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steam is leaving a pressure cooker whose operating pressure is 20 psia. it is observed that the amount of liquid in the cooker has decreased by 0.6 gal in 45 minutes after the steady operating conditions are established, and the cross-sectional area of the exit opening is 0.15 in2. determine:

Respuesta :

The steady operating conditions are established, and the cross-sectional area of the exit opening is 0.15 inc then the Exit velocity is 34.04 ft/s

The amount of liquid that has

evaporated

m = V/V6

Now V = 0 6 gal

V = 0. 6 * 0 .13 3 68 ft

V = 0 080208 4 3

m =0 . 080208/0.016ft

M= 4. 765 cbm

The mass flowrate of exiting steam liquid becomes

m=m/Δt

& Δt= 45 min

At = 45 x 60 sec

m = 4 . 765/45 x 60

n = 0.00 1 765 ( lbm/s )

Now the exit velocity is

V = m v/A

& A = 0 15 in 2

A=0 .15/( 12 )²ft²

A = 0.001042 ( ft² )

" . V = 0 . 001 7 65 x 20. 093/0.00 1 042

V = 34. 04 ( ft² )

.Exit velocity is 34.04 ( ( ft²/s )

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