5b. Konsep Aliran Air (Hidrodinamika)1

KONSEP ALIRAN KONSEP ALIRAN  AIR (HIDRODINAMIKA HIDRODINAMIKA))

Persamaan Aliran  Ada tiga persamaan dasar dalam Hidrolika  yang berkaitan dengan pengaliran air (aliran tertutup maupun terbuka), yaitu: Persamaan Kontinyuitas Persamaan Momentum Persamaan Energi Bernoulli • • •

The Continuity Equation Why does a hose with a nozzle shoot water further? Conservation of Mass: In a confined system, all of the mass that enters the system, must also exit the system at the same time. Flow rate = Q = Area x Velocity r1 A 1 V1(mass

 V1

inflow rate) =

r2 A 2 V2(



 A 1

 A 2  V2

Q2 = A 2 V2 Q1 = A 1 V1

 A 1 V1 = A 2 V2



mass outflow rate)

If the fluid at both points is the same, then the density drops out, and you get the continuity equation:  A 1 V1 =A 2 V2 Therefore If A 2 < A 1 then V2 > V1 Thus, water exiting a nozzle has a higher velocity

Persamaan Kontinyuitas Q  A1.V 1 



 A2 .V 2



konstan

 Q : debit aliran

  A : luas penampang aliran   V : kecepatan rerata aliran pada

penampang tersebut.  Indeks 1 dan 2 menunjukan titik penampang aliran yang ditinjau

Persamaan Momentum  F 



 r .Q(V 2



V 1 )

 F : gaya yang ditimbulkan oleh aliran

zat cair  r : rapat massa aliran

Persamaan Bernoulli  z 1 

 P  1





g  

 gz 1  

2

 P  1  r 

V 1

2 g 



 z 2



 P 2



2



 gz 2



g  

2

V 1

2



 P 2  r 

V 2

2 g  2



V 2

2

P/ =pressure head, V 2/2g=velocity head, z=elevation P/r as flow energy, V 2/2 as kinetic energy, and gz as potential energy 

The Energy Line and the Hydraulic Grade Line Lets first understand this drawing: Measures the Static Pressure

Measures the Total Head

12

12 EL

 V 2/2g

HGL

1: Static Pressure Tap Measures the sum of the elevation head and the pressure Head. 2: Pilot Tube Measures the Total Head EL : Energy Line

Q

P/γ

Total Head along a system HGL : Hydraulic Grade line

Z

Sum of the elevation and the pressure heads along a system

The Energy Line and the Hydraulic Grade Line Understanding the graphical approach of Energy Line and the Hydraulic Grade line is key to understanding what forces are supplying the energy that water holds.

 V 2/2g

EL  V 2/2g

2

P/γ

Q Z

1

Z

Majority of energy stored in the water is in the Pressure Head Point 2:

HGL

P/γ

Point 1:

Majority of energy stored in the water is in the elevation head If the tube was symmetrical, then the velocity would be constant, and the HGL would be level

Contoh Soal

Dengan Q konstan = 1,5 L/det, diameter pipa 1 = 50 mm, diameter pipa 2 = 40 mm, dan P1= 0,5 bar,  hitunglah tekanan di titik 2 (P2)  hitunglah gaya akibat aliran dari titik 1 ke titik 2

THE COMPLETE EXAMPLE Solve for the Pressure Head, Velocity Head, and Elevation Head at each point, and then plot the Energy Line and the Hydraulic Grade Line  Assumptions and Hints: P1 and P4 = 0 --- V 3 = V 4 same diameter tube We must work backwards to solve this problem γH2O= 62.4 lbs/ft3

1 4’ 

R = .5’  R = .25’ 

2

3 DATUM

4

1’ 

POINT 1: Pressure Head : Only atmospheric  Velocity Head : In a large tank   , V 1 = 0

P1/γ = 0 V 12/2g = 0

Elevation Head : Z1 = 4’ 

γH2O= 62.4 lbs/ft3

1

R = .5’  R = .25’ 

4’ 

2 DATUM

3

4

1’ 

POINT 4:  Apply the Bernoulli equation between 1 and 4 0 + 0 + 4 = 0 + V 42/2(32.2) + 1  V 4 = 13.9 ft/s Pressure Head : Only atmospheric

P4/γ = 0

 Velocity Head : V 42/2g = 3’  Elevation Head : Z4 = 1’  γH2O= 62.4 lbs/ft3

1 4’ 

R = .5’ 

R = .25’ 

2 DATUM

3

4

1’ 

POINT 3:  Apply the Bernoulli equation between 3 and 4 (V 3=V 4) P3/62.4 + 3 + 1 = 0 + 3 + 1 P3 = 0 Pressure Head : P3/γ = 0  Velocity Head : V 32/2g = 3’  Elevation Head : Z3 = 1’  γH2O= 62.4 lbs/ft3

1 4’ 

R = .5’ 

R = .25’ 

2

3

4

1’ 

POINT 2:  Apply the Bernoulli equation between 2 and 3 P2/62.4 + V 22/2(32.2) + 1 = 0 + 3 + 1  Apply the Continuity Equation (Π.52)V 2 = (Π.252)x13.9

V 2 = 3.475 ft/s

P2/62.4 + 3.4752/2(32.2) + 1 = 4

γH2O= 62.4 lbs/ft3

1 4’ 

P2 = 175.5 lbs/ft2

R = .5’ 

Pressure Head : P2/γ = 2.81’   Velocity Head :  V 22/2g = .19’ 

R = .25’ 

2 3 DATUM

4

1’ 

Elevation Head : Z2 = 1’ 

Plotting the EL and HGL Energy Line = Sum of the Pressure, Velocity and Elevation heads Hydraulic Grade Line = Sum of the Pressure and Elevation heads  V 2/2g=.19’  EL

P/γ =2.81’ 

 V 2/2g=3’   V 2/2g=3’ 

Z=4’ 

HGL Z=1’ 

Z=1’ 

Z=1’ 

DATUM

POLA ALIRAN

 Ada dua pola aliran dari fluida, yaitu:  aliran laminer  aliran turbulen

 ALIRAN LAMINER  Dalam aliran laminer partikel-partikel fluidanya bergerak di sepanjang lintasan-lintasan lurus, sejajar dalam lapisan-lapisan atau laminae.

 ALIRAN TURBULEN Dalam

aliran turbulen partikel-partikel

fluidanya bergerak secara acak ke semua arah.

BILANGAN REYNOLDS 

Bilangan Reynold digunakan untuk menyatakan pola aliran (laminer atau turbulen).



Laminer



Turbulen



Bilangan Reynold dapat dihitung dengan formula:

Re < 2000 Re > 2000

Bilangan Reynolds Re 

V  * d  * r  

 

V  :

kecepatan rata-rata, m/det d  : diameter pipa, m : viskositas kinematik fluida, m2 /dtk : rapat massa fluida, kg/m3 : viskositas mutlak, Pa dtk

atau

V  * d  