Potential Flow, Circular Loop. Neglecting viscosity for an air glow find VP, the radial acceleration a,, and the circulation I (m²/s) (eqn. 6.89) about the circular loop given by r = 5m, for the following: (a) rigid body fluid rotation with Ve=5r; (b) the potential free vortex given by Y = 5 In r; (c) for which case (s), a and /or/ b, can the pressure difference between the origin and any other point be determined using the Bernoulli equation? Note that for both flows, V, = 0. Ans OM: (a) 10² Pa/m; -10² m/s²; 10² m²/s; (b) 10¹ Pa/m; -10¹ m/s²; -10¹ m²/s y.

Elements Of Electromagnetics
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**Potential Flow, Circular Loop.** Neglecting viscosity for an air flow, find \( VP \), the radial acceleration \( a_r \), and the circulation \( \Gamma \) (m²/s) about the circular loop given by \( r = 5 \) m, for the following:

(a) Rigid body fluid rotation with \( V_\theta = 5r \)

(b) The potential free vortex given by \( \Psi = 5 \ln r \)

(c) For which case(s), a and/or b, can the pressure difference between the origin and any other point be determined using the Bernoulli equation? Note that for both flows, \( V_r = 0 \).

**Answer (Order of Magnitude):**

(a) \( 10^2 \, \text{Pa/m}; -10^2 \, \text{m/s}^2; 10^2 \, \text{m}^2/\text{s} \) 

(b) \( 10^{-1} \, \text{Pa/m}; -10^{-1} \, \text{m/s}^2; -10^1 \, \text{m}^2/\text{s} \)

**Diagram Description:**

The diagram shows a circular loop with a radius specified as \( r \). The center of the circle is labeled as the origin, and a radial vector \( \vec{r} \) is drawn from the origin to a point on the circumference of the circle. The angle \( \theta \) is indicated between the positive x-axis and the vector \( \vec{r} \). The x and y axes are shown, with the y-axis oriented vertically and the x-axis horizontally.
Transcribed Image Text:**Potential Flow, Circular Loop.** Neglecting viscosity for an air flow, find \( VP \), the radial acceleration \( a_r \), and the circulation \( \Gamma \) (m²/s) about the circular loop given by \( r = 5 \) m, for the following: (a) Rigid body fluid rotation with \( V_\theta = 5r \) (b) The potential free vortex given by \( \Psi = 5 \ln r \) (c) For which case(s), a and/or b, can the pressure difference between the origin and any other point be determined using the Bernoulli equation? Note that for both flows, \( V_r = 0 \). **Answer (Order of Magnitude):** (a) \( 10^2 \, \text{Pa/m}; -10^2 \, \text{m/s}^2; 10^2 \, \text{m}^2/\text{s} \) (b) \( 10^{-1} \, \text{Pa/m}; -10^{-1} \, \text{m/s}^2; -10^1 \, \text{m}^2/\text{s} \) **Diagram Description:** The diagram shows a circular loop with a radius specified as \( r \). The center of the circle is labeled as the origin, and a radial vector \( \vec{r} \) is drawn from the origin to a point on the circumference of the circle. The angle \( \theta \) is indicated between the positive x-axis and the vector \( \vec{r} \). The x and y axes are shown, with the y-axis oriented vertically and the x-axis horizontally.
Cauchy's Equation of Motion:

\[ \rho \frac{DV}{Dt} = \rho g + \nabla \cdot T \]

- (like \( \rho a = \Sigma F \))

Newtonian viscous stress relations by the tensor relation:

\[ T_{ij} = -p\delta_{ij} + \mu \left[ \frac{\partial v_j}{\partial x_i} + \frac{\partial v_i}{\partial x_j} \right] \]

- where \(\delta_{ij}\) is the Kronecker delta function (1 for \(i = j\); 0 for \(i \neq j\)).
- \(T\) includes pressure and viscous surface forces.

Into Cauchy's equation, and assuming constant viscosity, to get the Navier-Stokes vector equations:

\[ \rho \frac{DV}{Dt} = \rho g - \nabla p + \mu \nabla^2 V \]

The acceleration \(DV/Dt = \partial V/\partial t + (V \cdot \nabla)V\), which for steady-state flow gives \(DV/Dt = (V \cdot \nabla)V\).

Because \((V \cdot \nabla)V\) is a non-linear term on the LHS of the N-S equation.

Reynolds Number:

\[ Re = \rho VL/\mu \]

- A measure of the ratio of inertial to viscous forces.
  
### Given Values:
- \(P_{atm} = 10^5 \, \text{Pa}\)
- \(\rho_{water} \approx 1000\)
- \(\rho_{air} \approx 1.2\)
- \(\mu_{water} \sim 10^{-3} \, \text{Ns/m}^2\)
- \(\mu_{air} \sim 2 \times 10^{-5} \, \text{Ns/m}^2\)
- \(g = 9.8 \, \text{m/s}^2\)
Transcribed Image Text:Cauchy's Equation of Motion: \[ \rho \frac{DV}{Dt} = \rho g + \nabla \cdot T \] - (like \( \rho a = \Sigma F \)) Newtonian viscous stress relations by the tensor relation: \[ T_{ij} = -p\delta_{ij} + \mu \left[ \frac{\partial v_j}{\partial x_i} + \frac{\partial v_i}{\partial x_j} \right] \] - where \(\delta_{ij}\) is the Kronecker delta function (1 for \(i = j\); 0 for \(i \neq j\)). - \(T\) includes pressure and viscous surface forces. Into Cauchy's equation, and assuming constant viscosity, to get the Navier-Stokes vector equations: \[ \rho \frac{DV}{Dt} = \rho g - \nabla p + \mu \nabla^2 V \] The acceleration \(DV/Dt = \partial V/\partial t + (V \cdot \nabla)V\), which for steady-state flow gives \(DV/Dt = (V \cdot \nabla)V\). Because \((V \cdot \nabla)V\) is a non-linear term on the LHS of the N-S equation. Reynolds Number: \[ Re = \rho VL/\mu \] - A measure of the ratio of inertial to viscous forces. ### Given Values: - \(P_{atm} = 10^5 \, \text{Pa}\) - \(\rho_{water} \approx 1000\) - \(\rho_{air} \approx 1.2\) - \(\mu_{water} \sim 10^{-3} \, \text{Ns/m}^2\) - \(\mu_{air} \sim 2 \times 10^{-5} \, \text{Ns/m}^2\) - \(g = 9.8 \, \text{m/s}^2\)
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