Bernoulli's equation:

1. 

   w = 0 U _A = 0 (large cross section of tank)

   F = 28.5 ft given. ? _A = ? _B (incompressible fluid)

   

   Bernoulli equation reduces to Z _A - Z _B = F + = 28.5 + 1.5 = 30 ft

   Height from the bottom of tank = 30 - 10 = 20 ft

   Answer is (C)

2. 

   Bernoulli equation reduces to: w = F + (Z _B - Z _A) +

   Velocity through 2" pipe = 3(3.068/2.067) ² = 6.61 ft/s

   

   ? w = 12 + (40 - 10) + 0.68 = 42.68 ft

   Pressure developed = = 29.7 psia

   Therefore, pressure indicated by gage = 29.7 - 14.7 = 15 psig

   Answer is (D)

3. The sketch shows the conditions of the problem.

   

   P _A = P _B ? _A = ? _B U _A = 0 cross sectional area of tank very large

   ? The Bernoulli equation reduces to

   Z _A + W - F = Z _B +

   

   Z _A = Z _B + + F - W

   F = 1.5 ft Z _B = 20 ft

   U _B = 80/(7.48 x 60 x 0.0233) = 7.65 ft/s

   = 7.65 ²/64.4 = 0.91 ft

   = 0.25 m = = 11.12 lb/s

   W _P = 0.25(550)/11.12 = 12.36 ft

   Therefore, Z