The volume [tex]\( V \)[/tex] of liquid flowing in time [tex]\( t \)[/tex] through a pipe of radius [tex]\( r \)[/tex] is given by the equation

[tex]\[ \frac{V}{t} = \frac{\pi P r^4}{8 C l} \][/tex]

where [tex]\( P \)[/tex] is the pressure difference between the ends of the pipe of length [tex]\( t \)[/tex], and [tex]\( C \)[/tex] depends on the frictional effects of the liquid.

An experiment is performed to determine [tex]\( C \)[/tex]. The measurements made are shown in Fig. 1.1.

[tex]\[
\begin{tabular}{|c|c|c|c|}
\hline
\(\frac{V}{t} / 10^{-6} \, \text{m}^3 \text{s}^{-1}\) & \(P / 10^3 \, \text{Nm}^{-2}\) & \(r / \text{mm}\) & \(T / \text{m}\) \\
\hline
$1.20 \pm 0.01$ & $2.50 \pm 0.05$ & $0.75 \pm 0.01$ & $0.250 \pm 0.001$ \\
\hline
\end{tabular}
\][/tex]

Fig. 1.1

(a) Calculate the value of [tex]\( C \)[/tex].

[tex]\[ C = \][/tex]
[tex]\[ \qquad \][/tex]
[tex]\[ \text{Nsm}^{-2} \, [2] \][/tex]



Answer :

To find the value of [tex]\( C \)[/tex] from the given data, we need to use the provided equation:

[tex]\[ \frac{V}{t} = \frac{\pi P r^4}{8 C l} \][/tex]

Given the following measurements:
- [tex]\(\frac{V}{t} = 1.20 \times 10^{-6} \, \text{m}^3/\text{s}\)[/tex]
- [tex]\(P = 2.50 \times 10^{3} \, \text{N/m}^2\)[/tex]
- [tex]\(r = 0.75 \, \text{mm} = 0.75 \times 10^{-3} \, \text{m}\)[/tex]
- [tex]\(l = 0.250 \, \text{m}\)[/tex]

We first rearrange the equation to solve for [tex]\( C \)[/tex]:

[tex]\[ C = \frac{\pi P r^4}{8 \left( \frac{V}{t} \right) l} \][/tex]

Next, let's substitute the given values into the equation:

1. [tex]\(\pi \approx 3.14159\)[/tex]
2. [tex]\(P = 2.50 \times 10^{3} \, \text{N/m}^2\)[/tex]
3. [tex]\(r = 0.75 \times 10^{-3} \, \text{m}\)[/tex]
4. [tex]\(\left( \frac{V}{t} \right) = 1.20 \times 10^{-6} \, \text{m}^3/\text{s}\)[/tex]
5. [tex]\(l = 0.250 \, \text{m}\)[/tex]

Performing the substitution:

[tex]\[ C = \frac{\pi \times 2.50 \times 10^{3} \times (0.75 \times 10^{-3})^4}{8 \times 1.20 \times 10^{-6} \times 0.250} \][/tex]

Simplifying inside the parenthesis and exponentials:

[tex]\[ = \frac{3.14159 \times 2.50 \times 10^{3} \times (0.75)^4 \times 10^{-12}}{8 \times 1.20 \times 10^{-6} \times 0.250} \][/tex]

Continuing the calculation step by step:

1. Calculate [tex]\( (0.75)^4 \)[/tex]:

[tex]\[ (0.75)^4 = 0.31640625 \][/tex]

2. Multiply powers of ten:

[tex]\[ 0.31640625 \times 10^{-12} = 0.31640625 \times 10^{-12} \][/tex]

3. Multiply [tex]\( \pi \times 2.50 \times 10^3 \times 0.31640625 \)[/tex]:

[tex]\[ 3.14159 \times 2.50 \times 10^3 \times 0.31640625 = 2.480502625 \times 10^3 \][/tex]

4. Combine constants:

[tex]\[ = \frac{2.480502625 \times 10^3 \times 10^{-12}}{8 \times 1.20 \times 10^{-6} \times 0.250} \][/tex]

This expression simplifies to:

[tex]\[ C = \frac{2.480502625 \times 10^{-9}}{2.4 \times 10^{-6}} = \frac{2.480502625}{2.4} \times 10^{-3} = 1.0335428 \times 10^{-3} \approx 0.001035 \, \text{Ns/m}^2 \][/tex]

Thus, the value of [tex]\(C\)[/tex] is approximately:

[tex]\[ C = 0.001035 \, \text{Ns/m}^2 \][/tex]