Quiz+3+solutions

Quiz+3+solutions - V1 =(D2^2/D1^2 V2 then substitute in...

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1. 1) rho*V1*A1 = rho*V2*A2 => V1 = 2.6 m/s 2) P1 = P2 + 1/2*rho*(V2^2 - V1^2) = 28588 Pa (P2 = Patm = 101 kPa, but I set P2 = 0 because I calculate this with gage pressure.) 3) F = P1*A1 - rho*A1*V1*(V2 - V1) = 55.9 ~= 56 N 2. 1) manometer h = 0.66m 2) pressure in water at (1) is (rho*g*h)mercury - (rho*g*l)water where l is height of water column above the Hg in the manometer i.e., approx h + D1/2 3) P2 = 0 Pa g 4) Apply Bernoulli between 1 and 2 => P1 + 1/2 rho V1^2 = 1/2 rho V2^2 5) Apply continuity, assuming incompressible ±uid and SS => A1V1 = A2V2 so
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Unformatted text preview: V1 = (D2^2/D1^2)* V2 then substitute in equation above 6) solve for V2 3. By continuity, m_dot_1 = m_dot_2 => v2 = (A1/A2)*v1 = (D1/D2)^2*v1 = 9*v1 By Bernoulli's eqn (p2_gage = 0), p1/rho + (v1^2)/2 + gz1 = 0 + (v2^2)/2 + gz2 => (v2^2)/2 - (v1^2)/2 = p1/rho - g(z2 - z1) => 40*v1^2 = p1/rho + g*h => v1 = sqrt(1/40*(p1/rho + g*h) = sqrt(1/40*(51000/1000 - 9.91*0.7)) = 1.05 m/s => Q = v1*A1 = v1*pi/4*D1^2 = 1.05*pi/4*1.02^2 = 0.012 m^3/s => Q = 0.012(m^3/s)(1 gal/3.7854e-3 m^3)(60s/min) = 188 gal/min...
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This note was uploaded on 12/18/2011 for the course ENG 103 taught by Professor Chattot during the Fall '08 term at UC Davis.

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