Answer:
See explanations
Explanation:
(a) For steady flow we have
Q1 + Q2 + Q3 = Q4, or Fig. P3.8
VA VA VA VA 11 2 2 33 4 4 + + = (1)
Since 0.2Q3 = 0.1Q4, and Q4 = (120 m3/h)(h/3600 s) = 0.0333 m3/s,
3
4
3
2 3
(0.0333 m /s) (b) 2 (0.06 ) 2
Q V Ans.
A π == = 5.89 m/s
Substituting into (1),
22 2
1 (0.04 ) (5) (0.05 ) (5.89) (0.06 ) 0.0333 (a) 44 4
V Ans. ⎛⎞ ⎛⎞ ⎛⎞ ππ π
⎜⎟ ⎜⎟ ⎜⎟ ++ = ⎝⎠ ⎝⎠ ⎝⎠ V 5.45 m/s 1 =
From mass conservation, Q4 = V4A4
3 2
4 (0.0333 m /s) V ( )(0.06 )/4 (c) = π