Suppose the tube in the previous problem is kept vertical with A upward but the other conditions remain the same.
Suppose the tube in the previous problem is kept vertical with A upward but the other conditions remain the same. The separation between the cross-sections at A and B is 15/16 cm. Repeat parts (a), (b) and (c) of the previous problem. Take g = 10 m/s2.
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Since the discharge through the tube remains the same i.e. 1 cc/s, the speeds at A and B will be the same.
(a) The speed at A, v =1/0.04 cm/s =100/4 cm/s =25 cm/s
(b) The speed at B, v' =1/0.02 cm/s =100/2 cm/s =50 cm/s
(c) From the Bernoulli's theorem
Pₐ+ρgh+½ρv² = Pᵦ+½ρv'²
Taking the point B as a reference level for the height. Given that h=15/16 cm
→Pₐ-Pᵦ =½ρ(v'²-v²) -ρgh
=½*1000*(0.50²-0.25²) -1000*10*15/1600 N/m²
=½*1000*0.75*0.25 - 1500/16 N/m²
=93.75 - 93.75 N/m²
= zero
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