Question

At one point in a pipeline the water’s speed is 3.00 m/s and the gauge pressure is 5.00∙10⁴ Pa. Find the gauge pressure at a second point in the line, 11.0 m lower than the first, if the pipe diameter at the second point is twice that at the first.

Note: density of water is 1000 kg/m³.

Answer 1

Solution:

Given data

The speed of the water at point one,

A2 from the continuity equation AN=AQVA 2 v2=A vi Thi 41012 v2 = ¥ (3) = 0.75 ml da=ad, B = 2 / 782 = 281 Pa = 5x104+(1000) ( 320-454) +1000 (4-8) (11) 1.62x105 pa. Pat

and the Gauge pressure at point one,

P1 = 5.00 x 10Pa.

We have to find the Gauge pressure at a second point which is 11.0 m (h₁-h₂) below the first point, where the diameter of the tube is twice the diameter of the first point.

From this we have

$d_{1}=\frac{d_{2}}{2}\Rightarrow r_{2}=2r_{1}$

First, we calculate the velocity of the water at the second point using the equation of continuity

A101 A202

Where

A1 Area of the pipe at first point

$v_{1}=\mathrm{velocity\; of\; water\; at\; first \; point}$

$A_{2}=\mathrm{Area\; of\; the\; pipe\; at\; second \; point}$

$v_{2}=\mathrm{velocity\; of\; water\; at\; second \; point}$

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}$

$v_{2}=\frac{(\pi r_{1}^{2})v_{1}}{\pi r_{2}^{2}}$

$v_{2}=\frac{(r_{1}^{2})v_{1}}{(2r_{1})^{2}}$

$v_{2}=\frac{v_{1}}{4}=\frac{3}{4}=0.75\; \mathrm{m/s}$

Now, the gauge pressure at the second point can be calculated by using Bernoulli's equation.

$P+\frac{1}{2}\rho v^{2}+\rho gh=\mathrm{constant}$

$P_{1}+\frac{1}{2}\rho v_{1}^{2}+\rho gh_{1}=P_{2}+\frac{1}{2}\rho v_{2}^{2}+\rho gh_{2}$

$P_{2}=P_{1}+\frac{1}{2}\rho (v_{1}^{2}-v_{2}^{2})+\rho g(h_{1}-h_{2})$

Substitute the known values

$P_{2}=(5\times 10^{4})+\frac{1}{2}(1000) (3^{2}-(0.75)^{2})+(1000)(9.8)(11)$

$P_{2}=50000+4218.75+107800$

$P_{2}=1.62\times 10^{5}\; \mathrm{Pa}$

Answer 2

A2 from the continuity equation AN=AQVA 2 v2=A vi Thi 41012 v2 = ¥ (3) = 0.75 ml da=ad, B = 2 / 782 = 281 Pa = 5x104+(1000) ( 320-454) +1000 (4-8) (11) 1.62x105 pa. Pat

Answer 3

at let vze Let the point where pressure Pi=58104 Pa heigh he and water velocity as and where the pressure is unknown (P2) the height be hq and water velocity be from Bernoulês equation we have Pit h4 vie teghi =P + 1/2 var e tegha ex density of water P2 =Pit 12e (or-12") + eg (hi-he] (6) h, -ha=11m (given) and volume flow nate at A, and. Az are the crossectional at I and a. area A=TAN O2 = ALUI AIZTON, A2 29707 pandry are the radius on point at he and hq respectively. where two point Ari= Aqua Az

U22 Alvi A2 = Annu tra Dinos diameten DIL - = 'el Da 2 Uqa 뿍 r2 = 0 2 2 1/2 (given) = Dz -/ -ام DN D2^ -ام ; g =9081 m/sr using Ua in ean (i) P2= Pit eu (1 7 ) + eg (hi-ha) Pi = 5X104 Pa; e = 1000 kg/m3 he-he=um ; 0,23 m/s 5) P2 =[(5x109+{{ZX1000 * 3(4.35+ (1000X9.81*11)] Pq 21.62x105 Pa ... The gauge pressure at and point is P2 = 1.62x105 pa Ans

Answer 4

A2 from the continuity equation AN=AQVA 2 v2=A vi Thi 41012 v2 = ¥ (3) = 0.75 ml da=ad, B = 2 / 782 = 281 Pa = 5x104+(1000) ( 320-454) +1000 (4-8) (11) 1.62x105 pa. Pat

Answer 5

Concepts and reason

The concepts used to solve this problem are Bernoulli’s theorem and equation of continuity.

First, calculate the speed of the water at second point in the line by using the continuity equation. Finally, calculate the gauge pressure at a second point in the line by using the Bernoulli’s equation.

Fundamentals

The equation of continuity states that when the fluid is in motion, then the motion of the fluid must be in such a way that mass is conserved. The equation of rate of flow of liquid is given as follows:

${A_1}{v_1} = {A_2}{v_2}$

Here, ${A_1}$ is the area of cross section at point 1, ${v_1}$ is the speed of liquid at point 1, ${A_2}$ is the area of cross at point 2 and ${v_2}$ is the speed of liquid at point 2.

The Bernoulli equation is the conservation of energy principle for the flowing fluids. The equation of the Bernoulli principle is given as follows:

${P_1} + \frac{1}{2}\rho {v_1}^2 + \rho g{h_i} = {P_2} + \frac{1}{2}\rho {v_2}^2 + \rho g{h_2}$

Here, ${P_{1{\rm{ }}}}{\rm{and }}{P_2}$ is the pressure at point 1 and 2, $\frac{1}{2}\rho {v_1}^2{\rm{ and }}\frac{1}{2}\rho {v_2}^2$ is the kinetic energy per unit volume at point 1 and 2, $\rho g{h_1}{\rm{ and }}\rho g{h_2}$ are the potential energies at point 1 and 2, $\rho$ is the density, g is the acceleration due to gravity, and h is the height.

Rearrange the above equation for ${P_2}$ .

${P_2} = {P_1} + \frac{1}{2}\rho \left( {{v_1}^2 - {v_2}^2} \right) + \rho g\left( {{h_1} - {h_2}} \right)$

The diameter of the pipe at the second point is twice that at the first. So, the radius of the pipe at the second point is also twice that at the first.

${r_2} = 2{r_1}$

Here, ${r_2}$ is the radius of the pipe at the second point and ${r_1}$ is the radius of the pipe at the first point.

The cross-sectional area of the pipe at first point is,

${A_1} = \pi {r_1}^2$

The cross-sectional area of the pipe at second point is,

${A_2} = \pi {r_2}^2$

Substitute $2{r_1}$ for ${r_2}$ .

$\begin{array}{c}\\{A_2} = \pi {\left( {2{r_1}} \right)^2}\\\\ = 4\pi {r_1}^2\\\end{array}$

The equation of continuity is given as follows:

${A_1}{v_1} = {A_2}{v_2}$

Rearrange the above equation for ${v_2}$ .

${v_2} = \frac{{{A_1}}}{{{A_2}}}{v_1}$

Substitute $\pi {r_1}^2$ for ${A_1}$ and $4\pi {r_1}^2$ for ${A_2}$ .

$\begin{array}{c}\\{v_2} = \frac{{\pi {r_1}^2}}{{4\pi {r_1}^2}}{v_1}\\\\ = \frac{1}{4}{v_1}\\\end{array}$

Substitute 3.00 m/s for ${v_1}$ in the above equation.

$\begin{array}{c}\\{v_2} = \frac{1}{4}\left( {3.0{\rm{ m/s}}} \right)\\\\ = 0.75{\rm{ m/s}}\\\end{array}$

The expression for the gauge pressure at a second point in the line is,

${P_2} = {P_1} + \frac{1}{2}\rho \left( {{v_1}^2 - {v_2}^2} \right) + \rho g\left( {{h_1} - {h_2}} \right)$

Substitute $5.0 \times {10^4}{\rm{ Pa}}$ for ${P_1}$ , $1000{\rm{ kg/}}{{\rm{m}}^3}$ for $\rho$ , 3.0 m/s for ${v_1}$ , 0.75 m/s for ${v_2}$ and 11.0 m for $\left( {{h_1} - {h_2}} \right)$ in the above equation.

$\begin{array}{c}\\{P_2} = \left( {5.0 \times {{10}^4}{\rm{ Pa}}} \right) + \frac{1}{2}\left( {1000{\rm{ kg/}}{{\rm{m}}^3}} \right)\left( {{{\left( {3.0{\rm{ m/s}}} \right)}^2} - {{\left( {0.75{\rm{ m/s}}} \right)}^2}} \right) + \\\\\left( {1000{\rm{ kg/}}{{\rm{m}}^3}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {11.0{\rm{ m}}} \right)\\\\ = 1.62 \times {10^5}{\rm{ Pa}}\\\end{array}$

Ans:

The gauge pressure at a second point in the line is ${\bf{1}}{\bf{.62 \times 1}}{{\bf{0}}^{\bf{5}}}{\bf{ Pa}}$ .