How Do You Calculate Water Velocity and Pipe Diameter in Mechanical Engineering?

[Aaron9890]

Mechanical principles engineering question help??

Homework Statement

A boiler is supplied with water at a rate of 0.5 kg/s through a pipe of diameter 80mm

a) Calculate the velocity of the water in the pipe
b) Calculate the smaller internal diameter of this tapered section of pipe calculate the velocity to be 2.5 times greater than the entry velocity calculated earlier.

Assume density is 1000kg/m^3

Homework Equations

a) m= PVA
b) V1A1=V2A2

The Attempt at a Solution

a) Mass flow rate = 0.5kg/s
Velocity=?
Area = 5.03 x 10^-3
Density = 1000kg/m^3

so m=P1V1A1
V=P1XA1/m
V= (1000)x(5.03x10^-3) / 0.5kg/s = 10.06m/s


Thanks for anyone who attempts to help me, i really appreciate it as this is my last assignment of the year:)

 

[KataKoniK]

Hello! I would like to help you with your question. Let's start by looking at the given information and equations you have provided.

For part a), you are correct in using the equation m = PVA to solve for the velocity. However, there is a slight error in your calculation. The area of a circular pipe is calculated as A = πr^2, where r is the radius of the pipe. In this case, the diameter is given as 80mm, so the radius would be 40mm or 0.04m. Thus, the correct area would be A = π(0.04)^2 = 0.00503m^2. Plugging this into the equation, we get:

m = P1V1A1
0.5kg/s = (1000kg/m^3)(V1)(0.00503m^2)
V1 = 100m/s

So, the velocity of the water in the pipe is 100m/s.

For part b), you are on the right track in using the equation V1A1 = V2A2. However, this equation assumes that the mass flow rate remains constant. In this case, we are trying to find the smaller internal diameter that will result in a velocity 2.5 times greater than the entry velocity. This means that the velocity will change, and thus, the mass flow rate will also change.

To solve this, we can use the continuity equation, which states that the mass flow rate at any point in a pipe must remain constant. So, we can set up the following equation:

m1 = m2
P1V1A1 = P2V2A2

We know the values for P1, V1, and A1 from part a). We are also given the value for V2, which is 2.5 times greater than V1. So, V2 = 2.5(100m/s) = 250m/s. We also know the density, which is given as 1000kg/m^3. The only unknown in this equation is A2, which is the area of the smaller internal diameter.

Solving for A2, we get:

P1V1A1 = P2V2A2
(1000kg/m^3)(100m/s)(0.00503m^2) = (1000