Solving Water Nozzle Problem: Ratio of Plug Radius to Hose Radius

[curryman24]

THE PROBLEM
In an adjustable nozzle for a garden hose, a cylindrical plug is aligned along the axis of the hose and can be inserted into the hose opening. The purpose of the plug is to change the speed of the water leaving the hose. The speed of the water passing around the plug is to be 4 times greater than the speed of the water before it encounters the plug. Find the ratio of the plug radius to the inside hose radius.

The attempt at a solution:

I do know the equation of continuity and it is

(P1)(A1)(V1)= (P2)(A2)(V2)

and I used this to say that the pressure is going to be constant throughout the problem so it can be canceled.

So,
(A1)(V1)= (A2)(V2)
(πr1^2)(V1) = (πr2^2)(V2)
(r1^2)(V1) = (r2^2)(V2)
and V2= (4V1)
so (r1^2)(V1) = (r2^2)(4V1)
(r1^2) = (4r2^2)
(r2/r1) = 0.5

HOWEVER, this is wrong...please help.

 

[stewartcs]

You have found the ratio of the equivalent "hole" to the hose instead of the ratio of the "plug" to the hose.

 

[ogb p]

I would like to offer some suggestions to solve this water nozzle problem. Firstly, it is important to understand the concept of Bernoulli's principle which states that as the speed of a fluid increases, its pressure decreases. This principle is essential in solving this problem.

To find the ratio of the plug radius to the inside hose radius, we can use the equation of continuity as you have correctly identified. However, we also need to consider the fact that the water velocity changes as it passes through the plug. We can use the Bernoulli's equation to account for this change in velocity.

(P1 + 1/2ρV1^2) = (P2 + 1/2ρV2^2)

Where P1 and P2 are the pressures before and after the plug respectively, ρ is the density of water, and V1 and V2 are the velocities before and after the plug respectively.

Since the speed of the water passing around the plug is to be 4 times greater than the speed of the water before it encounters the plug, we can write:

V2 = 4V1

Substituting this into the Bernoulli's equation and simplifying, we get:

P1 + 3/2ρV1^2 = P2

Now, using the continuity equation (A1V1 = A2V2), we can express V2 in terms of V1 as:

V2 = (A1/A2)V1

Substituting this into the modified Bernoulli's equation, we get:

P1 + 3/2ρV1^2 = (A1/A2)^2P1

Solving for the ratio of the plug radius to the inside hose radius (r2/r1), we get:

(r2/r1) = √[(2/3)(A1/A2)^2 - 1]

Since we know that the speed of the water passing around the plug is 4 times greater than the speed of the water before it encounters the plug, we can write:

V2 = 4V1 = (A1/A2)V1

Solving for A1/A2, we get:

(A1/A2) = 1/4

Substituting this into the equation for the ratio of the plug radius to the inside hose radius, we get:

(r2/r1) = √[(2/3)(1