# Integral Form of the Momentum Equation - Reducer Question

1. Apr 4, 2013

### MrWinesy

1. The problem statement, all variables and given/known data

The internal volume of the reducer is 0.2m^3 and its mass is 25 kg. The fluid being pumped is oil (specific gravity of 0.72).
Evaluate the total force that must be provided to support the reducer.

d1 = 0.4m
d2 = 0.2m
u1 = 3m/s
p1 = 58.7 kPa
p2 = 49kPa (gauge)

2. Relevant equations

Qin=Qout

mdot=ρ*A*u

A=(∏*d^2)/4

3. The attempt at a solution

Tried. Failed. Help.

2. Apr 4, 2013

### MrWinesy

the answer is apparently F = −3.4xˆ +1.66yˆ kN but I am in need of the working please.

3. Apr 4, 2013

### rude man

What does a 'reducer" do?

4. Apr 4, 2013

### SteamKing

Staff Emeritus
A reducer couples a larger diameter pipeline to a smaller diameter pipeline, hence the two diameters specified in the OP.

5. Apr 4, 2013

### rude man

Thanks SK!

But - what is meant by its volume? Is it a tapered section of pipe going from the larger to the smaller diameter?

6. Apr 4, 2013

### SteamKing

Staff Emeritus
It can be. More often it is a cast fitting, either flanged or suited to welding, with a curved transition between the larger and smaller diameters, so that the overall length of the fitting is kept small. The volume, I believe, is just what it implies, the volume of the internal space of the reducer.

7. Apr 5, 2013

### MrWinesy

yea this is right. it can be basically thought of as a converging pipe and the question is based on the internal volume as this helpful fella said. and the mass is the mass of the actual reducer.

i have obtained the answer but not confident on my methods. anyone else had any luck?

8. Apr 5, 2013

### rude man

Thanks to my friend Chestermiller from a similar previous problem, the approach here is to consider the change in momentum per unit time of the oil and equate that to the (longitudinal) force exerted on the reducer.