Fluid Mechanics Notes for SSC JE & UPSSSC | Boundary Layer, Pipe Flow, Numericals (Hindi)
ЁЯФ╢ PART–1A : FLUID PROPERTIES (DEEP THEORY – JE EXAM)
рдпрд╣ рднाрдЧ UPSSSC / SSC JE рдкрд░ीрдХ्рд╖ा рдХो рдз्рдпाрди рдоें рд░рдЦрдХрд░ рддैрдпाрд░ рдХिрдпा рдЧрдпा рд╣ै। рдЗрд╕рдоें basic рд╕े рд▓ेрдХрд░ exam-trap рддрдХ рд╕рднी concepts рдХो detail рдоें рд╕рдордЭाрдпा рдЧрдпा рд╣ै।
1.1 Introduction to Fluid Mechanics
Fluid Mechanics engineering рдХी рд╡рд╣ рд╢ाрдЦा рд╣ै рдЬिрд╕рдоें fluids (liquid рдПрд╡ं gas) рдХे rest (Fluid Statics) рддрдеा motion (Fluid Dynamics) рдоें рд╡्рдпрд╡рд╣ाрд░ рдХा рдЕрдз्рдпрдпрди рдХिрдпा рдЬाрддा рд╣ै।
- Water Supply System
- Irrigation Canals
- Pumps & Turbines
- Hydraulic Press
- Aircraft & Automobile Design
JE Exam Focus: Applications рдкрд░ theory based рдк्рд░рд╢्рди рдкूрдЫे рдЬाрддे рд╣ैं।
1.2 Definition of Fluid
Fluid рд╡рд╣ рдкрджाрд░्рде рд╣ै рдЬो рдмрд╣ुрдд рдЫोрдЯे shear stress рдкрд░ рднी рд▓рдЧाрддाрд░ deform рдХрд░рддा рд╣ै рдФрд░ stress рд╣рдЯाрдиे рдкрд░ рдЕрдкрдиी original shape recover рдирд╣ीं рдХрд░рддा।
Concept: Shear stress ≠ 0 ⇒ Continuous deformation
Examples:- Water
- Oil
- Air
1.3 Difference Between Solid and Fluid
| Basis | Solid | Fluid |
|---|---|---|
| Shape | Fixed | Container dependent |
| Shear Stress | Resist рдХрд░рддा рд╣ै | Flow рдХрд░рддा рд╣ै |
| Deformation | Finite | Infinite |
Exam Tip: “Infinite deformation” JE рдоें important keyword рд╣ै।
1.4 Classification of Fluids
(A) Ideal Fluid
- Zero viscosity
- Incompressible
- Hypothetical (real life рдоें exist рдирд╣ीं рдХрд░рддा)
(B) Real Fluid
- Viscosity present
- Compressible рдпा incompressible
(C) Newtonian Fluid
Newton’s law of viscosity рдХो follow рдХрд░рддा рд╣ै:
╧Д = ╬╝ (du/dy)
Examples: Water, Air
JE Question: Water → Newtonian fluid
(D) Non-Newtonian Fluid
Viscosity constant рдирд╣ीं рд╣ोрддी। Stress рдФрд░ velocity gradient рдХा relation linear рдирд╣ीं рд╣ोрддा।
Examples: Toothpaste, Blood, Paint
1.5 Density (╧Б)
Density = mass per unit volume
╧Б = m / V
Unit: kg/m³
- Water = 1000 kg/m³
- Mercury = 13600 kg/m³
Trap: Density рдФрд░ specific weight рдХो mix рди рдХрд░ें।
1.6 Specific Weight (╬│)
Specific weight = weight per unit volume
╬│ = ╧Б g
Unit: N/m³
1.7 Specific Gravity (SG)
SG = ╧Бfluid / ╧Бwater
Properties:
• Dimensionless
• Water SG = 1
JE Shortcut: SG × 1000 = Density
1.8 Viscosity (рд╢्рдпाрдирддा)
Viscosity fluid рдХा internal resistance рд╣ै рдЬो flow рдХे opposite act рдХрд░рддा рд╣ै।
╧Д = ╬╝ (du/dy)
Dynamic Viscosity (╬╝): Pa·s
Kinematic Viscosity (╬╜): m²/s
High viscosity ⇒ Slow flow
1.9 Surface Tension (╧Г)
Surface tension liquid surface рдкрд░ acting force рд╣ै рдЬो surface area рдХो minimum рдХрд░рдиे рдХी рдХोрд╢िрд╢ рдХрд░рддा рд╣ै।
h = (4╧Г cos╬╕) / (╧Б g d)
╬╕ = 0° ⇒ Maximum capillary rise
1.10 Compressibility & Bulk Modulus
Liquids → almost incompressible
Gases → highly compressible
K = ╬ФP / (╬ФV / V)
1.11 Vapor Pressure & Cavitation
Pressure < Vapor Pressure ⇒ Cavitation
JE Exam рдоें pump problems рд╕े direct рд╕рд╡ाрд▓ рдЖрддे рд╣ैं।
✔ Definitions clear рд░рдЦें
✔ Units & dimensions рдпाрдж рд░рдЦें
✔ Density, SG, Viscosity рд╕े questions fix рд╣ोрддे рд╣ैं
ЁЯФ╢ PART–1B : FLUID PROPERTIES (NUMERICALS + PYQ – JE EXAM)
рдЗрд╕ рднाрдЧ рдоें UPSSSC / SSC JE рдкрд░ीрдХ्рд╖ा рдоें рдкूрдЫे рдЬाрдиे рд╡ाрд▓े numericals, PYQ concepts рдФрд░ exam traps рдХो detail рдоें рд╕рдордЭाрдпा рдЧрдпा рд╣ै।
1️⃣ Density – Important Numericals
Solution:
╧Б = m / V = 500 / 0.5 = 1000 kg/m³
Exam Tip: Water рдЬैрд╕ी density ⇒ fluid = water (approx).
2️⃣ Specific Gravity – PYQ Type
Solution:
Density = SG × 1000 = 0.8 × 1000 = 800 kg/m³
Exam Trap: рдХрдИ рдЫाрдд्рд░ SG рдХो unit рдХे рд╕ाрде рд▓िрдЦ рджेрддे рд╣ैं – SG рд╣рдоेрд╢ा unitless рд╣ोрддा рд╣ै।
3️⃣ Specific Weight – Numerical
Solution:
╬│ = ╧Б g = 1000 × 9.81 = 9810 N/m³
4️⃣ Viscosity – JE Favourite
Answer: High viscosity ⇒ Slow flow
PYQ Concept: Honey flows slower than water due to higher viscosity.
5️⃣ Newton’s Law of Viscosity – Concept
╧Д = ╬╝ (du/dy)
- ╧Д ∝ du/dy
- ╬╝ = constant ⇒ Newtonian fluid
JE Direct Question: Water, air → Newtonian fluids
6️⃣ Surface Tension – PYQ Numerical
Answer: Surface tension рдФрд░ adhesion force рдХे рдХाрд░рдг।
╬╕ = 0° ⇒ Maximum capillary rise
7️⃣ Compressibility – Concept
- Liquids → Nearly incompressible
- Gases → Highly compressible
JE PYQ: Hydraulic machines рдоें liquid use рдХिрдпा рдЬाрддा рд╣ै।
8️⃣ Vapor Pressure & Cavitation
Answer: рдЬрдм liquid pressure, vapor pressure рд╕े рдХрдо рд╣ो рдЬाрддा рд╣ै।
Exam Link: Pump suction problems
9️⃣ One-Line JE Questions
- SG of water = 1
- Viscosity рдХा SI unit = Pa·s
- Kinematic viscosity unit = m²/s
- Surface tension unit = N/m
✔ Numericals рдкрд╣рд▓े formula рд╕े рд╢ुрд░ू рдХрд░ें
✔ Units рдкрд░ extra рдз्рдпाрди рджें
✔ PYQ concepts repeat рд╣ोрддे рд╣ैं
ЁЯФ╖ PART–2 : HYDROSTATIC PRESSURE (Theory + Numericals – JE Exam)
рдЗрд╕ рднाрдЧ рдоें Hydrostatic Pressure рдХो concept + numericals + PYQ рдХे рд╕ाрде рд╡िрд╕्рддाрд░ рд╕े рд╕рдордЭाрдпा рдЧрдпा рд╣ै। рдпрд╣ рд╣िрд╕्рд╕ा UPSSSC / SSC JE рдкрд░ीрдХ्рд╖ा рдоें рдмрд╣ुрдд рдЬ़्рдпाрджा рдкूрдЫा рдЬाрддा рд╣ै।
2.1 Pressure in a Static Fluid
рдЬрдм fluid rest рдоें рд╣ोрддा рд╣ै, рддрдм рдЙрд╕рдХे рдХाрд░рдг рд▓рдЧрдиे рд╡ाрд▓े pressure рдХो Hydrostatic Pressure рдХрд╣рддे рд╣ैं।
Formula:
p = ╧Б g h
- p = pressure (N/m²)
- ╧Б = density (kg/m³)
- g = gravity (9.81 m/s²)
- h = depth (m)
JE Concept: Pressure depth рдкрд░ depend рдХрд░рддा рд╣ै, container shape рдкрд░ рдирд╣ीं।
2.2 Pascal’s Law
Pascal’s Law рдХे рдЕрдиुрд╕ाрд░, рдХिрд╕ी enclosed fluid рдкрд░ рд▓рдЧाрдпा рдЧрдпा pressure рд╣рд░ рджिрд╢ा рдоें рд╕рдоाрди рд░ूрдк рд╕े transmit рд╣ोрддा рд╣ै।
- Hydraulic press
- Hydraulic jack
- Hydraulic lift
JE PYQ: Hydraulic machines Pascal’s law рдкрд░ рдЖрдзाрд░िрдд рд╣ोрддी рд╣ैं।
2.3 Pressure Measurement Devices
- Piezometer – low pressure liquids
- U-tube Manometer
- Differential Manometer
Exam Tip: Gases рдХे рд▓िрдП piezometer рдЙрдкрдпोрдЧ рдирд╣ीं рд╣ोрддा।
2.4 Total Pressure on Submerged Surface
рдХिрд╕ी plane surface рдкрд░ fluid рдж्рд╡ाрд░ा рд▓рдЧाрдпा рдЧрдпा total force:
P = ╧Б g A h̄
рдЬрд╣ाँ h̄ = surface рдХा centroid depth
2.5 Center of Pressure
Center of Pressure рд╡рд╣ рдмिंрджु рд╣ै рдЬрд╣ाँ total pressure act рдХрд░рддा рд╣ै।
hcp = h̄ + ( IG / (A h̄) )
JE Rule: Center of pressure рд╣рдоेрд╢ा centroid рд╕े рдиीрдЪे рд╣ोрддा рд╣ै।
ЁЯУШ Numerical–1 (Pressure at Depth)
Solution:
p = ╧Б g h = 1000 × 9.81 × 5
p = 49,050 N/m²
ЁЯУШ Numerical–2 (Total Pressure)
Solution:
P = ╧Б g A h̄
P = 1000 × 9.81 × 2 × 4
P = 78,480 N
ЁЯУШ Numerical–3 (Center of Pressure – Concept)
Answer: рд╣рдоेрд╢ा centroid рд╕े рдиीрдЪे
ЁЯУЭ One-Line JE Questions
- Pressure ∝ Depth
- Hydrostatic pressure shape independent рд╣ोрддा рд╣ै
- Pascal’s law enclosed fluid рдкрд░ рд▓ाрдЧू рд╣ोрддा рд╣ै
- Center of pressure ≠ Centroid
✔ Formula рдпाрдж рд░рдЦें
✔ Numericals рдоें units check рдХрд░ें
✔ Center of pressure рдкрд░ direct рд╕рд╡ाрд▓ рдЖрддे рд╣ैं
ЁЯФ╖ PART–3 : ORIFICES, NOTCHES & WEIRS (JE Exam)
рдЗрд╕ рднाрдЧ рдоें Orifice, Notch рдФрд░ Weir рдХो theory + numericals + PYQ рдХे рд╕ाрде рд╕рдордЭाрдпा рдЧрдпा рд╣ै। рдпрд╣ topic SSC JE / UPSSSC JE рдоें рдмрд╣ुрдд important рд╣ै।
3.1 Orifice (рдУрд░िрдл़िрд╕)
Orifice рдПрдХ рдЫोрдЯा рд╕ा рдЫिрдж्рд░ рд╣ोрддा рд╣ै рдЬिрд╕рд╕े fluid рдмाрд╣рд░ рдиिрдХрд▓рддा рд╣ै।
- Small orifice
- Large orifice
Theoretical velocity:
v = √(2gh)
Real velocity = Cv √(2gh)
3.2 Coefficients of Orifice
- Cv – Coefficient of velocity
- Cc – Coefficient of contraction
- Cd – Coefficient of discharge
Cd = Cv × Cc
JE PYQ: Cd рд╣рдоेрд╢ा 1 рд╕े рдХрдо рд╣ोрддा рд╣ै।
3.3 Notch
Notch рдПрдХ sharp edged opening рд╣ोрддी рд╣ै рдЬिрд╕рд╕े fluid flow рдХрд░рддा рд╣ै।
- Rectangular notch
- Triangular (V-notch)
Rectangular Notch:
Q = (2/3) Cd b √(2g) h3/2
V–Notch:
Q = (8/15) Cd √(2g) tan(╬╕/2) h5/2
3.4 Weir
Weir рд╡ाрд╕्рддрд╡ рдоें рдПрдХ рдмрдб़ा notch рд╣ोрддा рд╣ै рдЬो channel рдоें рдмрдиाрдпा рдЬाрддा рд╣ै।
- Rectangular weir
- Triangular weir
- Broad crested weir
Difference: Notch → tank / small flow Weir → open channel / large flow
ЁЯУШ Numerical–1 (Orifice Velocity)
Solution:
v = √(2gh) = √(2 × 9.81 × 4)
v = 8.86 m/s
ЁЯУШ Numerical–2 (Rectangular Notch)
Solution:
Q = (2/3) × 0.62 × 1 × √(2 × 9.81) × (0.5)3/2
Q ≈ 0.46 m³/s
ЁЯУШ Numerical–3 (Concept PYQ)
Answer: V–Notch (small discharge рдХे рд▓िрдП)
ЁЯУЭ One-Line JE Questions
- Orifice tank рдоें use рд╣ोрддा рд╣ै
- Notch sharp edged opening рд╣ै
- Weir open channel рдоें use рд╣ोрддा рд╣ै
- Cd < 1
✔ Formulas рдпाрдж рд░рдЦें
✔ V–Notch small discharge рдХे рд▓िрдП best рд╣ै
✔ PYQ рдоें coefficients рдкूрдЫे рдЬाрддे рд╣ैं
ЁЯФ╖ PART–4 : PIPE FLOW & LOSSES (JE Exam)
рдЗрд╕ рднाрдЧ рдоें Pipe Flow рдФрд░ Losses in Pipe рдХो theory + numericals + PYQ рдХे рд╕ाрде рд╕рдордЭाрдпा рдЧрдпा рд╣ै। рдпрд╣ topic SSC JE / UPSSSC JE рдоें рдмрд╣ुрдд рдмाрд░ рдкूрдЫा рдЬाрддा рд╣ै।
4.1 Types of Flow
- Laminar Flow
- Turbulent Flow
Reynolds Number:
Re = (╧Б v d) / ╬╝
- Re < 2000 → Laminar
- Re > 4000 → Turbulent
- 2000–4000 → Transition
JE PYQ: Flow type рд╣рдоेрд╢ा Reynolds number рд╕े рддрдп рд╣ोрддा рд╣ै।
4.2 Darcy–Weisbach Equation
Pipe рдоें friction рдХे рдХाрд░рдг рд╣ोрдиे рд╡ाрд▓ा head loss:
hf = f (L/D) (v² / 2g)
- f = friction factor
- L = pipe length
- D = pipe diameter
4.3 Major Loss
Pipe рдХी рдкूрд░ी length рдоें friction рд╕े рд╣ोрдиे рд╡ाрд▓ा loss
Depends on: Length, diameter, velocity, roughness
4.4 Minor Losses
- Entry loss
- Exit loss
- Bend loss
- Sudden expansion
- Sudden contraction
Exit loss: h = v² / 2g
JE Concept: Minor losses ∝ velocity²
ЁЯУШ Numerical–1 (Reynolds Number)
Solution:
Re = (╧Б v d) / ╬╝
Re = (1000 × 1 × 0.05) / 0.001
Re = 50,000 → Turbulent flow
ЁЯУШ Numerical–2 (Head Loss)
Solution:
hf = f (L/D) (v² / 2g)
= 0.02 × (100 / 0.1) × (4 / 19.62)
= 4.08 m
ЁЯУШ Numerical–3 (Exit Loss)
Solution:
h = v² / 2g = 9 / 19.62 = 0.46 m
ЁЯУЭ One-Line JE Questions
- Major loss friction рдХे рдХाрд░рдг рд╣ोрддा рд╣ै
- Minor loss fittings рдХे рдХाрд░рдг рд╣ोрддा рд╣ै
- Exit loss = v² / 2g
- Reynolds number dimensionless рд╣ै
✔ Reynolds number рд╕े flow type рдкрд╣рдЪाрдиें
✔ Darcy formula exam favourite рд╣ै
✔ Minor losses рдХे formula рдпाрдж рд░рдЦें
ЁЯФ╖ PART–5 : BOUNDARY LAYER THEORY (JE Exam)
5.1 Boundary Layer рдХ्рдпा рд╣ै?
рдЬрдм рдХोрдИ fluid рдХिрд╕ी solid surface рдХे рдКрдкрд░ flow рдХрд░рддा рд╣ै, рддो surface рдХे рдкाрд╕ fluid рдХी velocity рд╢ूрди्рдп рд╣ोрддी рд╣ै। рдЗрд╕े No-Slip Condition рдХрд╣рддे рд╣ैं।
5.2 Boundary Layer Thickness (╬┤)
Boundary layer thickness рд╡рд╣ рджूрд░ी рд╣ै рдЬрд╣ाँ fluid рдХी velocity, free stream velocity рдХी 99% рд╣ो рдЬाрддी рд╣ै।
5.3 Types of Boundary Layer
- Laminar Boundary Layer
- Turbulent Boundary Layer
- Transition Zone
- Low velocity → Laminar
- High velocity → Turbulent
5.4 Laminar Boundary Layer
Laminar boundary layer рдоें fluid smooth layers рдоें flow рдХрд░рддा рд╣ै рдФрд░ mixing рдирд╣ीं рд╣ोрддी।
5.5 Turbulent Boundary Layer
Turbulent boundary layer рдоें fluid motion irregular рд╣ोрддा рд╣ै рдФрд░ mixing рдЕрдзिрдХ рд╣ोрддी рд╣ै।
5.6 Boundary Layer Separation
рдЬрдм fluid flow рдХрд░рддे рд╕рдордп pressure рдмрдв़рддा рд╣ै (Adverse Pressure Gradient), рддो boundary layer surface рд╕े рдЕрд▓рдЧ рд╣ो рдЬाрддी рд╣ै।
5.7 Effects of Boundary Layer
- Drag force рдЙрдд्рдкрди्рди рд╣ोрддा рд╣ै
- Energy loss рд╣ोрддा рд╣ै
- Lift force рдк्рд░рднाрд╡िрдд рд╣ोрддी рд╣ै
ЁЯУЭ One-Line JE Questions
- No-slip condition solid surface рдкрд░ рд▓ाрдЧू рд╣ोрддी рд╣ै
- Boundary layer thickness рдХो ╬┤ рд╕े рджрд░्рд╢ाрддे рд╣ैं
- Laminar BL рдоें friction рдХрдо рд╣ोрддा рд╣ै
- Boundary layer separation pressure increase рд╕े рд╣ोрддा рд╣ै
✔ 99% velocity concept рдпाрдж рд░рдЦें
✔ Separation рдХा рдХाрд░рдг рдЬрд░ूрд░ рдпाрдж рд░рдЦें
✔ Theory-based questions direct рдЖрддे рд╣ैं
ЁЯФ╖ PART–6 : FLOW MEASUREMENT DEVICES (JE Exam)
рдЗрд╕ рднाрдЧ рдоें Flow Measurement Devices рдХो theory + formulas + JE exam focus рдХे рд╕ाрде рд╕рдордЭाрдпा рдЧрдпा рд╣ै। рдпрд╣ topic SSC JE / UPSSSC JE рдоें direct рдкूрдЫा рдЬाрддा рд╣ै।
6.1 Flow Measurement Devices рдХ्рдпा рд╣ैं?
Flow measurement devices рдХा рдЙрдкрдпोрдЧ pipe рдпा channel рдоें flow rate (discharge) рдоाрдкрдиे рдХे рд▓िрдП рдХिрдпा рдЬाрддा рд╣ै।
- Venturimeter
- Orificemeter
- Pitot Tube
- Rotameter
6.2 Venturimeter
Venturimeter Bernoulli’s theorem рдкрд░ рдЖрдзाрд░िрдд device рд╣ै рдЬो pipe рдоें discharge measure рдХрд░рддा рд╣ै।
Discharge Formula:
Q = Cd × (A₁A₂ / √(A₁² − A₂²)) × √(2gh)
- Losses рдХрдо рд╣ोрддे рд╣ैं
- Accuracy рд╕рдмрд╕े рдЬ़्рдпाрджा
JE PYQ: Venturimeter рд╕рдмрд╕े accurate device рд╣ै।
6.3 Orificemeter
Orificemeter рднी Bernoulli principle рдкрд░ рдХाрдо рдХрд░рддा рд╣ै рд▓ेрдХिрди рдЗрд╕рдоें losses рдЕрдзिрдХ рд╣ोрддे рд╣ैं।
- Cheap device
- Losses рдЬ्рдпाрджा
- Accuracy рдХрдо
Exam Trap: Orificemeter ≠ Orifice
6.4 Pitot Tube
Pitot Tube velocity measurement рдХे рд▓िрдП use рд╣ोрддा рд╣ै।
Velocity Formula:
v = C √(2gh)
рдЬрд╣ाँ h = stagnation head
JE Concept: Pitot tube discharge рдирд╣ीं, velocity рдоाрдкрддा рд╣ै।
6.5 Rotameter
Rotameter рдПрдХ variable area flow meter рд╣ै।
- Vertical tube
- Float рдКрдкрд░–рдиीрдЪे move рдХрд░рддा рд╣ै
- Low pressure liquids рдХे рд▓िрдП
JE PYQ: Rotameter = Variable area meter
6.6 Comparison (Exam Favourite)
| Device | Measures | Accuracy |
|---|---|---|
| Venturimeter | Discharge | High |
| Orificemeter | Discharge | Low |
| Pitot Tube | Velocity | Medium |
| Rotameter | Discharge | Medium |
ЁЯУЭ One-Line JE Questions
- Venturimeter рд╕рдмрд╕े accurate рд╣ोрддा рд╣ै
- Pitot tube velocity measure рдХрд░рддा рд╣ै
- Orificemeter рдоें head loss рдЕрдзिрдХ рд╣ोрддा рд╣ै
- Rotameter variable area meter рд╣ै
✔ Bernoulli theorem рдмрд╣ुрдд important рд╣ै
✔ Comparison questions рдмाрд░-рдмाрд░ рдЖрддे рд╣ैं
✔ Pitot vs Venturi confuse рди рдХрд░ें
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ЁЯСЙ UPSSSC Civil JE – Structural Engineering Complete Notes (Hindi)

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