Question

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 cans, 0.355 - l each, per minute. at point 2 in the pipe, the gauge pressure is 152 kpa and the cross-sectional area is 8.00 cm2. at point 1, 1.35m above point 2, the cross-sectional area is 2.00 cm2. find the gauge pressure at point 1 (in kpa).

Answer 1

The pressure of point 1 on a pipe is $\fbox{\\120.9\,{\text{kPa}}}$ or $\fbox{121\text{ kPa}}$ .

Further Explanation:

Bernoulli’s principle states that for a steady flow of fluids, sum of all the forms of energy along a stream is same for all points on the stream, i.e. the sum of kinetic, potential, and internal energy is same for all points on the stream.

Bernoulli’s equation can be stated as:

$\fbox{\begin\\H=z+\dfrac{p}{{\rho g}}+\dfrac{{{v^2}}}{{2g}}\end{minispace}}$

 

Here, $H$ is the energy head or total head or constant, $z$ is the height or elevation head, $\dfrac{p}{{\rho g}}$ is the pressure head and $\dfrac{{{v^2}}}{{2g}}$ is the dynamic head.

Given:

Density of water is $1000\text{ kg}/\text{m}^3$.

The number of cans is $220$.  

The flow rate of fluid through each can is $0.355\text{ l}/\text{min}$.

The area of cross-section of point 1 is $2\,{\text{c}}{{\text{m}}^{\text{2}}}$.

The height of pipe of point 1 is $1.35\,{\text{m}$.

The area of cross-section of point 2 is $8{\text{ cm}}^{\text{2}}}$.

The pressure of pipe of point 2 is $152\,{\text{kPa}}$.

The point 2 is at the datum line. Therefore, the height of point is $0\,{\text{m}}$.

Concept:

Total flow rate of fluid in pipe:

$F=n\times{f}$

Here, $n$ is the number of cans, $f$ is the flow of rate of fluid through each can, and $F$ is the total flow rate of fluid.

Substitute $220$ for $n$ and $0.355\text{ l}/\text{min}$ for $f$.

$\begin{aligned\\F&= 220 \times 0.355\text{ litre}/\text{min}\\&=78.1\text{ litre}/\text{min}\\&=1.3\text{ litre}/\text{sec}\\&=0.0013\text{ m}^3/\text{s}}\end{aligned}$

The velocity of fluid at point 1:

${{\text{v}}_1} = \dfrac{F}{{{a_1}}}$

Here, $F$ is the total flow rate of fluid, ${a_1}$ is the cross-section area of point 1 and ${v_1}$ is the velocity of fluid at point 1.

Substitute $0.0013\text{ m}^3/\text{s}$ for $F$ and $0.0002\,{{\text{m}}^{\text{2}}}$ for ${a_1}$ in above expression.

$\begin{aligned}{{\text{v}}_1}&=\frac{{0.0013\,{{{{\text{m}}^{\text{3}}}} \mathord{\left/{\vphantom{{{{\text{m}}^{\text{3}}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}}{{0.0002\,{{\text{m}}^{\text{2}}}}}\\&=6.5\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

The velocity of fluid at point 2:

${{\text{v}}_2} = \dfrac{F}{{{a_2}}}$

Here, ${a_2}$ is the cross-section area of point 2 and ${v_2}$ is the velocity of fluid of point 2.

Substitute $0.0008\,{{\text{m}}^{\text{2}}}$ for ${a_2}$

$\begin{aligned}{{\text{v}}_2}=\frac{{0.0013\text{ m}^3/\text{s}}}{0.0008\,\text{m}^{2}}}\\=1.625\text{ m}/\text{s}\\\end{aligned}$

Applying Bernoulli’s principle at point 1 and point 2:

$\fbox{\begin\\{z_1}+\dfrac{{{p_1}}}{{\rho g}}+\dfrac{{v_1^2}}{{2g}}={z_2}+\dfrac{{{p_2}}}{{\rho g}}+\dfrac{{v_2^2}}{{2g}}\end{minispace}}$

Rearrange the above equation for ${p_{_1}}$.

${p_1} = \rho g\left( {{z_2} - {z_1}} \right) + \dfrac{\rho }{2}\left( {v_2^2 - v_1^2} \right) + {p_2}$

Here, $\rho$  is the water density, $g$ is the acceleration due to gravity, ${z_1}$ is the height of point 1, ${z_2}$ is the height of point 2, ${p_1}$ is the pressure of point 1 and ${p_2}$ is the pressure of point 2.

Substitute all the values in above equation

$\begin{aligned}p_1&=(1000\times{9.81}\times{(0-1.35)})+\dfrac{1000}{2}(1.625^2-6.5^2)+154000\\&=-13243.5+(-19804.68)+154000\\&=120951.82\text{ Pa}\\&\approx{121\text{ kPa}\end{aligned}$

Thus, the pressure of point 1 on a pipe is  $\fbox{\\120.9\,{\text{kPa}}}$ or $\fbox{121\text{ kPa}}$ .

Learn more:

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2.  A 700 kg car driving at 29m/s brainly.com/question/9484203

3. A 30kg box being pulled across a carpeted floor brainly.com/question/7031524

Answer Details:

Grade: College

Subject: Physics

Chapter: Fluid Mechanics

Keywords:

Bernoulli’s theorem, pipe, flow rate, pressure head, gauge pressure, point, fluid, height, 0.355 l/min, cross section area is 8.00 cm2 or 8.00 cm^2, and 1.35m above point 2.

Answer 2

Flow rate = 220*0.355 l/m = 78.1 l/min = 1.3 l/s = 0.0013 m^3/s

Point 2:
A2= 8 cm^2 = 0.0008 m^2
V2 = Flow rate/A2 = 0.0013/0.0008 = 1.625 m/s
P1 = 152 kPa = 152000 Pa

Point 1:
A1 = 2 cm^2 = 0.0002 m^2
V1 = Flow rate/A1 = 0.0013/0.0002 = 6.5 m/s
P1 = ?
Height = 1.35 m

Applying Bernoulli principle;
P2+1/2*V2^2/density = P1+1/2*V1^2/density +density*gravitational acceleration*height
=>152000+0.5*1.625^2*1000=P1+0.5*6.5^2*1000+1000*9.81*1.35
=> 153320.31 = P1 + 34368.5
=> P1 = 1533210.31-34368.5 = 118951.81 Pa = 118.95 kPa