Recent content by angelala

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    Siphon pressure and velocity problem

    Thanks @Chestermiller and @haruspex !
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    Siphon pressure and velocity problem

    okay, but does point O work as well?
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    Siphon pressure and velocity problem

    So rho*g is gamma = 62.4 lbf / ft^3 right? I was just converting from lbm to lbf using the gravitational constant g_c
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    Siphon pressure and velocity problem

    P/rho*g = -2 rho is 62.4 lbm/ft^3 g = 32.2 ft / s^2 g_c = 32.2 P = -2 * (62.4/32.2) *32.2 = -4 psf.
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    Siphon pressure and velocity problem

    Bernoulli P/rho*g + v^2/2g + z = P/rho*g + v^2/2g + z For point of water surface O and D, P_atm/rho*g + 0/2g + 8 = P_atm/rho*g + v^2/2g + -2 8 = v^2/2g -2 v = sqrt ( (8+2) *2g ) = 25.4 ft/sFor point O and A P/rho*g + v^2/2g + z = P/rho*g + v^2/2g + z P_atm/rho*g + 0 + 8 = P/rho*g + 25.4^2/2g...
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    Siphon pressure and velocity problem

    Homework Statement Homework Equations Bernoulli equation. Continuity equation The Attempt at a Solution Area of pipe is constant, so v_a = v_b = v_c = v_d.[/B] Using point on water surface and point D. Point on water surface: z = 8, v = 0, P = 0 (P_atm) Point D: z = -2, P = 0 (P_atm) v =...