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## derive equation for bernoulis theorem?

6 years ago

## Consider a fluid of negligible viscosity moving with laminar flow, as shown in Figure 1.

Let the velocity, pressure and area of the fluid column be v_{1}, P_{1}and A_{1}at Q and v_{2}, P_{2}and A_{2}at R. Let the volume bounded by Q and R move to S and T where QS = L_{1}, and RT = L_{2}. If the fluid is incompressible:

A_{1}L_{1}= A_{2}L_{2}

The work done by the pressure difference per unit volume = gain in k.e. per unit volume + gain in p.e. per unit volume. Now:

Work done = force x distance = p x volume

Net work done per unit volume = P_{1}- P_{2}

k.e. per unit volume = ½ mv^{2}= ½ V? v^{2}= ½?v^{2}(V = 1 for unit volume)

Therefore:

k.e. gained per unit volume = ½ ?(v_{2}^{2}- v_{1}^{2})

p.e. gained per unit volume = ?g(h_{2}– h_{1})

where h_{1}and h_{2}are the heights of Q and R above some reference level. Therefore:

P_{1}- P_{2}= ½ ?(v_{1}^{2}– v_{2}^{2}) + ?g(h_{2}- h_{1})

P_{1}+ ½ ?v_{1}^{2}+ ?gh_{1}= P_{2}+ ½ ?v_{2}^{2}+rgh_{2}thanks and regardssunil kraskIItian faculty

6 years ago

The Bernoulli Equation is a statement derived from conservation of energy and work-energy ideas that come from Newton`s Laws of Motion. Statement: It states that the total energy (pressure energy, potential energy and kinetic energy) of an incompressible and non–viscous fluid in steady flow through a pipe remains constant throughout the flow, provided there is no source or sink of the fluid along the length of the pipe. This statement is based on the assumption that there is no loss of energy due to friction. To prove Bernoulli’s theorem, we make the following assumptions: 1. The liquid is incompressible. 2. The liquid is non–viscous. 3. The flow is steady and the velocity of the liquid is less than the critical velocity for the liquid. Proof of Bernoulli’s Theorem: Imagine an incompressible and non–viscous liquid to be flowing through a pipe of varying cross–sectional area as shown in Fig. The liquid enters the pipe with a normal velocity v1 and at a height h1 above the reference level (earth’s surface). It leaves the pipe with a normal velocity v2 at the narrow end B of cross–sectional area a2 and at a height h2 above the earth’s surface. Fluid flow to the right. We examine a fluid section of mass m traveling to the right as shown in the schematic above. The net work done in moving the fluid is Net work done = work 2 - work 1. Eq.(1) where F denotes a force and an x a displacement. The second term picked up its negative sign because the force and displacement are in opposite directions. Pressure is the force exerted over the cross-sectional area, or P = F/A. Rewriting this as F = PA and substituting into Eq.(1) we find that Net work done = (PFA)_2 - (PFA)_1. Eq.(2) The displaced fluid volume V is the cross-sectional area A times the thickness x. This volume remains constant for an incompressible fluid, so Volume is constant for an incompressible fluid. Eq.(3) Using Eq.(3) in Eq.(2) we have dW = (P_1 - P_2) V. Eq.(4) Since work has been done, there has been a change in the mechanical energy of the fluid segment. This energy change is found with the help of the next diagram. Diagram for derivation of Bernoulli`s equation for an imcompressible unrestricted steady fluid flow. The energy change between the initial and final positions is given by Energy change = E_2 - E_1. Eq.(5) Here, the the kinetic energy K = mv²/2 where m is the fluid mass and v is the speed of the fluid. The potential energy U = mgh where g is the acceleration of gravity, andh is average fluid height. The work-energy theorem says that the net work done is equal to the change in the system energy. This can be written as Net work = energy change. Eq.(6) Substitution of Eq.(4) and Eq.(5) into Eq.(6) yields Expansion of dW = dE. Eq.(7) Dividing Eq.(7) by the fluid volume, V gives us Expansion of dW = dE. Eq.(8) where Density = mass / volume. Eq.(9) is the fluid mass density. To complete our derivation, we reorganize Eq.(8). Expansion of dW = dE. Eq.(10) Finally, note that Eq.(10) is true for any two positions. Therefore, P + mgh + mv^2/2 = Constant. Eq.(11) Equation (11) is commonly referred to as Bernoulli`s equation. Keep in mind that this expression was restricted to incompressible fluids and smooth fluid flows. Please approve my answer by clicking approved if you liked it.

6 years ago

Bernoulli theorem when a non-viscous,incompressible fluid flows steadily in a tube of varying cross-section, then the sum of the pressure energy,potential energy and the kinetic energy remains constant per unit volume at any point in its flow. Prof:consider the fluid moving with a velocity(v1),area(A2),a pressure of P1with density (d)at a height of h1 at a point A. Then also consider the fluid moving with velocity(V2),area(A2),a pressure of P2with a density(d)at a height of h2 at a point B. Therefore the resultant height h=h1-h2. (1)Work done by the gravity, Wg=m*g*h Wg=m*g*(h1-h2) where wg is the work done due to gravity. (2) Work done at A due to pressure P1 W=F*s WP1=A1*P1*V1*t P=F/A>F=P*A V=S/t>S=V*t Work done at B due to pressure P2 WP2=A2*P2*V2*t Total work done due to pressure difference, WP=P1*A1*V1*t-A2*P2*V2*t Total work done, W=Wg+WP W=mg(h1-h2)+P1*A1*V1*t-P2*A2*V2*t According to the work energy theorem, W=change in kinetic energy. W=1/2m(V2^2-V1^2) mg(h1-h2)+P1*A1*V1*t-P2*A2*V2*t=1/2m(V2^2-V1^2) we know that m=m1=m2=density(d)*V1*A1*t=density(d)*V2*A2*t. divide with m on both sides, mg(h1-h2)/m+P1*V1*A1*t/d*A1*V1*t-P2*A2*V2*t/d*A2*V2*t=1/2m/m*(V2^2-V1^2). g(h1-h2)+P1/d-P2/d=1/2(V2^2-V1^2) h1g+P1/d+1/2V1^2=h2g+P2/d+1/2V2^2 h*g+P/d+1/2V^2=constant d*g*h+P+1/2dV^2

6 years ago

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