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,

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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).

Physics

2 answers:

Softa [3K] · Answer 1 · 2026-02-23T11:14:34+03:00

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

Further Details:

According to Bernoulli’s principle, for fluids in steady flow, the sum of all energy forms remains consistent across points in the stream, meaning the total of kinetic, potential, and internal energy equals a constant value at each latitude along the stream.

The Bernoulli equation can be expressed as:

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

In this equation, $H$ is referred to as the energy head (total head), $z$ represents the elevation head, $\dfrac{p}{{\rho g}}$ indicates the pressure head, and $\dfrac{{{v^2}}}{{2g}}$ corresponds to the dynamic head.

Provided Information:

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

There are $220$ cans in total.

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

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

The height of the pipe at point 1 measures $1.35\,{\text{m}$ .

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

The pressure at point 2 in the pipe is $152\,{\text{kPa}}$ .

Point 2 is on the datum line, so its height is $0\,{\text{m}}$ .

Understanding:

Total fluid flow rate through the pipe:

$F=n\times{f}$

In this context, $n$ refers to the number of cans, $f$ denotes the fluid flow rate through each can, and $F$ is the cumulative fluid flow rate.

Replace $220$ with $n$ and $0.355\text{ l}/\text{min}$ with $f$ .

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

Fluid velocity at point 1:

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

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

In this expression, substitute $0.0013\text{ m}^3/\text{s}$ for $F$ and $0.0002\,{{\text{m}}^{\text{2}}}$ for ${a_1}$ .

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

Fluid velocity at point 2:

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

In this relation, ${a_2}$ is the cross-sectional area of point 2, while ${v_2}$ denotes the fluid velocity at that point.

Make the substitution: $0.0008\,{{\text{m}}^{\text{2}}}$ for ${a_2}$

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

Utilizing Bernoulli’s principle for points 1 and 2:

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

Rearranging the equation to solve 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, $\rho$ refers to water density, $g$ is the gravitational acceleration, ${z_1}$ is the height at point 1, ${z_2}$ is the height at point 2, ${p_1}$ denotes the pressure at point 1, and ${p_2}$ the pressure at point 2.

Insert all known values into the previous equation.

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

Consequently, the pressure at point 1 in the pipe is $\fbox{\\120.9\,{\text{kPa}}}$ or $\fbox{121\text{ kPa}}$ .

Learn more:

1. Motion of a ball under gravity

2. A car with a mass of 700 kg moving at 29 m/s

3. Box weighing 30 kg being dragged over a carpeted surface

Answer Summary:

Grade: College

Field: Physics

Chapter: Fluid Mechanics

Key Terms:

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

Keith_Richards [3.2K] · Answer 2 · 2026-02-23T11:14:14+03:00

Flow rate calculations yield 220 cans, each with a volume of 0.355 l, leading to 78.1 l/min or 1.3 l/s or 0.0013 m³/s.

At Point 2:
A2 = 8 cm² = 0.0008 m²
V2 = Flow rate/A2 = 0.0013/0.0008 = 1.625 m/s
P1 = 152 kPa = 152000 Pa

At Point 1:
A1 = 2 cm² = 0.0002 m²
V1 = Flow rate/A1 = 0.0013/0.0002 = 6.5 m/s
P1 =?
Height = 1.35 m

Using Bernoulli’s principle;
P2 + 1/2 * V2² / density = P1 + 1/2 * V1² / 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