Question

A 30 reducing elbow is shown below. The fluid is water. Evaluate the x-y coordinate velocity components . V,,VV, and V2), the gage pressure (in unit of kPa) at the section 2, and the resultant forces (R and Ry in unit of kN) that must be provided by the adjacent pipes to keep the elb from moving. Assume steady state. Give an extra attention on the flow directions. Problem 1 110 points] pelooo k Elbow mass M-10 k Internal volume, vingnorable Section 2 Section -0.0182 m 35 Pa (gage) A0.0081 m -1wis 2yo. V 7 V and Hydraulics Gage pressure at section 2 (in unit of kPa)

Answer 1

Ans) We know,

Q = A x V

=> u1 = (Q / A1) Cos (30)

= (0.11 / 0.0081) Cos(30)

= - 11.77 m/s ( Negative sign due to flow direction)

=> v1 = (Q / A1) Sin (30)

= (0.11 / 0.0081) Sin(30)

= 6.8 m/s

= -11.77 i + 6.8 j

V1

Now,

=> u2 = Q / A2

= 0.11 / 0.0182

= - 6.04 m/s

=> v2 = Q/A2

Since, here in no inclination v2 = 0

   = - 6.04 i

V2

Now, sum of horizontal force = 0

=> - V2 Cos(30) u2 P A2 + V1 P1 u1 A1 + R_x = P2A2 Cos(30) - P1 A1        

=> R_x = P2A2 Cos(30) - P1 A1 + V2 Cos(30) u2 P A2 - V1 P1 u1 A1

=> R_x = P2 (0.0182)(0.866) - 15000(0.0081) + 9724.37

=> R_x = 0.0158 P2 - 9602.9

Sum of vertical forces = 0

=> R_y = P1A1Sin(30) + W_w + W_elbow - v1 P1 A1 u1 + v1 Sin(30)P A2 v2

=> R_y = 15000(0.0081)(0.5) + 0 + (10 x 9.81) + 9724.4

=> R_y = 9.83 kN

To determine Gauge pressure P1 apply Bernoulli equation,

P1 / $\rho$g + V1²/2g + Z1 = P2 /$\rho$g + V2²/2g + Z2

Since elevation is same, Z1 =Z2 =0

=> P2 /$\rho$ = P1/g + V₁² / 2 - V₂²/ 2

V1 = 0.11 / 0.0081 = 13.6 m/s

V2 = 6.04 m/s

=> P₂ = [(15 / 9.81) + 13.6² / 2 - 6.04²/2 ]

= (1.53 + 92.48 - 18.06) kPa

= 75.94 kPa

Hence, R_x = 0.0158(75940) - 9602.9

= - 8.4 kN

R_y = 9.83 kN

P2 =75.94 kPa