A cylinder has a diameter of 15 cm. The water level in the cylinder is maintaine

Question

A cylinder has a diameter of 15 cm. The water level in the cylinder is maintained at a constant height of 0.48 m. If the diameter of the spout pipe is 0.60 cm, how high is h, the vertical stream of water? (Assume the water to be an ideal fluid.) Express your answer using two significant figures or you will not receive points.

A cylinder has a diameter of 15 cm. The water level in the cylinder is maintained at a constant height of 0.48 m. If the diameter of the spout pipe is 0.60 cm, how high is h, the vertical stream of water? (Assume the water to be an ideal fluid.) Express your answer using two significant figures or you will not receive points.

Explanation / Answer

Then the radius of cylinder is R = D /2 = 0.075 m Diameter of the spout pipe is d = 0.6 m Then the radius of the spout pipe is r = 0.3 cm = 0.3x10-2 m   Height of the water level is h = 0.48 m ------------------------------------------------------------------------------- Since the height of the water level in the cylinder is constant, the velocity of the water at surface of water is u1 = 0 Let the speed of the water at spout pipe is u2 Apply Bernoulli's equation at both surfaces,         P1 + (1/2)u12 + gh = P2 + (1/2)u22                  (1/2)u12 + gh = (1/2)u22        (Since both surfaces are exposed to atmosphere P1 = P2)                         u12 + 2gh = u22                                           2gh = u22                                       u2 = 2gh                                   = 2*9.8m/s2*0.48 m                                   = 3.0672 m/s Then the maximum height reached by the water i s                     H = u22 / 2g                         = (3.0672 m/s)2 / 2(9.8 m/s2)                         = 0.48 m                                          (1/2)u12 + gh = (1/2)u22        (Since both surfaces are exposed to atmosphere P1 = P2)                         u12 + 2gh = u22                                           2gh = u22                                       u2 = 2gh                                   = 2*9.8m/s2*0.48 m                                   = 3.0672 m/s Then the maximum height reached by the water i s                     H = u22 / 2g                         = (3.0672 m/s)2 / 2(9.8 m/s2)                         = 0.48 m                        

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.