p0698 - 6.98. CHAPTER 6, PROBLEM 98 865 6.98 Chapter 6,...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 6.98. CHAPTER 6, PROBLEM 98 865 6.98 Chapter 6, Problem 98 Problem: A water jet of density ρ with initial velocity W and diameter do issues vertically from a wall and supports a cone of half angle φ that rises to height h as shown. You can assume the flow is steady, irrotational and that velocities are constant across all cross sections. Also, you can ignore the weight of the fluid in the control volume. NOTE: Neither the upflow velocity, w, nor the area of the upflowing fluid at the base of the cone are given, you must solve for them. (a) Using the indicated control volume, determine the weight of the cone, M g . (b) Compute M g for ρ = 1.94 slug/ft3 , W = 20 ft/sec, do = 2 in, h = 1 ft and φ = 30o . w w Mg ... ......... . . ......... .... .................... ......................... ................... . .. .. .. .. . .......................................................................................................... ...................................................................................................... ................................................................................................. . . ....... ...................... ............................................................................................. ......................................................................................... ....................................................................................... .... .................. .... ............. ............................................................................... .................................................................................. ........................................................................... ......................................................................... .................................................................... .................................................................. ....... ...... . .... .... . .... ................................................................. .............................................................. ................................................... . .................................................... ...................................................... .................................................. . . ............................................... ............................................. ........................................... .. ......................................... ...................................... ...................................... ..... . .. ... .................................... ............................ .. . ................................ .............................. ............................. ..... ......... ........................... .......................... ......................... ......................... ......................... ....................... ....................... ...................... ... ....... .......................... .................. ..................... ................ ..................... o ..................... ..... ...................... ..................... ... ................................................................................................................................................................................................................................. .................... ..................................................................................................................................................................................................................................... .. . . . ................................................................................................................................................................................................................................. ................................................ . .. .............................................................................................................................................................................................................................. ................................................................................................................................. . .. . .. .. 2φ g = −g k h z W d Solution: (a) We use the stationary control volume indicated in the figure, which includes only fluid. Mass Principle: In general, the mass principle tells us that d dt ρ dV + V S ρ ure · n dS = 0 The flow is incompressible and the control volume’s size is constant. Thus, the instantaneous rate of change of the mass in the control volume vanishes. Then, since ure = u, we have ρ u · n dS = 0 S Evaluating the closed-surface integral yields (with A denoting upflow area) π ρ −W d2 + 4o In et ρ (wA) Upflow stream =0 =⇒ wA = π W d2 o 4 866 CHAPTER 6. CONTROL-VOLUME METHOD z -Momentum Principle: Turning now to z momentum, we have S ρw (u · n) dS = −k · S (p − pa ) n dS − ρg dV V First, consider the net flux of z momentum out of the control volume. Fluid enters at the ground plane and exits in the upflow stream. Hence, π π ρw (u · n) dS = ρW −W d2 + (ρw cos φ) (wA) = − ρW 2 d2 + ρw2 A cos φ o o 4 4 S Upflow stream Inlet Turning to the pressure integral, since the pressure is atmospheric at all points on the surface bounding the control volume other than on the cone surface, −k · S (p − pa ) n dS = −k · C one (p − pa ) n dS = −Mg Also, we are given that the weight of the fluid contained in the control volume can be neglected, wherefore ρg dV ≈ 0 V So, the z -momentum equation simplifies to π − ρW 2 d2 + ρw2 A cos φ = −M g o 4 We can simplify further by using our result from the mass principle, so that Mg = π π w π ρW 2 d2 − ρW wd2 cos φ = ρW 2 d2 1 − cos φ o o o 4 4 4 W Bernoulli’s Equation: There are 3 unknowns in the problem, viz., w, A and M . The mass and z -momentum principles provide 2 of the required equations. To close our system of equations, we now appeal to Bernoulli’s equation. We use it to relate conditions at the surface and in the upflow area, noting that pressure is atmospheric at both points. 1 1 pa + ρW 2 = pa + ρw2 + ρgh 2 2 Therefore, we conclude that w2 = W 2 − 2gh =⇒ w= W 2 − 2gh Substituting for w in the momentum equation, we find ⎡ π M g = ρW 2 d2 ⎣1 − o 4 ⎤ 2gh 1 − 2 cos φ ⎦ W 6.98. CHAPTER 6, PROBLEM 98 867 (b) We are given ρ = 1.94 slug/ft3 , W = 20 ft/sec, do = 2 in, h = 1 ft and φ = 30o . First, note that 2 32.174 ft/sec2 (1 ft) 2gh = = 0.161 W2 400 ft2 /sec2 Thus, the weight of the cone is slug π 1.94 3 Mg = 4 ft ft2 400 sec2 12 ft 36 1− √ 1 − 0.161 (0.866) = 3.50 lb ...
View Full Document

This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

Ask a homework question - tutors are online