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 calculate the force required to hold the reducer in place, we need to consider the change in momentum of the fluid as it flows through the reducing section. This change in momentum will exert a force on the reducer.
1. Calculate the mass flow rate:
The mass flow rate (ṁ) can be calculated using the equation:
ṁ = ρ * Q
where ρ is the density of the fluid and Q is the volumetric flow rate.
Given:
Volumetric flow rate (Q) = 3 ft^3/s
To get the volumetric flow rate in terms of cubic inches per second, we need to convert the given value:
1 ft^3 = (12 in)^3 = 1728 in^3
3 ft^3/s = 3 * 1728 in^3/s = 5184 in^3/s
Now, we can calculate the mass flow rate:
ρ (density) = specific gravity (SG) * density of water
Since the specific gravity (SG) of oil is given as 0.8,
ρ = 0.8 * density of water
The density of water can be found from reference tables and is approximately 62.4 lb/ft^3.
ρ = 0.8 * 62.4 lb/ft^3
To convert lb/ft^3 to lb/in^3:
1 ft = 12 in
(1 ft)^3 = (12 in)^3
1 ft^3 = 1728 in^3
Density ρ = (0.8 * 62.4 lb/ft^3) / 1728 in^3/ft^3
Now, we can calculate the mass flow rate:
ṁ = ρ * Q
2. Calculate the velocities at the two sections:
We can use the principle of continuity to relate the velocities at the two sections.
A1 * V1 = A2 * V2
where A is the cross-sectional area of the pipe and V is the velocity of the fluid.
Given:
Diameter of the pipe at section 1 = 12 in
Diameter of the pipe at section 2 = 2.5 in
The cross-sectional area A can be calculated using the equation:
A = π * (diameter/2)^2
Calculate the cross-sectional areas A1 and A2 using the given diameters.
Now, assuming the fluid is incompressible, we can equate the volumetric flow rate to the product of the cross-sectional area and the velocity:
Q = A * V
Rearrange and solve for the velocities V1 and V2 using the known values of Q and A1, A2.
3. Calculate the change in momentum:
The change in momentum (∆P) is given by:
∆P = m * ∆v
where m is the mass flow rate and ∆v is the change in velocity of the fluid.
The change in velocity ∆v can be calculated as:
∆v = |V1 - V2|
Calculate the change in momentum ∆P using the values of m and ∆v.
4. Calculate the force required:
The force required to hold the reducer in place is equal to the change in momentum:
Force = ∆P
Calculate the force required using the value of ∆P.
Note: Conversion factors and density values may vary depending on specific units used. Make sure to use the appropriate conversions and values for accurate calculations.