Pa p 02 101200 p 02 p atm and hence pa p atm 101200 p

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Pa P 02 101200 = P 02 P atm := and hence Pa P atm 101200 := P 01 P atm 1 2 ρ 0 2 + = P atm = - therefore, the total pressure in the test section is the same as the total pressure in the room - but for 1 well away from the bellmouth, V 1 = 0 and P 1 = P atm - thus, P 01 P 02 = or P 1 1 2 ρ V 1 2 + P 2 1 2 ρ V 2 2 + = - thus, (1) reduces to (since the working fluid is a gas, the potential energy change would be small even if elevation changed between 1 and 2) y 1 y 2 = taking m/s 2 g 9.81 := (1) P 1 1 2 ρ V 1 2 + ρ g y 1 + P 2 1 2 ρ V 2 2 + ρ g y 2 + = - consider the streamline from 1 to 2 - Bernoulli is applicable (steady, incompressible flow with negligible friction) (a) Total pressure in the test section: P atm = 101,200 Pa T atm = 18C Water 15 cm P = 95,000 Pa 1 2 4.1 Flow in open circuit wind tunnel: MAAE2300/4.1
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