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 consider the change in momentum of the oil flowing through the reducer.

First, let's find the mass flow rate of the oil entering and leaving the pipe.

The mass flow rate (ṁ) can be calculated using the following equation:

ṁ = ρ * A * V

where:
ṁ = mass flow rate (in kg/s)
ρ = density of the oil (in kg/m^3)
A = cross-sectional area of the pipe (in m^2)
V = velocity of the oil (in m/s)

Given that the specific gravity of the oil (sp. gr.) is 0.8, we can find the density (ρ) using the following relationship:

ρ = sp. gr. * ρ_water

where ρ_water is the density of water at standard conditions (1000 kg/m^3).

The cross-sectional area (A) of a circular pipe is calculated using the formula:

A = π * r^2

where r is the radius of the pipe.

Let's calculate the mass flow rate for both the entering and leaving sections:

For the entering section:
The diameter of the pipe is given as 12 inches, so the radius (r) can be calculated as half the diameter: r = 12 in / 2 = 6 in = 0.1524 m.
Using the formula for the area of a circle, we can find the cross-sectional area (A).
A = π * (0.1524 m)^2

For the leaving section:
The diameter of the pipe is given as 2.5 inches, so the radius (r) can be calculated as r = 2.5 in / 2 = 1.25 in = 0.03175 m.
Using the formula for the area of a circle, we can find the cross-sectional area (A).

Now, let's calculate the mass flow rates:

For the entering section:
ṁ_entering = ρ * A * V

For the leaving section:
ṁ_leaving = ρ * A * V

Next, let's find the change in momentum of the oil as it flows through the reducer. The change in momentum (∆p) is given by the following equation:

∆p = m * ∆v

where:
∆p = change in momentum (in kg·m/s)
m = mass flow rate (in kg/s)
∆v = change in velocity (in m/s)

Since the velocity profiles are uniform for both the entering and leaving sections, the change in velocity (∆v) will be zero. So, the change in momentum will also be zero (∆p = 0).

To hold the reducer in place, an opposing force is required to counteract the change in momentum (∆p = 0). Therefore, the force required to hold the reducer in place is zero.

In conclusion, no force is required to hold the reducer in place.