Respuesta :
According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by
[tex]\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )[/tex]
Here, [tex]v_{1}[/tex] is the velocity of water through wide ends of cylindrical pipe and [tex]v_{2}[/tex] is the velocity of water through narrow ends of cylindrical pipe.
Given, [tex]v_{1} =1.4 m/s[/tex]
Now from equation continuity,
[tex]v_{1} A_{1} = v_{2} A_{2}[/tex].
Here, [tex]A_{1}[/tex] and [tex]A_{2}[/tex] are cross- sectional areas of wide and narrow ends of cylindrical pipe.
As pipe is circular, so
[tex]v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}[/tex].
At the second point, the diameter is halved, which means the radius is also halved. Therefore,
[tex]v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}[/tex]
[tex]v_{2} = 4 \times 1.4 = 5.6 m/s[/tex]
Substituting these values with the density of water is [tex]1000 \ kg/m^3[/tex] in pressure difference formula we get.
[tex]\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa[/tex]
The pressure difference between the wide and the narrow ends of the pipe is [tex]\boxed{14700\,{\text{Pa}}}[/tex] .
Further Explanation:
According to the equation of continuity, the amount of water entering at one end of the cylindrical pipe is always equal to the amount of water coming out from the other end of the pipe.
[tex]{A_1}{v_1}={A_2}{v_2}[/tex]
Here, [tex]{A_1}\,\&\,{A_2}[/tex] are the areas of cross-section of the two ends and [tex]{v_1}\,\&\,{v_2}[/tex] are the velocities of the flow at the two ends.
The area of the cross section of the pipe at the starting end is.
[tex]{A_1}=\pi{r^2}[/tex]
Since the diameter of the pipe is reduced to half at the other end then its radius will also become half. Therefore, the area of cross section of the pipe will be reduced by a factor of 4.
[tex]\begin{aligned}{A_2}&=\pi{\left({\frac{r}{2}}\right)^2}\\&=\frac{{\pi{r^2}}}{4}\\&=\frac{{{A_1}}}{4}\\\end{aligned}[/tex]
The speed of the water flow at the entering point is [tex]1.4\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] .
Substitute the values in equation (1).
[tex]\begin{aligned}{A_1} \times 1.4&=\frac{{{A_1}}}{4}\times{v_2}\\{v_2}&=5.6\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}[/tex]
Now according to the Bernoulli’s equation, the change in pressure between the two ends of the cylindrical pipe is given as:
[tex]\Delta P=\frac{1}{2}\rho\left({v_2^2 - v_1^2}\right)[/tex]
The density of the water is [tex]1000\,{{{\text{kg}}}\mathord{\left/{\vphantom{{{\text{kg}}}{{{\text{m}}^{\text{3}}}}}}\right.\kern-\nulldelimiterspace} {{{\text{m}}^{\text{3}}}}}[/tex] .
Substitute the values in above expression:
[tex]\begin{aligned}\Delta P&=\frac{1}{2}\times1000\times\left({{{5.6}^2}-{{1.4}^2}} \right)\\&=500\times29.4\,{\text{Pa}}\\&{\text{ = 14700}}\,{\text{Pa}}\\\end{aligned}[/tex]
Thus, the pressure difference between the wide and the narrow ends of the pipe is [tex]\boxed{14700\,{\text{Pa}}}[/tex] .
Learn More:
1. A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill https://brainly.com/question/9805263
2. For flowing water what is the magnitude of the velocity gradient https://brainly.com/question/5181841
Answer Details:
Grade: High School
Subject: Physics
Chapter: Fluid Mechanics
Keywords:
Water flows, horizontal cylindrical pipe, diameter is halved, pressure difference, wide and narrow ends, equation of continuity, Bernoulli’s equation, A1v1=A2v2.
