / 4 )( 0.04 m ) Q / Ain = Q / (π D 2 / 4 ) = (V ' )in = Vin − Vvane Vvane = Vin − (V ' )in = (22.6 − 27.7) m/s = −5.1 m/s x • • p V2 p p V2 p V2 Q2 = + z − + z = Δ = +z+ = +z+ γ 2 g 1 γ 2 g 2 2 g 1 2 gA12 γ γ 2 1 V2 Δ p 2 − p a = −γ ( z 2 − z a ) p p Q2 pa − pb ≈ 0 Δ = + z − + z = za − zb = h = 2 gA12 γ 2 γ 1 pb − p1 = −γ ( zb − z1 ) Δ = (γ m / γ ) − 1 h A = π Dh FA,z, FA,z = 3.52 kN xx x − Rx + ( pA )in − ( pA )out = ( ρ QV )out − ( ρ QV )in x x Rx = ( pA )in − ( pA)out + ( ρ QV )in − ( ρ QV )out x x x x = (γ H 2 / 2 ) − (γ h 2 / 2 ) + ρ Q ( Q / H ) − ( Q / h ) Q = VinH hs hp p p V2 V2 + z +α + hp = + z + α + hL 2 g in 2 g out γ γ pin = 0 = Vin = Vout h p = Wpump / ( mg ) = W pump / (γ Q ) = 3 hp/ ( 62.4 lbf/ft 3 ) (100 gpm) = 118.7 ft p 2 atm hL = hp + zin − zout − out = 118.7 ft − 20 ft − = 30.9 ft γ 62.4 lbf/ft 3 h p ≥ 0 pout hL = −3.1 ft pout = 3 atm y R y x x R ) QV (ρ p =0 θ − Ry + [( pA)in ] y + [ ( pA)out ] y = ( ρ QV )out − ( ρ QV )in = ρ Q (Vout ) y − (Vin ) y y y (ρQV) ( ρ Q )in = ( ρ Q )out pout = 0 y out out in (Vout ) y = ( Q / Aout )( − sinθ ) Q Ry + ρ Q − sinθ − Vin Aout = 189.6 lbf/ft 2 pin = Ain (pA)in x p p V2 V2 + z +α = + z +α + hL 2 g in γ 2 g out γ pout 2 p V 2 − Vout hL = in + in = 2.36 ft γ 2g ...
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## This note was uploaded on 01/20/2012 for the course CIVIL ENGI 340 taught by Professor Troy during the Spring '11 term at Purdue.

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