Ethylene glycol at 300 K flows through the section of bent p

Ethylene glycol at 300 K flows through the section of bent pipe shown in the figure. At location 1 of the pipe (which represents a cut through the pipe), the pipe diameter is 6 cm and there is a pressure sensor. Location 2 of the pipe is open to atmosphere and has a reduced diameter of 3 cm. The angle is 45 degrees from horizontal. The ethylene glycol flows at a weight flux of 100 N/s. Assume no friction. Neglect changes in potential and kinetic energy. Known from homework set #5 are the inlet and exit velocities, the mass flow rate and the gage pressure at the inlet, the latter of which is 87, 500 Pa. Determine the x force (R_x) and the y force (R_y) needed to hold the pipe in place. Use the control volume and coordinate system provided.

Solution

Given:

Temperature, T = 300 K

Diameter at section 1, d₁ = 6 cm = 0.06 m

Diameter at section 2, d₂ = 3 cm = 0.03 m

Pressure at inlet, P₁ = 87500 Pa = 87.5 KPa

Pressure at outlet, P₂ = Atmospheric = 101 KPa

Weight Flux, = 100 N/s = 100 Kg. m/s

Angle of bend, = 45°

From the continuity equation,

                                ₁ V₁ A₁ = ₂ V₂ A₂ =                                                                    Eq. 1

Since only one fluid is being used, i.e. ethylene glycol density remains constant.

Therefore,          ₁ = ₂

So, from Equation 1                        ₁ V₁ A₁=

                                                                V₁ = / ₁ A₁

Assuming density of ethylene glycol at NTP as 1097 Kg/m³

                                                                V₁ = 100 / 1097 x (/4 x 0.06²)

                                                                V₁ = 32.25 m/s

Now, from Equation 1,                  V₂ = V₁ A₁ / A₂

                                                                V₂ = 32.25 x (0.06)² / 0.03²

                                                                V₂ = 129 m/s

Carrying out force balancing in x-direction, we get

                P₁ A₁ – P₂ A₂Cos + R_x = ₂ V₂ A₂ V₂Cos – ₁ V₁ A₁ V₁

OR          P₁ A₁ – P₂ A₂Cos + R_x = V₂Cos - V₁

OR          P₁ A₁ – P₂ A₂Cos + R_x = (V₂Cos - V₁)

Similarly, carrying out force balancing in y-direction, we get

                0 – P₂A₂Sin + R_y = ₂ V₂ A₂ V₂Sin – 0

OR          P₂A₂Sin + R_y = V₂Sin

Thus the magnitude of force components R_x and R_y can be given as:

R_x = P₂ A₂Cos – P₁ A₁ – (V₂Cos - V₁)                                                 Eq. 2

R_y = – P₂A₂Sin + V₂Sin                                                                           Eq. 3

By filling appropriate values in Equation 2 and 3, we get

                R_x = 101 x /4 x 0.03²Cos45° - 87.5 x /4 x 0.06² – 100(129Cos45° - 32.25)

                R_x = 3154.08 N                    Ans.

Similarly, R_y = - 101 x /4 x 0.03² Sin45° + 100 x 129 x Sin45°

                R_y = 9121.62 N                 Ans.