Water enters a turbo-machine at a velocity 12 ft/s. The flow occurs in the horizontal plane. The inside diameters of the inlet and exit pipes are 1 ft and 2 ft, and the gage pressures at the inlet and exit are 10 psi and 5 psi, respectively. The machine is anchored at the lower right corner, which is 18 in. to the right of the inlet pipe centerline and also 20 in. below the exit pipe centerline. Determine:

(a) The velocity at the exit and the flow rate.
(b) The x and y components of the reaction force at O.
(c) The reaction moment at O.
(d) The shaft power of the machine. Is it a pump or a turbine?

To determine the various parameters for the given turbo-machine, we can apply the principles of fluid mechanics and use the equations for flow analysis. Let's break down the problem and solve it step by step:

(a) The velocity at the exit and the flow rate:
We can use the principle of continuity to find the velocity at the exit, which states that the mass flow rate is constant in an incompressible flow. Mathematically, it can be expressed as:

A1V1 = A2V2

where A1 and A2 are the cross-sectional areas of the inlet and exit pipes, and V1 and V2 are the velocities at the inlet and exit.

Given:
Inlet pipe diameter = 1 ft
Exit pipe diameter = 2 ft
Inlet velocity, V1 = 12 ft/s

We need to find V2 (velocity at the exit) and the flow rate.

To calculate V2, we can start by finding the cross-sectional areas:

A1 = π(0.5)^2 = π/4 sq ft
A2 = π(1)^2 = π sq ft

Using the principle of continuity, we have:
A1V1 = A2V2
(π/4)(12) = πV2

Simplifying the equation:
3V2 = 12
V2 = 4 ft/s

So, the velocity at the exit is 4 ft/s.

To find the flow rate, we can multiply the cross-sectional area of the exit pipe by the velocity at the exit:
Q = A2 * V2
Q = π * (4)
Q = 4π ft³/s

(b) The x and y components of the reaction force at O:
To determine the reaction force at point O, we need to consider the change in momentum of the fluid as it flows through the turbo-machine. Since the system is anchored, the net force should be zero.

The x-component of the reaction force can be found using the equation:

Fx = (Δp1 + Δp2) * A2

where Δp1 and Δp2 are the pressure differences at the inlet and exit, and A2 is the cross-sectional area of the exit pipe.

Given:
Δp1 (pressure at the inlet) = 10 psi
Δp2 (pressure at the exit) = 5 psi
A2 = π sq ft

Converting pressure to force using the equation 1 psi = 144 lb/ft², we have:

Δp1 = 10 * 144 lb/ft²
Δp2 = 5 * 144 lb/ft²

Substituting the values, we get:
Fx = ((10 * 144) + (5 * 144)) * π lb

Calculating Fx:
Fx = (1440 + 720) * π lb
Fx = 2160π lb

The y-component of the reaction force at O can be found using the equation:

Fy = ρ * Q * (V1 - V2) + (Δp1 - Δp2) * A2

where ρ is the density of water.

Given:
Density of water, ρ = 62.4 lb/ft³

Substituting the values, we get:
Fy = (62.4) * (4π) * (12 - 4) + (10 - 5) * π
Fy = 1872π + 5π
Fy = 1877π lb

Therefore, the x-component of the reaction force at O is 2160π lb, and the y-component is 1877π lb.

(c) The reaction moment at O:
The reaction moment about point O can be calculated using the equation:

Mo = y * Fx - x * Fy

Given:
Distance from the inlet pipe centerline to point O, x = 18 in = 1.5 ft
Distance from the exit pipe centerline to point O, y = -20 in = -1.67 ft (negative since it is below)

Substituting the values, we get:
Mo = (1.5) * (2160π) - (-1.67) * (1877π)
Mo = 3240π + 3134.59π
Mo ≈ 6374.59π lb-ft

Therefore, the reaction moment at O is approximately 6374.59π lb-ft.

(d) The shaft power of the machine and its nature (pump or turbine):
The shaft power of the machine can be determined using the equation:

Power = Q * Δp2 / 550

Given:
Q (flow rate) = 4π ft³/s
Δp2 (pressure difference at the exit) = 5 psi

Converting pressure to force: Δp2 = 5 * 144 lb/ft²

Substituting the values, we get:
Power = (4π) * (5 * 144) / 550 hp

Calculating Power:
Power = (2880π/550) hp

To determine whether it's a pump or a turbine, we need to consider the sign of the power. If the power is positive, it's a pump (input energy); if negative, it's a turbine (output energy).

In this case, the power is positive (since π is positive), so it's a pump.

Therefore, the shaft power of the machine is (2880π/550) hp, and it is a pump.