This preview shows pages 1–8. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: (1) The force acting on a window F = p 2 A (2) where A = area of the window. Step5: Known parameters z 1 = z 2 , p 1 = p atm = 0, u 1 = 27.8 m/s, u 2 = 0 m/s, A = 1m × 2m = 2 m 2 . Step6: Calculations Substitution of the known values into Eq. (1) yields 2 2 2 1 2 N/m 464 2 (27.8) 1.2 2 = × = = V p ρ Substituting p 2 = 464 N/m 2 and A = 2 m 2 into Eq. (2) gives F = 464 × 2 = 928 N against window . 2 1 g V z p g V z p 2 2 2 2 2 2 2 1 1 1 + + = + + γ Problem3: Solution Problem4: Solution...
View
Full
Document
This note was uploaded on 12/09/2009 for the course CE 2200 taught by Professor Deng during the Spring '09 term at LSU.
 Spring '09
 DENG

Click to edit the document details