Oil (sp. gr.= 0.8) flows smoothly through the circular reducing section shown at 3 ft^3/s. If the entering and leaving velocity profiles are uniform, estimate the force which must be applied to the reducer to hold it in place.
When Fluid is entering the pipe: P= 50 psig
Diameter of the pipe= 12 in.
Fluid leaving the pipe: P= 5 psig
Diameter of the pipe= 2.5 in.
To estimate the force required to hold the reducer in place, we need to calculate the change in momentum of the fluid as it passes through the reducer.
1. Calculate the mass flow rate of the fluid:
The mass flow rate (ṁ) can be calculated using the equation:
ṁ = ρ * A * V
Where:
ρ is the density of the fluid (given as specific gravity),
A is the cross-sectional area of the pipe, and
V is the velocity of the fluid.
Given:
Specific gravity (SG) = 0.8
Density of water (ρw) = 62.4 lbs/ft^3
To find the density of the fluid, we use the formula:
ρ = SG * ρw
Now, let's calculate the mass flow rate:
ρ = 0.8 * 62.4 lbs/ft^3 = 49.92 lbs/ft^3
For the entering pipe:
Diameter (De) = 12 in = 1 ft
Area (Ae) = (π/4) * (De)^2 = (π/4) * (1 ft)^2 = π/4 ft^2
Velocity (Ve) = ṁ / (ρ * Ae) = 3 ft^3/s / (49.92 lbs/ft^3 * π/4 ft^2) = 0.1914 ft/s
For the leaving pipe:
Diameter (Dl) = 2.5 in = 0.2083 ft
Area (Al) = (π/4) * (Dl)^2 = (π/4) * (0.2083 ft)^2 = 0.0342 ft^2
Velocity (Vl) = ṁ / (ρ * Al) = 3 ft^3/s / (49.92 lbs/ft^3 * 0.0342 ft^2) = 1.078 ft/s
2. Calculate the change in momentum of the fluid:
The change in momentum (Δp) can be calculated using the equation:
Δp = ρ * (Vl - Ve)
Δp = 49.92 lbs/ft^3 * (1.078 ft/s - 0.1914 ft/s) = 49.92 lbs/ft^3 * 0.8866 ft/s = 44.2 lbs/ft^2
3. Convert the force from area to total force:
The force required to hold the reducer in place can be calculated by multiplying the change in momentum by the cross-sectional area of the outlet pipe.
Area (Al) = 0.0342 ft^2
Force = Δp * Al = 44.2 lbs/ft^2 * 0.0342 ft^2 = 1.509 lbs
Therefore, the force required to hold the reducer in place is approximately 1.509 pounds.