ME 262 BASIC FLUID MECHANICS Assistant Professor Neslihan Semerci Lecture 6 (Bernoulli’s Equation) 1 19. CONSERVATION OF ENERGY- BERNOULLI’S EQUATION Law of Conservation of Energy: “energy can be neither created nor destroyed. It can be transformed from one form to another.” Potential energy Kinetic energy Pressure energy In the analysis of a pipeline problem accounts for all the energy within the system. Inner wall of the pipe V P Centerline z element of fluid Reference level An element of fluid inside a pipe in a flow system; - Located at a certain elevation (z) - Have a certain velocity (V) - Have a pressure (P) The element of fluid would possess the following forms of energy; 1. Potential energy: Due to its elevation, the potential energy of the element relative to some reference level PE = Wz W= weight of the element. 2. Kinetic energy: Due to its velocity, the kinetic energy of the element is KE = Wv2/2g 2 3. Flow energy(pressure energy or flow work): Amount of work necessary to move element of a fluid across a certain section aganist the pressure (P). PE = W P/γ Derivation of Flow Energy: L P F = PA Work = PAL = FL=P∀ ∀= volume of the element. Weight of element W = γ ∀volume W Volume of element ∀ = γ Total amount of energy of these three forms possessed by the element of fluid; E= PE + KE + FE P E = Wz + Wv2/2g + W γ 3 Figure 19.1. Element of fluid moves from a section 1 to a section 2, (Source: Mott, R. L., Applied Fluid Mechanics, Prentice Hall, New Jersey) 2 V1 P1 Total energy at section 1: E = Wz + W +W 1 1 2g γ 2 V2 P2 Total energy at section 2: E = Wz + W + W × 2 2 2g γ If no energy is added to the fluid or lost between sections 1 and 2, then the principle of conservation of energy requires that; E1 = E2 2 2 V1 P1 V2 P2 Wz + W +W = Wz + W + W × 1 2g γ 2 2g γ The weight of the element is common to all terms and can be divided out. BERNOULLI’S EQUATION v2 P v 2 P z + + 1 = z + 2 + 2 1 2g γ 2 2g γ 4 19.1 INTERPRETATION OF BERNOULLI’S EQUATION 2 2 v P1 v2 P2 z + + = z + + 1 2g γ 2 2g γ Each term in Bernoulli’s equation is one form of the energy possessed by the fluid per unit weight of fluid flowing in the system. The units in each term are “energy per unit weight”. In the SI system the units are N.m/N and in the U.S. Customary System the units are Ib.ft/Ib. The force unit appears in both the numerator and denominator and it can be cancelled. The resulting unit is simply the meter (m) or foot (ft) and can be interpreted to be a height. In fluid flow analysis, the terms are typically expressed as “head” referring to a height above a reference level. P/γ pressure head z elevation head Summation of the terms called total head V2/2g velocity head Figure 19.2.Pressure head, elevation head, velocity head, and total head in a pipe flow. (Source: Mott, R. L., Applied Fluid Mechanics, Prentice Hall, New Jersey) 5 19.2. Restrictions on Bernoulli’s Equation 1. It’s valid only for incompressible liquids since the specific weight of the fluid is assumed to be the same at two sections of interest. 2. There can be no mechanical devices between section 1 and 2 that would add energy to or remove energy from the system, since the equation states that the total energy of fluid is constant. 3. There can be no heat transfered into or out of the fluid. 4. There can be no energy loss due to friction. Example 19.1. In Figure 19.2, water at 100C is flowing from section 1 to section 2. At section 1, which is 25 mm in diameter, the gage pressure is 345kPa and the velocity of flow is 3.0 m/s. Section 2, which is 50 mm in diameter, is 2.0 m above section 1. Assuming there are no energy losses in the system, calculate the pressure P2. Knowns; P1 = 345 kPa P2 =? V1 = 3 m/s V2 =? ∅1 = 25 mm ∅2 = 50 mm Twater = 10 ℃ z2 − z1 = 2.0 m Q = A1 × V1 25 mm 2 π ( ) × 3 m/s = 1.47 × 10−3 m3/s 1000 m 4 From continuity Equation Q = A1 × V1 = A2 × V2 25 mm 2 π 50 mm 2 π ( ) × 3 m/s = ( ) × V 1000 m 4 1000 m 4 2 V2 = 0.75 m/s 2 P1 = 345 kPa = 345 kN/m 2 2 v1 P1 v2 P2 z + + = z + + 1 2g γ 2 2g γ 6 2 2 2 345 kN/m (3.0) (0.75) P2 z + + = z + + 1 9.81 kN/m3 2×(9.81) 2 2×(9.81) γ 2 2 2 P2 345 kN/m (3.0) (0.75) = + − + z − z γ 9.81 kN/m3 2×(9.81) 2×(9.81) 1 2 P2 = 35.1 m + 0.458 −0.0286 – 2 m γ P2 = 33.5 m Pressure Head γ 3 P2 = (9.81 kN/m ) × (33.5 m) = 328.9 kPa 19.3. Tanks, Reservoirs, and Nozzles Exposed to the Atmosphere A siphon draws fluid from a tank and delivers it through a nozzle at the end of the pipe. Point A and F are exposed to atmosphere (they are not confined with a solid boundry).Gage pressure of point A and F are zero. (Source: Mott, R. L., Applied Fluid Mechanics, Prentice Hall, New Jersey) Pressure Head at points open to the atmosphere :When the fluid at reference point is exposed to the atmosphere, the pressure is zero and the pressure head term can be cancelled from Bernoulli’s equation. Velocity Head at points inside the tank or reservoir :The velocity head at the surface of the tank or reservoir is considered to be zero and it can be cancelled from Bernoulli’s equation. 7 Velocity Heads at points with the same c/s area:Points B,C, D and E are inside the pipes and has a uniform flow area. The velocity will be same througout the pipe. When the two points of reference for Bernoulli’s equation are both inside a pipe of the same size, the velocity head terms at both sides of the equation are equal and can be cancelled. Elevation Head at points on the same level:When two points of reference for Bernoulli’s equation are both at the same elevation, the elevation head terms, z1 and z2 are equal and can be cancelled. Example: (Example 6.10 from Mott, R. L., Page 164) A siphon draws fluid from a tank and delivers it through a nozzle at the end of the pipe. Calculate the volume flow rate through the siphon and the pressure at points B, C, D, and E. Diameter of siphon = 40 mm Diameter of nozzle = 25 mm Assume that there are no energy losses in the system. Reference points A and F (most convenient points) 2 2 vA PA vF PF z + + = z + + A 2g γ F 2g γ A is open to the atmosphere PA = 0 Velocity of point A is small, velocity head will be negligible. Elevation difference between A and F = (1.8+1.2) = 3.0 m. F is open to the atmosphere, PF = 0 zA − zF = 3.0 m zF − zA = −3.0 m 8 2 vF z = z + A F 2g VF = √(zA − zF) × 2g VF = √(3.0) × (2 × 9.81) VF = 7.67 m/s Q = A × V 25 mm 2 π = ( ) × 7.67 m/s Q = 3.77 × 10−3 m3/s 1000 m 4 c/s area of nozzle Determine the pressures at points B, C, D and E Write Bernoulli’s Equation between A and B 2 2 vA PA vB PB z + + = z + + A 2g γ B 2g γ zA = zB Q = AB × VB 40 mm 2 π 3.77 × 10−3 m3/s = ( ) × V 1000 m 4 B VB = 3.0 m/s 2 (3.0) PB 0 = + 2×(9.81) γ PB = −0.459 m 9.81 kN/m3 PB = −4.50 kPa Negative sign indicates the pressure is below atmospheric pressure. 9 Write Bernoulli’s Equation between A and C 2 2 vA PA vC PC z + + = z + + A 2g γ C 2g γ vC = vB since pipe diameter is same. 2 (3.0) PC 3.0 m = 4.2 m + + 2×(9.81) 9.81 kN/m3 PC = −16.27 kPa Write Bernoulli’s Equation between A and D 2 2 vA PA vD PD z + + = z + + A 2g γ D 2g γ zA = zD 2 3 PD 0 = + 2×(9.81) 9.81 kN/m3 PD = −4.50 kPa Write Bernoulli’s Equation between A and E 2 2 vA PA vE PE z + + = z + + A 2g γ E 2g γ 2 (3.0) PE 3.0 m + 0 + 0 = 0 + + 2×(9.81) 9.81 kN/m3 PE = 24.93 kPa Summary of the Results of Example 1. Velocity from nozzle and therefore the volume flow rate delivered by siphon depends on the elevation difference between the free surface of fluid and the outlet of nozzle. 2. The pressure at point B is below atmospheric pressure even though it is on the same level as point A, which is exposed to the atmosphere.