Bernoulli Equation is derived from _______. 

This question was previously asked in
JKSSB JE Mechanical 02 Nov 2021 Shift 1 Official Paper
View all JKSSB JE Papers >
  1. Kepler
  2. Laplace
  3. Euler
  4. Poisson

Answer (Detailed Solution Below)

Option 3 : Euler
Free
JKSSB JE Civil RCC Structures Mock Test
1.2 K Users
20 Questions 20 Marks 20 Mins

Detailed Solution

Download Solution PDF

 

Explanation:

Bernoulli's equation is obtained by integrating Euler's equation of motion involves the assumption that velocity potential exists and that the flow condition do not change with time at any point. :

\(\int {\frac{{\rm{P}}}{{\rm{\rho }}}} {\rm{ + }}\int {{\rm{gdz + }}\int {{\rm{VdV = const}}} } \)

If the flow is incompressible, ρ is constant and

\(\frac{P}{\rho}+gz+\frac{V^2}{2}=const\)

\(\frac{P}{\rho g}+z+\frac{V^2}{2g}=const\)

Bernoulli’s Equation is known as the conservation of energy principle and states that in a steady, ideal flow of an incompressible fluid, the total energy at any point of the fluid is constant. The total energy consists of pressure energy, kinetic energy and potential energy or datum energy.

\(\frac{P_1}{\rho_1 g}+\frac{V_1^2}{2g}+Z_1=\frac{P_2}{\rho_2 g}+\frac{V_2^2}{2g}+Z_2\)

All the terms of Bernoulli’s equation:

\(\frac{P}{\gamma } + \frac{{{v^2}}}{{2g}} + z = C\)

P/γ = P/ρg = Pressure energy per unit weight of fluid or pressure head

v2/2g = Kinetic energy per unit weight or kinetic head

z = Potential energy per unit weight or potential head

Latest JKSSB JE Updates

Last updated on Jul 1, 2025

-> JKSSB Junior Engineer recruitment exam date 2025 for Civil and Electrical Engineering has been rescheduled on its official website. 

-> JKSSB JE exam will be conducted on 31st August (Civil), and on 24th August 2025 (Electrical).

-> JKSSB JE application form correction facility has been started. Candidates can make corrections in the JKSSB recruitment 2025 form from June 23 to 27. 

-> JKSSB JE recruitment 2025 notification has been released for Civil Engineering. 

-> A total of 508 vacancies has been announced for JKSSB JE Civil Engineering recruitment 2025. 

-> JKSSB JE Online Application form will be activated from 18th May 2025 to 16th June 2025 

-> Candidates who are preparing for the exam can access the JKSSB JE syllabus PDF from official website of JKSSB.

-> The candidates can check the JKSSB JE Previous Year Papers to understand the difficulty level of the exam.

-> Candidates also attempt the JKSSB JE Mock Test which gives you an experience of the actual exam.

More Bernoulli’s Equation Questions

More Fluids Questions

Get Free Access Now
Hot Links: teen patti club teen patti game - 3patti poker teen patti 3a teen patti win