Answer to Question #156863 in Mechanics | Relativity for haze

Question

Water flowing along a horizontal pipe has a speed of 2.27 m/s and a pressure of 4.86 atm. Further along, the pipe narrows so that the cross-sectional area decreases by a factor of 6.44. What is the pressure in the narrow section?

Answer 1

Explanations & Calculations

• To workout this, it is necessary to assume water flows in laminar flow such that Bernoulli's equation could be applied.
• Since the pipe is horizontal, the regular Bernoulli equation reduces down to,

$\qquad\qquad\small P_1+\frac{1}{2}\rho v^2_1=P_2+\frac{1}{2}\rho v^2_2$

• Subjecting the final pressure: $\small P_2$ gives,

$\qquad\qquad\small P_2=P_1+\frac{1}{2}\rho(v_1^2-v_2^2)$

• Then it is about finding the final speed:$\small v_2$.

• The initial pressure is given in atm which needs to be converted into Pascal before substituting,. Then,

$\qquad\qquad\small 4.86atm=4.86\times 1.01325\times 10^5Pa=492439.5Pa$

• Since the water was assumed incompressible, volumetric flow is consistent thoughout. Then,

$\qquad\qquad\small Q=A_1v_1=A_2v_2$

• Now $\small A_2=\large\frac{1}{6.44}\small A_1\to\large\frac{A_1}{A_2}=\small6.44$
• Rearranging for Q gives,

$\qquad\qquad\large\frac{A_1}{A_2}=\large\frac{v_2}{v_1}=\small6.44$

• Then the final speed is

$\qquad\qquad\small v_2= 6.44\times 2.27ms^{-1}=14.6188ms^{-1}$

• Therefore, the final pressure is,

$\qquad\qquad \begin{aligned} \small P_2&= \small 492439.5Pa+\frac{1}{2}\times 1000kgm^{-3}\times(2.27^2-14.6188^2)\\ &= \small 338461.3Pa\\ &= \small \frac{338461.3Pa}{1.01325\times 10^5}\\ &=\small \bold{3.34atm} \end{aligned}$