MOVING FLUID IDEAL FLUID CONTINUITY PRINCIPLES BERNOULLI EQUAL APPLICATION BERNOULLI MOVING FLUID IDEAL FLUID CONTINUITY PRINCIPLES BERNOULLI EQUAL APPLICATION BERNOULLI MOVING FLUID IDEAL FLUID CONTINUITY PRINCIPLES BERNOULLI EQUAL APPLICATION BERNOULLI MOVING FLUID

IDEAL FLUIDCONTINUITYPRINCIPLES BERNOULLIEQUAL APPLICATION BERNOULLI













BY: 1. ADE RADIYAS Ponda2. AMIN efficacious3. Tengku Muhammad Falah Miftahul

MOVING FLUID

IDEAL FLUIDIn order for the discussions that we will do more simple and easy to understand, then the fluid is a fluid dikmaksud special called ideal fluid. Ideal fluid is someone actually one moddel, so it's not real fluid as an ideal fluid in reality does not exist.Ideal fluid properties such as:



non-compressible fluidthat the intent is that the density of the compressible fluid is independent of pressure. In general, the fluid (mainly gas) is compressible, the density of the fluid depends on pressure. When the gas pressure in the larger, for example by reducing vplumenya, masstype of gas will change.Fluid flow is turbulent (steamline)Simply that in turbulent flow is meant by the swirling flow, such as a puff of cigarette smoke is a turbulent flow. Opponents of the turbulent flow is laminar flow (streamline)
Fluid flow is stationary (steady)Understanding stasione the ideal gas is almost the same as understanding the wave stasione the flow is stationary if the velocity at any arbitrary point is always constant. This does not mean that kecepatn fluid flow at point A with the titk Saima B. The flow velocity at a point is always constant, ie vA = vB (see the figure)
The fluid is viscous (non-viscous)As we discussed earlier about the viscosity, the viscosity of the fluid friction on the fluid menyuebabkan. In the ideal fluid, we ignore all the friction that arises, which means ignoring the symptoms of viscosity.The line is also called stream-lined flow (laminar flow laminar flow =. Speed ​​of fluid particles at each point on the line of direct current with a tangent at that point. Thus, the flow lines never intersect.When it exceeds a certain pace, the fluid flow becomes turbulent. Marked by turbulent flow in a rotating aliaran. There is a particle-particle has a different motion directions and even, in contrast with the overall direction of motion of the fluid. To determine whether a fluid flow or turbulent flow is a line, you simply drop a bit of ink dye into the liquid substance. If the shape of the vortex, it was a alirn fluid flow lines. However, when the ink around in circles and eventually spread, including the fluid flow is turbulent flow.
 Well, the fluid will be discussed in this chapter is fluid in view of the ideal fluid is the fluid of steady-state, not compressed, not thick, and the stream line (straight line).
CONTINUITYIn previous subab we have assumed that the fluid is ideal to have a stationary flow, the fluid flowing through a given point have the same speed. In addition, has the ideal fluid flow is laminar (streamline).Because the fluid flow is stationary, then we can conclude that the mass of the fluid element that passes a given point is always the same every time. Many elements of the mass of fluid passing a given surface area in a given time, the image diair in a dam is x liter, or discharge of water from the tap is 0.1 liters / second. Q discharge equation can be scaled with the following car. Discharge volume is a measure of the amount of fluid flowing per unit waktu.ΔxQ = ΔV / ΔtKarenaΔv = A. Δx, where A is the cross-sectional area, and Δx is the distance traveled in the fluid, then:Q = (A. Δx) / ΔtIs not (Δx) / Δt = v, so that the general equation debitQfluida vmelaluipenampang pipe with a flow rate of A isQ = vA
In the picture show a moving fluid in a pipe with a changing cross-sectional area of ​​A1dan A2 along the pipe. In this case we will only review the dualokasi the right end of the pipe and the pipe left end of the pipe. Based on the concept satsioner flow, we can conclude that a lot of fluid that enters the left end of the pipe equal to the amount of fluid that comes out of the right end of the pipe. it is in because of the previous agreement that the flowing fluid is considered an ideal fluid, the fluid mass in this eternal.Suppose keceptan fluid at the left end was v1 and fluid velocity at the right end is v2, while the fluid density is ρ. In time Δt, the fluid in the far left has moved so far v1Δt. The amount of fluid mass that has moved past the left end of the pipe is the same as rasterized volume multiplied by its density. Volume equal to v1 Δtdi multiply by the surface area A1.Δm = ρVΔm = ρv1 ΔtA_1Then the fluid velocity passing through the left end of the pipe is〖〗 _1/Δt Δm = ρv1A_1At the right end of the pipe, the fluid velocity is v2, while the density ρ remains fluid because the fluid is assumed ideal. In time Δt, the amount of mass that comes out of the right end of the pipe which has a surface area A2 is:Δm = ρv2 ΔtA_2Mass velocity of fluid flowing through the right end of the pipe is:〖〗 _2/Δt == Ρv2A_2 ΔmWould not have been explained previously, because the fluid flow is stationary, then the total mass of fluid flowing through the pipe at both ends of the large. Even so not with:〖〗 _1/Δt Δm = Δm〗 〖_2/Δtρv1A_1 = ρv2A_2Fluid used is not the same, then the density of the fluid is equal tov1A_1 v2A_2 or v = A = constantyou already know that Q = v A, where Q is the discharge of fluid. Therefore, the continuity equation for the fluid is compressed it can also be expressed as equations kostan discharge.
Kotinuitas equation we can change to forms:v1A_1 = v2A_2v_1/v_2 = A_2/A_1From the above equation states that if a larger cross section of pipe, speed of the fluid at that point is smaller.
PRINCIPLES BERNOULLI15.1 presents a picture of a vessel flow (or an actual pipe, for that matter) that the pass by an ideal fluid flowing with a steady pace. In the time interval Δt, suppose that a fluid volume Δv, which is characterized by the I region in the image 15.1a, entering the vessel at the left end (or input) and a similar volume, which is on the mark with the region III in the figure 15.1b , exit on the right end (out-put). Out to be the same volume of incoming sepeerti volume because fluid can not be compressed, with a mass density ρ is assumed constant.
 Note the picture shows a pipe containing flowing fluid with varying heights and cross-sectional area. We review the motion diarsir.mula first fluid, the fluid keadanaan looks like in the picture, where the diameter of the fluid volume element located at an altitude y1 A1.After the lapse of time, this vluida element moves to the right so that the fluid under consideration shifts to the right, resulting in an element diameter of A2 which is at a height y2. In each section A1 and A2 work force F1 and F2 are the opposite direction as the show in the picture.Now we will calculate the work done each style F1 and F2. From the definition of business as a force in the displacement multiply, much work is done F1 style is:W1 = F1Δl1Semnentara it, because the direction opposite the direction of F2 to F1 so much effort in lakukanoleh force F2 isW2 = F2Δl2
Total need is carried out by forces F1 and F2 areWtot = W1 + W2= F1Δl1 - F2Δl2By using the relationship F = PA and ΔV = AΔl, where p is pressure, A cross-sectional area, volume change ΔV and Δl the displacement changes:Wtot = F1Δl1 - F2Δl2
              Δl1 = pA - pA Δl2
              ΔV1 = P_1 - P_2 ΔV2according to the equation of continuity, then V1 = V2 = ΔV, soW = (P1-P2) ΔVThis equation states that the total force by force F1 and F2. work done by F1 and F2 will eventually lead to changes in kinetic and potential energy in the fluid milkiki. ΔEK kinetic energy changes occur along with changes in the speed of v1 and v.2.ΔEK = 1/2 Δmv_22 - 1/2 Δmv_12ΔEK = 1/2 Δm (v_22 - v_12)Semntara, the potential energy perubhan ΔEP occur along with changes in the position of the fluid from y1 to y2ΔEP = Δmgy2 - Δmgy1Δmg ΔEP = (y2 - y1)Does not change the mechanical eneri ΔEM which is owned by an object similar toΔEM = + ΔEP ΔEKΔEM = 1/2 Δm (v_22 - v_12) + Δmg (y2 - y1)Based on the theorem of effort and energy is not the work done by an object is the change of energy possessed by the object. Thus it is obtained:

W = ΔEM(P1-P2) ΔV = 1/2 Δm (v_22 - v_12) + Δmg (y2 - y1)P1-P2 == 1/2 Δm / ΔV (v_22 - v_12) + Δm / ΔV g (y2 - y1)We know that Δm / ΔV = ρ, the density of the fluid. Thus, it can be in perolehpersamaanP1-P2 == 1/2 ρ (v_22 - v_12) + ρg (y2 - y1)Ρg y1 + P1 + 1/2 ρv_12 = P2 + + ρgy2 half ρv_22This equation is known as the Bernoulli equation. Just as the equation of continuity, Bernoulli also available on persaanP + ρg y + 1/2 = constant ρv2In the Bernoulli effect persaaan strictly only to the extent that fluidanya ideal. If the viscous forces arise, thermal energy will be involved we do not see this in the next drop.
EQUAL APPLICATION BERNOULLIWe have used this principle to explain some qualitative secaara events in everyday life. In this section we will discuss some of the application of Bernoulli's Law.Determining the Hydrostatic Pressure
From the Bernoulli equationΡgh P_1 + +1 / 2 ^ 2 = P_2 ρV_1 ρgh_2 + +1 / 2 ^ 2 ρV_2
because the fluid can not flow (hydrostatic)
⇒ V_1 = V_2 = 0P_0 = P_2 + +0 +0 +0 ρgh_2P_0 = P_2 + ρgh_2 → (h = negative)P_0 = P_2-ρgh_2 → (h = below the surface)-Ρgh_2 P_2 = P⇒ P = P_0 + PhPh = ρgh_2
Determining the Speed ​​of a Flowing Water Hole Vessel

From the Bernoulli equationΡg_h + P +1 / 2 ρV ^ 2 = constantΡgh P_1 + +1 / 2 ^ 2 = P_2 ρV_1 ρgh_2 + +1 / 2 ^ 2 ρV_2+0 +0 = P_0 + P_0 ρgh_2 +1 / 2 ^ 2 ρV_20 =-ρgh_2 +1 / 2 ^ 2 ρV_2ρgh_2 = 1/2 ^ 2 ρV_2

V_2 = √ 2ghVelocity of water that must be removed from a hole at a height h of the surface water is to water the speed of free fall from a height h.
Speed ​​determines the effect of narrowing the pipe hole (vinturi effect)









From the Bernoulli equation
Ρg_h + P +1 / 2 ρV ^ 2 = constantΡgh_1 P_1 + +1 / 2 ^ 2 = P_2 ρV_1 ρgh_2 + +1 / 2 ^ 2 ρV_2
h_1 = h_2 = h
P_1 +1 / 2 ^ 2 = P_2 ρV_1 +1 / 2 ^ 2 ρV_2P +1 / 2 ρV ^ 2 = constant
When the fluid moves into the constriction (cross-sectional area becomes smaller) then the speed of V because penyampitan becomes larger (V ↗ ↘ ⇒ A)if V increases greater then P must be smaller (P ↘ ↗ ⇒ V).
Speed ​​determines Liquids (fluid) in the venturi meter openVenturi without monometer
 remember! meter venturi effectP +1 / 2 ρV ^ 2 = constant
quantitatively↘ ↗ ⇒ P V, P ↗ ↘ ⇒ V, V ↗ ↘ ⇒ A, V ↘ ↗ ⇒ A
of the image V_2> V_1, because A_2 <A_1 then P_2 <P_1
at position 1: P_1 = P_0 +〗 〖Ph ⇒ _1 + P_1 = P_atm ρgh_1at position 2: P_2 = P_0 + 〖〗 _2 Ph ⇒ + P_2 = P_atm ρgh_2 -
  P_1-P_2 = ρg 〖(h〗 _2-h_1)
 P_1-P_2 = ρgy
of the effects of venturi meterP +1 / 2 ρV ^ 2 = constantΡgh_1 P_1 + +1 / 2 ^ 2 = P_2 ρV_1 ρgh_2 + +1 / 2 ^ 2 ρV_2P_1-P_2 = 1/2 ρ (V_1 ^ 2-V_2 ^ 2)
⇒ P_1-P_2 = ρgy
To search V_1 or V_2 ^ ^legal discharge of Q = AV = constantV_1 A_1 = A_2 V_2V_1 = V_2 A_2/A_1
⇒ V_2 = V_1 A_1/A_2
Of 1/2 ρ (V_1 ^ 2-V_2 ^ 2) = ρgy
V_1 ^ 2-V_2 ^ 2 = 2gy

V_2 = .....〖〗 V_2 ^ 2 = 2gy + (V_1) ^ 2〖〗 V_2 ^ 2 = 2gy + (A_2/A_1) ^ 2 〖〗 ^ 2 V_2〖〗 V_2 ^ 2 - (A_2/A_1) ^ 2 〖〗 ^ 2 = V_2 2gy〖〗 V_2 ^ 2 [-1 (A_2/A_1) ^ 2] = 2gy〖〗 ^ 2 = V_2 2gy / (-1 (A_2/A_1) ^ 2)V_2 = √ (2gy / (-1 (A_2/A_1) ^ 2)) ⇒ V_1 = √ (2gy / (-1 (A_1/A_2) ^ 2))

Venturi without the monometer



 
Basic law Hidristatika
PA = PBPhAPhA = P_2 + ρg (h_2 y) + ρ ^ 'gyΡgh_1 = P_1 +-ρgy ρgh_2 + ρ ^ 'gyP_1-P_2 = ρgh_2-ρgh_1-ρgy + ρ ^ 'gyP_1-P_2 = ρg (h_2-h_1-y) + ρ ^ 'gy
h_2 = h_1P_1-P_2 =-ρ + ρgy ^ 'gy
of the effects of venturi meterP +1 / 2 ρV ^ 2 = constant
P_1 +1 / 2 ^ 2 = P_2 ρV_1 +1 / 2 ^ 2 ρV_2P_1-P_2 = 1/2 ρ (V_2-V_1 ^ 2 ^ 2)P_1-P_2 = (ρ ^ '-ρ) gy = 1/2 ρ (V_2-V_1 ^ 2 ^ 2)V_2-V_1 ^ 2 ^ 2 = 2 (ρ ^ '-ρ) gy / ρ

PROBLEMS AND DISCUSSIONAhmad fill a bucket that has a capacity of 20 liters with water from a faucet as shown below!If the sectional area of ​​the faucet with a diameter D2 is a 2 cm2 and velocity of water flow at the faucet is 10 m / s find:a) Debit waterb) The time required to fill the bucketDiscussionData:A2 = 2 cm 2 = 2 x 10-4 m2v2 = 10 m / sa) Debit waterQ = A2v2 = (2 x 10-4) (10)Q = 2 x 10-3 m3 / s
b) The time required to fill the bucketData:V = 20 liters = 20 x 10-3 m3Q = 2 x 10-3 m3 / st = V / Qt = (20 x 10-3 m3) / (2 x 10-3 m3 / s)t = 10 second
Underground plumbing has a shape like the picture below!If the cross-sectional area of ​​the pipe is 5 m2, small pipe cross-sectional area is 2 m2 and velocity of water flow in the pipe is 15 m / s, determine the velocity of the water as it flows in small pipes!DiscussionThe continuity equationA1v1 = A2v2(5) (15) = (2) v2v2 = 37.5 m / s

Water tank with leak holes are shown in the following picture!Hole spacing is 10 m to the ground and surface water into the hole spacing is 3.2 m.Specify:a) The speed of the water dischargeb) reached the farthest horizontal distance the waterc) The time required for ground water leaksDiscussiona) The speed of the water dischargev = √ (2gh)v = √ (2 x 10 x 3.2) = 8 m / s
b) reached the farthest horizontal distance the waterX = 2 √ (mm)X = 2 √ (3.2 x 10) = 8 √ 2 m
c) The time required for ground water leakst = √ (2H / g)t = √ (2 (10) / (10)) = √ 2 second
To measure the velocity of water flow in a horizontal pipe used tool as shown image below!
If the cross-sectional area of ​​the pipe is 5 cm2 and a small pipe cross-sectional area was 3 cm2 and different water levels on the two vertical pipes are 20 cm find:a) The velocity of water as it flows in large pipesb) the speed of the water as it flows in small pipes
Discussiona) The velocity of water as it flows in large pipes√ v1 = A2 [(2gh): (A12 - A22)]v1 = (3) √ [(2 x 10 x 0.2): (52-32)]v1 = 3 √ [(4): (16)]v1 = 1.5 m / sTips:A unit let in cm2, g and h must be in m/s2 and m. v will have the units m / s.
b) the speed of the water as it flows in small pipesA1v1 = A2v2(3/2) (5) = (v2) (3)v2 = 2.5 m / s
Pipeline to deliver water attached to a wall as shown in the picture below! Comparison of cross-sectional area of ​​the pipe and small pipe is 4: 1.

The position of the pipe is 5 m above ground and a small pipe 1 m above the ground.Velocity of water flow in large pipes is 36 km / h with 9.1 x 105 Pa pressure. Specify:a) The speed of water in small pipesb) The excess pressure on the pipec) Pressure on the small pipe(Ρair = 1000 kg/m3)
DiscussionData:h1 = 5 mh2 = 1 mv1 = 36 km / h = 10 m / sP1 = 9.1 x 105 PaA1: A2 = 4: 1
a) The speed of water in small pipesContinuity equation:A1v1 = A2v2(4) (10) = (1) (v2)v2 = 40 m / s
b) The excess pressure on the pipeFrom the Bernoulli equation:P1 + 1/2 + ρv12 ρgh1 = P2 + 1/2 + ρv22 ρgh2P1 - P2 = 1/2 ρ (V22 - v12) + ρg (h2 - h1)P1 - P2 = 1/2 (1000) (402-102) + (1000) (10) (1-5)P1 - P2 = (500) (1500) - 40000 = 750000-40000P1 - P2 = 710 000 Pa = 7.1 x 105 Pa
c) Pressure on the small pipeP1 - P2 = 7.1 x 1059.1 x 105 - P2 = 7.1 x 105P2 = 2.0 x 105 Pa

PROBLEM SOLVEDWater flowing through a horizontal pipe which has two distinct parts, as shown in FIG. If the cross section of 8.0 cm2 X which covers the water flow rate was 3.0 cm / s, calculate:Water flow rate on the extent of cross-section Y 2.0 cm2;Pressure difference between the X and Y
a homogeneous sphere with the type 2/3 the type of water falling vertically into a pool from a height of 10 m above the surface of the pond. How deep under the surface of the pond, the ball before it could sink the buoyant force pressed it back up? (Abajkan water barriers on the movement of the ball). Take g = 9.8 m / s
In the picture shown how the flow of water from one faucet to bend down, when the water falls cross-sectional area A0 = 1.2 cm2, and A = 0.35 cm2. Both tiers were separated by a vertical distance of h = 45 mm. at what speed drove the tap water?
REFERENCESFoster. Bob. Of 1999. Integrated Physics SMU. New York: KPG (literature).Haliday and Resnik. In 1991. Physics Volume 1 (translation). London: publisher.Kanginan, Marthen. Of 1996. High school physics. London: publisher.Kanginan, Marthen. Of 1996. High school physics vol 2B. London: publisher.Based on the lecture Dr. Tomo Djudin