Notes : Full Derivation Bernoulli's Principle , Equation - Class 11 Physics JEE NEET
Learn Bernoulli's Principle Class 11 Physics Chapter 8 mechanical properties of fluids NCERT notes, derivation, formula, assumptions, limitations, applications, MCQs, FAQs, numericals, and question answers JEE NEET - Physicsfund
Bernoulli's Principle (Bernoulli's Theorem)
Bernoulli's Principle is one of the most important principles of Fluid Mechanics. It explains the relationship between the pressure, velocity, and height of a moving fluid. The principle is based on the Law of Conservation of Energy.
1. Introduction
According to Bernoulli's Principle:
For the steady flow of an ideal fluid, the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline.
Bernoulli's Equation:
$$ P+\frac{1}{2}\rho v^2+\rho gh=\text{Constant} $$Between two points:
$$ P_1+\frac12\rho v_1^2+\rho gh_1 = P_2+\frac12\rho v_2^2+\rho gh_2 $$2. Meaning of Each Term
| Term | Meaning |
|---|---|
| $P$ | Pressure Energy per Unit Volume |
| $\frac12\rho v^2$ | Kinetic Energy per Unit Volume |
| $\rho gh$ | Potential Energy per Unit Volume |
(A) Pressure Energy per Unit Volume : P
It is the energy possessed by the fluid due to the pressure exerted on it.
- SI Unit = Pascal (Pa)
- Also equal to Joule per cubic metre (J/m³)
(B) Kinetic Energy per Unit Volume : $ \frac12\rho v^2 $
where
- $\rho$ = Density of fluid
- $v$ = Velocity of fluid
It is the energy due to the motion of the fluid.
(C) Potential Energy per Unit Volume : $ \rho gh $
where
- $\rho$ = Density of fluid
- $g$ = Acceleration due to gravity
- $h$ = Height above the reference level
It is the gravitational potential energy of the fluid.
3. Physical Meaning of Bernoulli's Principle
The total mechanical energy of an ideal fluid remains constant throughout its flow. If one form of energy increases, another decreases so that the total energy remains unchanged.
- Higher velocity → Lower pressure
- Higher pressure → Lower velocity
- Higher height → Lower pressure or lower velocity
- Total mechanical energy always remains constant.
4. Derivation of Bernoulli's Equation
The derivation of Bernoulli's equation is based on the Work-Energy Theorem.
Work-Energy Theorem:
$$ W=\Delta K+\Delta U $$5. Consider a Flowing Fluid
Consider an ideal fluid flowing through a pipe of varying cross-sectional area.
| Section 1 | Section 2 |
|---|---|
| Pressure = $P_1$ | Pressure = $P_2$ |
| Velocity = $v_1$ | Velocity = $v_2$ |
| Height = $h_1$ | Height = $h_2$ |
During a small time interval $\Delta t$, the fluid moves:
- Distance $x_1$ at Section 1
- Distance $x_2$ at Section 2
6. Volume of Fluid Passing
The volume of fluid entering Section 1 is equal to the volume leaving Section 2.
$$ \Delta V=A_1x_1=A_2x_2 $$where
- $A_1$ = Area of Section 1
- $A_2$ = Area of Section 2
- $x_1$ = Distance travelled at Section 1
- $x_2$ = Distance travelled at Section 2
7. Mass of the Fluid Element
Since the fluid is incompressible,
$$ \Delta m=\rho\Delta V $$The same small fluid element is followed throughout the derivation. Therefore,
$$ m_1=m_2=\rho\Delta V $$Why is the Mass the Same?
- The same fluid element is considered.
- The flow is steady.
- The fluid is incompressible.
- Density remains constant.
- Therefore, the mass does not change while moving from Section 1 to Section 2.
8. Work Done at Section 1 (Inlet)
At Section 1, the pressure of the surrounding fluid pushes the fluid element forward.
Pressure at Section 1:
$$ P_1 $$
Cross-sectional Area:
$$ A_1 $$Pressure Force:
$$ F_1=P_1A_1 $$Distance travelled:
$$ x_1 $$Work Done:
$$ W_1=F_1x_1 $$Substituting the value of force,
$$ W_1=P_1A_1x_1 $$Since,
$$ A_1x_1=\Delta V $$Therefore,
$$ \boxed{W_1=P_1\Delta V} $$This work is positive because the pressure force acts in the direction of motion.
9. Work Done at Section 2 (Outlet)
At Section 2, the fluid pushes against the fluid ahead. Hence, work is done by the fluid on the surroundings.
Pressure at Section 2:
$$ P_2 $$Pressure Force:
$$ F_2=P_2A_2 $$Work Done:
$$ W_2=-F_2x_2 $$Substituting,
$$ W_2=-P_2A_2x_2 $$Since,
$$ A_2x_2=\Delta V $$Therefore,
$$ \boxed{W_2=-P_2\Delta V} $$The negative sign indicates that the work is done by the fluid against the pressure of the surrounding fluid.
10. Total Work Done on the Fluid
The total work done is the sum of the work done at the inlet and outlet.
$$ W=W_1+W_2 $$Substituting the values,
$$ W=P_1\Delta V-P_2\Delta V $$Taking common factor,
$$ \boxed{W=(P_1-P_2)\Delta V} $$11. Change in Potential Energy
Mass of the fluid element,
$$ \Delta m=\rho\Delta V $$Initial Potential Energy,
$$ U_1=m_1 g h_1 $$
$$ U_1=\rho\Delta Vgh_1 $$Final Potential Energy,
$$ U_2=m_2 g h_2 $$
$$ U_2=\rho\Delta Vgh_2 $$Therefore,
$$ \Delta U=U_2-U_1 $$Substituting the values,
$$ \Delta U=\rho\Delta Vgh_2-\rho\Delta Vgh_1 $$Taking common factor,
$$ \boxed{\Delta U=\rho\Delta Vg(h_2-h_1)} $$12. Change in Kinetic Energy
Initial Kinetic Energy,
$$ K_1=\frac12(\rho\Delta V)v_1^2 $$Final Kinetic Energy,
$$ K_2=\frac12(\rho\Delta V)v_2^2 $$Hence,
$$ \Delta K=K_2-K_1 $$Substituting,
$$ \Delta K= \frac12\rho\Delta Vv_2^2 - \frac12\rho\Delta Vv_1^2 $$Taking common factor,
$$ \boxed{\Delta K= \frac12\rho\Delta V(v_2^2-v_1^2)} $$13. Apply the Work-Energy Theorem
According to the Work-Energy Theorem,
$ W=\Delta K+\Delta U $Substituting all expressions,
$ (P_1-P_2)\Delta V= \frac12\rho\Delta V(v_2^2-v_1^2) +\rho\Delta Vg(h_2-h_1) $Dividing the entire equation by \(\Delta V\),
$ P_1-P_2= \frac12\rho(v_2^2-v_1^2) +\rho g(h_2-h_1) $14. Rearranging the Equation
Expand the right-hand side,
$$ P_1-P_2= \frac12\rho v_2^2 - \frac12\rho v_1^2 +\rho gh_2 -\rho gh_1 $$Move all terms of Point 1 to the left side and Point 2 to the right side,
$$ P_1+\frac12\rho v_1^2+\rho gh_1 = P_2+\frac12\rho v_2^2+\rho gh_2 $$15. Final Bernoulli's Equation
$$ \boxed{ P_1+\frac12\rho v_1^2+\rho gh_1 = P_2+\frac12\rho v_2^2+\rho gh_2 } $$Or, along any streamline,
$$ \boxed{ P+\frac12\rho v^2+\rho gh=\text{Constant} } $$16. Special Case: Fluid at Rest
If the fluid is at rest, then its velocity at every point is zero.
$$ v_1=v_2=0 $$Substituting into Bernoulli's equation,
$$ P_1+\rho gh_1=P_2+\rho gh_2 $$or,
$$ P+\rho gh=\text{Constant} $$This is known as the Hydrostatic Pressure Equation.
17. Assumptions of Bernoulli's Principle
Bernoulli's equation is valid only under the following conditions:
- The fluid is non-viscous (ideal).
- The fluid is incompressible (density remains constant).
- The flow is steady.
- The flow is streamline (laminar).
- No external work is done on or by the fluid.
- No heat is added to or removed from the fluid.
- The equation is applied along the same streamline.
18. Limitations of Bernoulli's Principle
Bernoulli's equation is not applicable when:
- The fluid is viscous.
- The flow is turbulent.
- The density changes significantly (compressible fluids).
- Pumps or turbines add or remove energy.
- Heat transfer takes place.
- Frictional losses are significant.
19. Applications of Bernoulli's Principle
- Venturimeter
- Pitot Tube
- Carburetor
- Atomizer
- Perfume Spray
- Aeroplane Wing Lift
- Bunsen Burner
- Chimney Draft
- Blood Flow Measurement
- Roof Uplift During Storms
20. Important Formulae
Bernoulli's Equation
$$ P+\frac12\rho v^2+\rho gh=\text{Constant} $$Between Two Points
$$ P_1+\frac12\rho v_1^2+\rho gh_1 = P_2+\frac12\rho v_2^2+\rho gh_2 $$Hydrostatic Equation
$$ P+\rho gh=\text{Constant} $$Mass of Fluid Element
$$ m=\rho\Delta V $$Frequently Asked Questions (FAQs)
Q1. What is Bernoulli's Principle?
It states that the sum of pressure energy, kinetic energy and potential energy per unit volume remains constant along a streamline for an ideal fluid.
Q2. On which law is Bernoulli's Principle based?
Law of Conservation of Energy.
Q3. Why does pressure decrease when velocity increases?
An increase in kinetic energy is balanced by a decrease in pressure energy so that total mechanical energy remains constant.
Q4. When is Bernoulli's equation valid?
It is valid for steady, incompressible, non-viscous flow along the same streamline.
Very Short Answer Questions
- What is Bernoulli's Principle?
It is the principle of conservation of mechanical energy in a flowing ideal fluid. - Which law is used to derive Bernoulli's equation?
Work-Energy Theorem. - What is the SI unit of pressure?
Pascal (Pa). - What is the SI unit of pressure energy per unit volume?
J/m³ or Pa. - Name one application of Bernoulli's Principle.
Venturimeter.
Short Answer Questions
- State Bernoulli's Principle.
- Write Bernoulli's equation.
- What are the assumptions of Bernoulli's equation?
- Why is Bernoulli's equation based on the conservation of energy?
- Explain why pressure decreases when fluid velocity increases.
Long Answer Questions
- Derive Bernoulli's equation using the Work-Energy Theorem.
- Explain the assumptions and limitations of Bernoulli's Principle.
- Discuss the practical applications of Bernoulli's Principle.
Multiple Choice Questions (MCQs)
-
Bernoulli's Principle is based on:
A. Newton's Second Law
B. Conservation of Momentum
C. Conservation of Energy
D. Hooke's Law
Answer: C -
The SI unit of pressure is:
A. Joule
B. Pascal
C. Newton
D. Watt
Answer: B -
When the velocity of an ideal fluid increases, its pressure:
A. Increases
B. Remains Constant
C. Decreases
D. Becomes Zero
Answer: C -
Bernoulli's equation is applicable for:
A. Turbulent Flow
B. Viscous Flow
C. Steady Flow
D. Compressible Flow
Answer: C
True / False
- Bernoulli's Principle is based on conservation of energy. (True)
- Bernoulli's equation is valid for viscous fluids. (False)
- Pressure decreases as velocity increases. (True)
- Bernoulli's equation applies to turbulent flow. (False)
- Pressure energy is measured in J/m³. (True)
Fill in the Blanks
- Bernoulli's Principle is based on the law of Conservation of Energy.
- The SI unit of pressure is Pascal.
- For an incompressible fluid, ______ remains constant. Density
- Higher velocity corresponds to ______ pressure. Lower
- Bernoulli's equation is valid for ______ flow. Steady
Numerical-Based Questions
- State Bernoulli's equation and identify each term.
- A fluid flows through a horizontal pipe where its speed increases from 2 m/s to 5 m/s. Explain what happens to the pressure.
- Why does an aeroplane wing experience lift according to Bernoulli's Principle?
Key Points for Revision
- Bernoulli's Principle is based on the Law of Conservation of Energy.
- Total mechanical energy per unit volume remains constant.
- Higher velocity means lower pressure.
- The equation is valid only for an ideal, incompressible, non-viscous fluid in steady streamline flow.
- All three terms of Bernoulli's equation have the SI unit Pascal (Pa) or J/m³.
- Important applications include Venturimeter, Pitot Tube, Carburetor, Atomizer, Aeroplane Wings, and Bunsen Burner.
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