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ch13 - Problem 13.1 Given Air extracted from a large tank...

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Problem 13.1 [2] Given: Air extracted from a large tank Find: Mass flow rate Solution: Basic equations: m rate ρ V A = h 1 V 1 2 2 + h 2 V 2 2 2 + = p ρ k const = T p 1 k ( ) k const = Given or available data T 0 70 273 + ( ) K = p 0 101 kPa = p 25 kPa = D 15 cm = c p 1004 J kg K = k 1.4 = R 286.9 J kg K = The mass flow rate is given by m rate ρ A V = A π D 2 4 = A 0.0177m 2 = We need the density and velocity at the nozzle. In the tank ρ 0 p 0 R T 0 = ρ 0 1.026 kg m 3 = From the isentropic relation ρ ρ 0 p p 0 1 k = ρ 0.379 kg m 3 = We can apply the energy equation between the tank (stagnation conditions) and the point in the nozzle to find the velocity h 0 h V 2 2 + = V 2 h 0 h ( ) = 2 c p T 0 T ( ) = Fot T we again use insentropic relations T T 0 p 0 p 1 k ( ) k = T 230.167K = T 43.0 °C = Then V 2 c p T 0 T ( ) = V 476 m s = The mass flow rate is m rate ρ A V = m rate 3.18 kg s = Note that the flow is supersonic at this point c k R = c 304 m s = M V c = M 1.57 = Hence we must have a converging-diverging nozzle
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