I need full steps please A water jet is discharged from a 6 inch diameter pipe a

Question

I need full steps please

A water jet is discharged from a 6 inch diameter pipe as shown below. The water impinges on a 90degree turning vane where it is turned vertically upward. It reaches a maximum height of 20 feet before it falls back to the ground, rho = 1.94 sl/ft3 What is the velocity of the water jet at point 1? What is the volumetric flow rate through the pipe (assume the velocity in the pipe the same as the velocity at point 1)? What is the mass flow rate through the pipe? What is the horizontal anchoring force required to hold the turning vane in

Explanation / Answer

Using P1 + 1/2*rho*V1^2 + rho*g*h1 = P2 + 1/2*rho*V2^2 + rho*g*h2

We have, P1 = P2 = Pambient and V2 = 0

Thus, 1/2*rho*V1^2 = rho*g*(h2 - h1)

1/2*V1^2 = g*20

V1 = (2*32.2*20)^0.5

V1 = 35.89 ft/s

b)

Volumetric flow rate Q = Area*V = (3.14/4*(6/12)^2)*35.89 = 7.04 ft^3 /s

c)

Mass flow rate m = density*Q = 1.94*7.04 = 13.66 slug /s = 13.66*32.2 lb/s = 440 lb/s

d)

Horizontal force = Change in Horizontal component of mometum = m*V1 - 0 = 440*35.89 = 15790.6 lb-ft/s^2 = 15790.6 / 32.2 lb = 490.4 lbs

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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