Water in an enclosed tank is subjected to a gauge pressure of 2x10^4 Pa, applied by a compressed air introduced into the top of the tank. There is a small hole (diameter = 4 cm) in the side of the tank 5 m below the level of the water. Calculate the discharge rate.
According to the Bernoulli equation
"(p_A+p_M)+\\rho gh\\approx p_A+\\rho v^2\/2 \\to"
"p_M+\\rho gh\\approx \\rho v^2\/2 \\to v=\\sqrt{2(p_M+\\rho gh)\/\\rho}="
"=\\sqrt{2\\cdot(20000+1000\\cdot9.8\\cdot5)\/1000}=11.75\\ (m\/s)" . Answer
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