A Ysplitter for a water system is shown rhow 1000 kgm3 The

A Y-splitter for a water system is shown (rho_w = 1000 kg/m^3). The Y-splitter has mass 6 kg and internal volume 0.004 m^3. Section I (inlet): V_1 = 6.00 m/s. p_1 = 50 kPa, D_1 = 12 cm. Section 2 (outlet): V_2 = 4.00 m/s, p_1 = 30 kPa. D_2 = 8.0 cm. Section 3 (outlet): V_3 = 3.00 m/s, p_3 = 20 kPa, D_3 = 6.0 cm.

Solution

To solve this problem we need to draw control volume:

In question the shaded portion available can be considered as control volume.

Force acting on the control volume is:

1. Body forces:

   It is due to gravity
   
   It is assymed that that the elevation of the three branches are same, so body forces = 0.

2. Pressure forces
    
   It is becuase of the pressure carrying the fluid.
   
   F=P*A

3. Force due to velocity and the mass

   It is because of the fluid motion.

   F = dot(m)*velocity

4. Force due to structure

   We need to find this force out from the equilibrium.

   To find,
   
x- component of outside force,

   Total force in x direction = 0
   
P1*A1*sind(15)+P2*A2*cosd(60)-P3*A3*cosd(40)+Fx = m1*v1*sind(15)-m2*v2*cosd(60)+m3*v3*cosd(40)

Fx = m1*v1*sind(15)-m2*v2*cosd(60)+m3*v3*cosd(40)-(P1*A1*sind(15)+P2*A2*cosd(60)-P3*A3*cosd(40)) -----(1)

now,

A1=(pi/4)*(d1^2)= (pi/4)*(120^2) = 11309.73 mm^2

A2=(pi/4)*(d2^2)= (pi/4)*(80^2) = 5026.54 mm^2

A3=(pi/4)*(d3^2)= (pi/4)*(60^2) = 2827.43 mm^2

m1=rho*A1*v1 = 1000*11309.73*(10^-6)*6 = 67.85 kg/s

m2=rho*A2*v2 = 1000*5026.54*(10^-6)*4 = 20.10 kg/s

m3=rho*A1*v1 = 1000*2827.43 *(10^-6)*3 = 8.48 kg/s

After putting this value in equation 1,

Fx = -93.78 N

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 A Y-splitter for a water system is shown (rho_w = 1000 kg/m^3). The Y-splitter has mass 6 kg and internal volume 0.004 m^3. Section I (inlet): V_1 = 6.00 m/s.
 A Y-splitter for a water system is shown (rho_w = 1000 kg/m^3). The Y-splitter has mass 6 kg and internal volume 0.004 m^3. Section I (inlet): V_1 = 6.00 m/s.

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