Question

10 points Sive Answer A fluid of specific gravity 1.5 enters along two inlets and exit along one outlet as shown. Determine the velocity V (in units m/s) which makes an angle 60° with horizontal as indicated in the figure. Here 8 = 60°, diameter at section 3 is 0.15 m and density of water = 1000 kg/m3 0.12 m (3) V2 = 6 m/s 90 (2) 0.10 m V = 4 m/s BO & 7 9 8 $ 4 % 5 # 3 P Y T E R.

Answer 1

Given Information :

• 
  V1 = 4 m/s
  
• 
  di = 0.1 m
  
• 
  V2 = 6 m/s
  
• 
  d2 = 0.12 m
  
• 
  d3 = 0.15 m
  

Solution :

Calculating the velocity of fluid along the direction of pipe 3.

Using the law of conservation of mass :

min = mout

p(41 * V1 + A2 *V2) = p(43 * 13)

(41 * V1 + A2* V2) = (A3 *V3)

(d * Vi + d2 * V2) = *(d* V3)

di * V1 + d *V2 = d * V3

0.12 * 4+0.122 *6 = 0.152 * V3

Finding the velocity V :

Now that we have calculated the Veloctiy of fluid flow along the direction of pipe, we can calculate velocity V which is the horizontal component :

V = V3* * COSI os(@

$V=5.61*\cos(60)$

Therefore the OPTION 2 is the right answer.