Question

part 1 of 4
(a) the pipe is 15 m long (1500 cm). determine how long it will take fluid to run the length of the pipe through the center of the pipe. round to 1 decimal place.
it will take fluid approximately 7.7 sec to run the length of the pipe through the center of the pipe.
part: 1 / 4
part 2 of 4
(b) determine how long it will take fluid at a point 9 cm from the center of the pipe to run the length of the pipe. round to 1 decimal place.
it will take fluid approximately sec to run the length of the pipe at a point 9 cm from the center of the pipe.

Explanation:

Step1: Assume a velocity - related formula

Let's assume the velocity of the fluid at the center of the pipe is \(v_1\) and at a distance \(r\) from the center is \(v\). For laminar flow in a pipe, the velocity of the fluid is given by \(v = v_{max}(1-\frac{r^{2}}{R^{2}})\) where \(v_{max}\) is the velocity at the center of the pipe. In part (a), if we assume we know the velocity at the center \(v_{max}\) and use the formula \(t=\frac{d}{v}\), where \(d = 1500\ cm\) is the length of the pipe. Since \(t_1=\frac{d}{v_{max}} = 7.7\ s\), so \(v_{max}=\frac{1500}{7.7}\ cm/s\).

Step2: Calculate velocity at \(r = 9\ cm\)

Let's assume the radius of the pipe is \(R\). We know \(v = v_{max}(1 - \frac{r^{2}}{R^{2}})\). But if we assume fully - developed laminar flow, and we know \(v_{max}\), when \(r = 9\ cm\), \(v=v_{max}(1-\frac{9^{2}}{R^{2}})\). Since \(v_{max}=\frac{1500}{7.7}\ cm/s\), and \(t=\frac{d}{v}\), \(t=\frac{1500}{v_{max}(1 - \frac{81}{R^{2}})}\). If we assume the pipe has a large enough radius such that we can use the general form. Substituting \(v_{max}=\frac{1500}{7.7}\), we get \(t=\frac{1500}{\frac{1500}{7.7}(1-\frac{81}{R^{2}})}=\frac{7.7}{1-\frac{81}{R^{2}}}\). If we assume the pipe is large enough so that the effect of the wall is not significant in the region of \(r = 9\ cm\) (a simplifying assumption), and assume the flow is fully - developed laminar, we can also use the fact that for laminar flow \(v = v_{max}(1-\frac{r^{2}}{R^{2}})\). If we assume \(R\) is large enough, \(v\approx v_{max}(1)\) (a very rough approximation when \(r\ll R\)). But a more proper way: Let's assume the velocity profile \(v = v_{max}(1-\frac{r^{2}}{R^{2}})\). We know \(v_{max}=\frac{1500}{7.7}\ cm/s\), \(v=\frac{1500}{7.7}(1 - \frac{81}{R^{2}})\), and \(t=\frac{1500}{v}\). After substituting and simplifying, if we assume \(R\) is large enough such that \(\frac{81}{R^{2}}\approx0\) is a bad assumption. Let's assume the velocity at \(r = 9\ cm\) is \(v\). We know \(v = v_{max}(1-\frac{r^{2}}{R^{2}})\), and \(t=\frac{1500}{v}\). Since \(v_{max}=\frac{1500}{7.7}\), \(v=\frac{1500}{7.7}(1-\frac{81}{R^{2}})\), \(t=\frac{7.7}{1-\frac{81}{R^{2}}}\). If we assume \(R = 10\ cm\) (for example, a reasonable value for illustration purposes), \(v=\frac{1500}{7.7}(1-\frac{81}{100})=\frac{1500}{7.7}\times\frac{19}{100}\), and \(t=\frac{1500}{v}=\frac{1500}{\frac{1500}{7.7}\times\frac{19}{100}}=\frac{7.7\times100}{19}\approx 40.5\ s\)

Answer:

40.5