Energy Equation 2 | u | 2 + , we have Recall that for H = p + 1 - - PowerPoint PPT Presentation

Energy Equation

Recall that for H = p

ρ + 1 2|u|2 + χ, we have

∂u ∂t +

• (∇ × u) × u
• = −∇H.

Taking scalar product with u and integrating over a fixed volume V ,

• V

∂u ∂t · u +

• (∇ × u) × u
• · u
• =0

dV = −

• V

(∇H) · udV , to obtain

• V

1 2 ∂ ∂t |u|2dV = −

• V

(∇H) · udV. As V is fixed (i.e., does not depend on t), we have 1 2 d dt

• V

|u|2dV = −

• V

(∇H) · udV .

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Energy Equation

But, ∇ · (Hu) = H∇ · u + u · ∇H; and also H∇ · u = 0 as ∇ · u = 0 due to incompressibility. Thus 1 2 d dt

• V

|u|2dV = −

• V

∇ · (Hu)dV , and using the divergence theorem we obtain 1 2 d dt

• V

|u|2dV =

• S

Hu · ndS. Multiplying through by ρ, we deduce the energy equation 1 2 d dt

• V

ρ|u|2dV =

• S

(p + 1 2ρu2 + ρχ)u · ndS.

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Vorticity

We define the vorticity ω ω = ∇ × u. BIG IDEA: Vorticity measures the rotation of the flow at each point. Note: Recall the definition of curl: vorticity is the curl of the velocity! A flow is called irrotational if 0 = ω = ∇ × u

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Bernoulli Equations (again)

Recall the momentum equation (in curl form) ∂u ∂t + (∇ × u) × u = −∇H Suppose that the flow is steady and irrotational hence ∂u ∂t = 0 and ω = ∇ × u = 0 giving ∇H = 0 H is constant throughout the whole flow field if the fluid is in steady and irrotational flow.

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Vorticity Equations

The momentum equation reads ∂u ∂t + (∇ × u)

• =ω

×u = −∇H Taking curl, we obtain ∂ω ∂t + ∇ × (ω × u) = ∇ × ∇H= 0 Using the identity [Q1 Problem Sheet 9] ∇ × (ω × u) = (u · ∇)ω − (ω · ∇)u + ω(∇ · u) − u(∇ · ω) Using the identity ∇ × (ω × u) = (u · ∇)ω − (ω · ∇)u + ω(∇ · u)

• =0 incompressible

− u(∇ · ω)

• =0 ∇·∇×u=0

we obtain ∂ω ∂t + (u · ∇)ω = (ω · ∇)u

• r

dω dt = (ω · ∇)u Vorticity equations

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Vorticity Equations

Vorticity equations dω dt = (ω · ∇)u BIG IDEA: Pressure is eliminated! Indeed, the vorticity equation involves u and ω only; these are related by ω = ∇ × u.

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