Answer to Question #156863 in Mechanics | Relativity for haze

Question #156863

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?


1
Expert's answer
2021-01-22T15:05:40-0500

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,

P1+12ρv12=P2+12ρv22\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: P2\small P_2 gives,

P2=P1+12ρ(v12v22)\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:v2\small v_2.


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

4.86atm=4.86×1.01325×105Pa=492439.5Pa\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,

Q=A1v1=A2v2\qquad\qquad\small Q=A_1v_1=A_2v_2

  • Now A2=16.44A1A1A2=6.44\small A_2=\large\frac{1}{6.44}\small A_1\to\large\frac{A_1}{A_2}=\small6.44
  • Rearranging for Q gives,

A1A2=v2v1=6.44\qquad\qquad\large\frac{A_1}{A_2}=\large\frac{v_2}{v_1}=\small6.44

  • Then the final speed is

v2=6.44×2.27ms1=14.6188ms1\qquad\qquad\small v_2= 6.44\times 2.27ms^{-1}=14.6188ms^{-1}


  • Therefore, the final pressure is,

P2=492439.5Pa+12×1000kgm3×(2.27214.61882)=338461.3Pa=338461.3Pa1.01325×105=3.34atm\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}




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