Question

activity 4-1: pipe velocity
a 2\ id pipe that is 10 in length supplies pressure from a hydraulic power unit must be replaced because it was damaged by a fork-lift truck. the only pipe available is 1.5\ id pipe and the company is losing thousands of dollars during this downtime. the pump is supplying 12 gpm. the head engineer has asked for the difference in the velocity of the fluid flowing through the pipes to determine if the 1.5\ diameter pipe would work in the system.

1. determine the difference in velocity of the two pipes.

(note: both pipes are 10 foot in length, flow rate stays the same in both pipes, only velocity changes, remember to convert from feet to inches for ( l_1 ))

1. determine the optimal velocity for the system.

1. explain (2-3 sentences) your findings. how close to optimal is the 1.5 inch pipe velocity? will that damage the system over time? why or why not?

helpful tips:
step 1, find the velocity of the 2\ pipe.
(note: use the ( v = \frac{l_2}{\frac{a \times l_1}{231} \times \frac{60}{q}} ) equation)

step 2, use the same formula to find the velocity in the 1.5\ pipe.

step 3, find the optimal velocity for a 1.5\ pipe using the ( v = \frac{q \times 0.3208}{a} ) equation

Answer:

Part 1: Difference in Velocity of the Two Pipes

Step 1: Find the velocity of the 2" Pipe

First, we need to find the cross - sectional area \(A\) of a pipe. The formula for the area of a circle is \(A=\pi r^{2}\), where \(r\) is the radius. For a 2" ID pipe, the diameter \(d = 2\) inches, so the radius \(r=\frac{d}{2}=1\) inch. Then \(A=\pi(1)^{2}=\pi\) square inches. The length of the pipe \(l_{2} = 10\) feet. We need to convert the length from feet to inches, so \(l_{1}=10\times12 = 120\) inches. The flow rate \(Q = 12\) gpm.

The velocity formula is \(v=\frac{l_{2}}{\frac{A\times l_{1}}{231}\times\frac{60}{Q}}\)

First, calculate the denominator: \(\frac{A\times l_{1}}{231}\times\frac{60}{Q}=\frac{\pi\times120}{231}\times\frac{60}{12}\)

\(\frac{\pi\times120}{231}\times5=\frac{600\pi}{231}\approx\frac{600\times3.1416}{231}=\frac{1884.96}{231}\approx8.16\)

Then \(v=\frac{10}{\frac{\pi\times120}{231}\times\frac{60}{12}}=\frac{10}{8.16}\approx1.225\) ft/s (Wait, let's re - do the formula. The correct formula manipulation:

\(v=\frac{l_{2}\times231\times Q}{A\times l_{1}\times60}\)

Substitute the values: \(l_{2} = 10\) ft, \(A=\pi(\frac{2}{2})^{2}=\pi\) in², \(l_{1}=10\times12 = 120\) in, \(Q = 12\) gpm.

\(v=\frac{10\times231\times12}{\pi\times120\times60}=\frac{27720}{7200\pi}=\frac{27720}{22619.47}\approx1.225\) ft/s

Step 2: Find the velocity of the 1.5" Pipe

For a 1.5" ID pipe, the radius \(r=\frac{1.5}{2}=0.75\) inches. The area \(A=\pi(0.75)^{2}=\pi\times0.5625\approx1.767\) in².

Using the same velocity formula \(v=\frac{l_{2}\times231\times Q}{A\times l_{1}\times60}\)

Substitute \(l_{2}=10\) ft, \(A = 1.767\) in², \(l_{1}=120\) in, \(Q = 12\) gpm.

\(v=\frac{10\times231\times12}{1.767\times120\times60}=\frac{27720}{12722.4}\approx2.18\) ft/s

Step 3: Find the difference in velocity

The difference \(\Delta v=v_{1.5}-v_{2}=2.18 - 1.225 = 0.955\) ft/s (approximate value, more precise calculation below)

Let's do the calculation more precisely:

For 2" pipe:

\(A_1=\pi(\frac{2}{2})^2=\pi\) in²

\(v_1=\frac{10\times231\times12}{\pi\times120\times60}=\frac{27720}{7200\pi}=\frac{27720}{22619.4671}\approx1.225\) ft/s

For 1.5" pipe:

\(A_2=\pi(\frac{1.5}{2})^2=\pi\times0.5625\) in²

\(v_2=\frac{10\times231\times12}{\pi\times0.5625\times120\times60}=\frac{27720}{\pi\times0.5625\times7200}=\frac{27720}{12723.45}\approx2.18\) ft/s

\(\Delta v=2.18 - 1.225 = 0.955\) ft/s (or more precisely, \(\frac{10\times231\times12}{\pi\times0.5625\times120\times60}-\frac{10\times231\times12}{\pi\times1\times120\times60}\)

Factor out \(\frac{10\times231\times12}{\pi\times120\times60}\):

\(\frac{10\times231\times12}{\pi\times120\times60}(\frac{1}{0.5625}-1)=\frac{27720}{7200\pi}(\frac{1 - 0.5625}{0.5625})=\frac{27720}{7200\pi}\times\frac{0.4375}{0.5625}\)

\(\frac{27720\times0.4375}{7200\pi\times0.5625}=\frac{12127.5}{4050\pi}\approx\frac{12127.5}{12723.45}\approx0.953\) ft/s

Part 2: Optimal Velocity for the 1.5" Pipe

The formula for optimal velocity is \(V=\frac{Q\times0.3208}{A}\)

First, find the area \(A=\pi(\frac{1.5}{2})^2=\pi\times0.5625\approx1.767\) in²

\(Q = 12\) gpm

\(V=\frac{12\times0.3208}{1.767}=\frac{3.8496}{1.767}\approx2.18\) ft/s (Wait, that's the same as the velocity of the 1.5" pipe? Wait, no, let's check the formula again. Maybe the units: 0.3208 is a conversion factor for gpm to ft³/s or something. Wait, the formula \(V=\frac{Q\times0.3208}{A}\), where \(Q\) is in gpm, \(A\) is in square inches, and \(V\) is in ft/s.

Let's re - calculate \(A\) for 1.5" pipe: \(d = 1.5\) inches, \(r = 0.75\) inches, \(A=\pi r^{2}=\pi\times(0.75)^{2}\approx1.767\) in²

\(Q = 12\) gpm

\(V=\frac{12\times0.3208}{1.767}=\frac{3.8496}{1.767}\approx2.18\) ft/s

Part 3: Explanation

The velocity of the 2" pipe is approximately \(1.225\) ft/s and the velocity of the 1.5" pipe is approximately \(2.18\) ft/s, so the difference is about \(0.955\) ft/s. The optimal velocity for the 1.5" pipe is also approximately \(2.18\) ft/s, which means the velocity of the 1.5" pipe is at the optimal value. Since the velocity is at the optimal level, it should not damage the system over time as the flow is within the recommended velocity range for the pipe size.

Final Answers

1. The difference in velocity is approximately \(\boldsymbol{0.95}\) ft/s (more precisely \(\approx0.953\) ft/s)
2. The optimal velocity for the 1.5" pipe system is approximately \(\boldsymbol{2.18}\) ft/s
3. The 1.5" pipe velocity is at the optimal level. It should not damage the system over time because the velocity matches the optimal velocity for the pipe size, ensuring that the fluid flow is within the safe and efficient range for the pipe.