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Water is moving with speed of 5m/S through the pipe with cross sectional area of 4cm^2 .the water gradually descents 10m as the pipe inc. in area to 8cm^2 .if the pressure at the upper end is 1.5×10^5Pa, the pressure at lower level will be approximately. Explain in detail?
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Water is moving with speed of 5m/S through the pipe with cross section...
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Water is moving with speed of 5m/S through the pipe with cross section...
Answer:
Introduction:
In this problem, we have given the velocity of water, the cross-sectional area of the pipe at two different points, and the height difference between the two points. We need to find the pressure at the lower point.
Formula:
The Bernoulli's principle states that the total energy of a fluid flowing through a pipe is constant. The principle can be expressed mathematically as:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
Where,
P1 is the pressure at point 1
v1 is the velocity of fluid at point 1
h1 is the height of fluid at point 1
P2 is the pressure at point 2
v2 is the velocity of fluid at point 2
h2 is the height of fluid at point 2
ρ is the density of the fluid
g is the acceleration due to gravity
Solution:
Given,
Velocity of water, v = 5 m/s
Cross-sectional area of the pipe at point 1, A1 = 4 cm^2
Cross-sectional area of the pipe at point 2, A2 = 8 cm^2
Height difference, h = 10 m
Pressure at point 1, P1 = 1.5 × 10^5 Pa
We need to find the pressure at point 2, P2.
Step 1:
First, we need to convert the cross-sectional areas from cm^2 to m^2.
A1 = 4 cm^2 = 4 × 10^-4 m^2
A2 = 8 cm^2 = 8 × 10^-4 m^2
Step 2:
We can find the velocity of water at point 2 using the continuity equation:
A1 * v1 = A2 * v2
v2 = (A1 * v1) / A2
v2 = (4 × 10^-4 * 5) / (8 × 10^-4)
v2 = 2.5 m/s
Step 3:
We can find the density of water using the formula:
ρ = m / V
Where,
m is the mass of water
V is the volume of water
The density of water at room temperature is approximately 1000 kg/m^3.
Step 4:
We can now apply the Bernoulli's principle between points 1 and 2.
P1 + (1/2 * ρ * v1^2) + (ρ * g * h1) = P2 + (1/2 * ρ * v2^2) + (ρ * g * h2)
We can ignore the kinetic energy term (1/2 * ρ * v1^2) at point 1 as the velocity is negligible due to the small cross-sectional area.
P1 + (ρ * g * h1) = P2 + (1/2 * ρ * v2^2) + (ρ * g * h2)
P2 = P1 + (1/
Introduction:
In this problem, we have given the velocity of water, the cross-sectional area of the pipe at two different points, and the height difference between the two points. We need to find the pressure at the lower point.
Formula:
The Bernoulli's principle states that the total energy of a fluid flowing through a pipe is constant. The principle can be expressed mathematically as:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
Where,
P1 is the pressure at point 1
v1 is the velocity of fluid at point 1
h1 is the height of fluid at point 1
P2 is the pressure at point 2
v2 is the velocity of fluid at point 2
h2 is the height of fluid at point 2
ρ is the density of the fluid
g is the acceleration due to gravity
Solution:
Given,
Velocity of water, v = 5 m/s
Cross-sectional area of the pipe at point 1, A1 = 4 cm^2
Cross-sectional area of the pipe at point 2, A2 = 8 cm^2
Height difference, h = 10 m
Pressure at point 1, P1 = 1.5 × 10^5 Pa
We need to find the pressure at point 2, P2.
Step 1:
First, we need to convert the cross-sectional areas from cm^2 to m^2.
A1 = 4 cm^2 = 4 × 10^-4 m^2
A2 = 8 cm^2 = 8 × 10^-4 m^2
Step 2:
We can find the velocity of water at point 2 using the continuity equation:
A1 * v1 = A2 * v2
v2 = (A1 * v1) / A2
v2 = (4 × 10^-4 * 5) / (8 × 10^-4)
v2 = 2.5 m/s
Step 3:
We can find the density of water using the formula:
ρ = m / V
Where,
m is the mass of water
V is the volume of water
The density of water at room temperature is approximately 1000 kg/m^3.
Step 4:
We can now apply the Bernoulli's principle between points 1 and 2.
P1 + (1/2 * ρ * v1^2) + (ρ * g * h1) = P2 + (1/2 * ρ * v2^2) + (ρ * g * h2)
We can ignore the kinetic energy term (1/2 * ρ * v1^2) at point 1 as the velocity is negligible due to the small cross-sectional area.
P1 + (ρ * g * h1) = P2 + (1/2 * ρ * v2^2) + (ρ * g * h2)
P2 = P1 + (1/
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Water is moving with speed of 5m/S through the pipe with cross sectional area of 4cm^2 .the water gradually descents 10m as the pipe inc. in area to 8cm^2 .if the pressure at the upper end is 1.5×10^5Pa, the pressure at lower level will be approximately. Explain in detail?
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Water is moving with speed of 5m/S through the pipe with cross sectional area of 4cm^2 .the water gradually descents 10m as the pipe inc. in area to 8cm^2 .if the pressure at the upper end is 1.5×10^5Pa, the pressure at lower level will be approximately. Explain in detail? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Water is moving with speed of 5m/S through the pipe with cross sectional area of 4cm^2 .the water gradually descents 10m as the pipe inc. in area to 8cm^2 .if the pressure at the upper end is 1.5×10^5Pa, the pressure at lower level will be approximately. Explain in detail? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Water is moving with speed of 5m/S through the pipe with cross sectional area of 4cm^2 .the water gradually descents 10m as the pipe inc. in area to 8cm^2 .if the pressure at the upper end is 1.5×10^5Pa, the pressure at lower level will be approximately. Explain in detail?.
Water is moving with speed of 5m/S through the pipe with cross sectional area of 4cm^2 .the water gradually descents 10m as the pipe inc. in area to 8cm^2 .if the pressure at the upper end is 1.5×10^5Pa, the pressure at lower level will be approximately. Explain in detail? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Water is moving with speed of 5m/S through the pipe with cross sectional area of 4cm^2 .the water gradually descents 10m as the pipe inc. in area to 8cm^2 .if the pressure at the upper end is 1.5×10^5Pa, the pressure at lower level will be approximately. Explain in detail? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Water is moving with speed of 5m/S through the pipe with cross sectional area of 4cm^2 .the water gradually descents 10m as the pipe inc. in area to 8cm^2 .if the pressure at the upper end is 1.5×10^5Pa, the pressure at lower level will be approximately. Explain in detail?.
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