Explanation:

Given:
- The velocity is zero over half of the cross-sectional area.
- The velocity is uniform over the remaining half.

To find:
- The momentum correction factor.

Solution:
The momentum correction factor is a dimensionless quantity that accounts for the difference in velocity profiles between the actual flow and an idealized flow with uniform velocity.

Velocity distribution:
The given velocity distribution can be visualized as follows:

```
_______
v=0 | |
_| |_
v=V | |
-------
```

Where:
- v = velocity
- V = uniform velocity
- The velocity is zero over half of the cross-sectional area and uniform over the remaining half.

Momentum correction factor:
The momentum correction factor (C_m) is given by the equation:

C_m = 1 + (A_2/A_1) * (v_m/V)^2

Where:
- A_1 = Total area of the cross-section
- A_2 = Area over which the velocity is uniform
- v_m = Mean velocity of the flow

Calculating the momentum correction factor:
In this case, since the velocity is zero over half of the cross-sectional area, we can divide the cross-section into two parts:
- Part 1: Area with zero velocity (half of the cross-sectional area)
- Part 2: Area with uniform velocity (remaining half of the cross-sectional area)

Let's assume that the total area of the cross-section is A.
- Area with zero velocity (Part 1) = A/2
- Area with uniform velocity (Part 2) = A/2

The mean velocity of the flow (v_m) can be calculated by assuming equal mass flow rates through both parts of the cross-section:
v_m = (0 * (A/2) + V * (A/2)) / A
= V/2

Substituting the values into the equation for the momentum correction factor:
C_m = 1 + ((A/2) / A) * ((V/2) / V)^2

Simplifying the equation:
C_m = 1 + (1/2) * (1/2)^2
C_m = 1 + (1/2) * (1/4)
C_m = 1 + 1/8
C_m = 1.125

Rounding the answer up to the nearest integer, the momentum correction factor is 2.

Answer:
The momentum correction factor is 2.