IPhO 2001 · Theory problem 3

Magnetohydrodynamic (MHD) Generator

Download PDFIPhO 2001

A horizontal rectangular plastic pipe of width \(w\) and height \(h\), which closes upon itself, is filled with mercury of resistivity \(\rho\). An overpressure \(P\) is produced by a turbine which drives this fluid with a constant speed \(v_0\). The two opposite vertical walls of a section of the pipe with length \(L\) are made of copper.

Figure 1

The motion of a real fluid is very complex. To simplify the situation we assume the following:

  • Although the fluid is viscous, its speed is uniform over the entire cross section.
  • The speed of the fluid is always proportional to the net external force acting upon it.
  • The fluid is incompressible.

These walls are electrically shorted externally and a uniform, magnetic field \(\mathbf B\) is applied vertically upward only in this section. The set up is illustrated in the figure above, with the unit vectors \(\hat{x}\), \(\hat{y}\), \(\hat{z}\) to be used in the solution.

a) Find the force acting on the fluid due to the magnetic field (in terms of \(L\), \(B\), \(h\), \(w\), \(\rho\) and the velocity \(v\)) [2.0 pts]

b) Derive an expression for the new speed \(v\) of the fluid (in terms of \(v_0\), \(P\), \(L\), \(B\) and \(\rho\)) after the magnetic field is applied. [3.0 pts]

c) Derive an expression for the additional power that must be supplied by the turbine to increase the speed to its original value \(v_0\). Copy your result onto the answer form. [2.0 pts]

d) Now the magnetic field is turned off and mercury is replaced by water flowing with speed \(v_0\). An electromagnetic wave with a single frequency is sent along the section with length \(L\) in the direction of the flow. The refractive index of water is \(n\), and \(v_0\ll c\). Derive an expression for the contribution of the fluid’s motion to the phase difference between the waves entering and leaving section \(L\). Copy your result onto the answer form. [3.0 pts]

Official English text of the problem.

Answers

  • a) The magnetic force opposes the flow: \[ \boxed{\mathbf F_B=-\frac{LB^2hw}{\rho}\,v\,\hat{x}.} \]

  • b) \[ \boxed{v=\frac{v_0}{1+\dfrac{LB^2v_0}{\rho P}}.} \]

  • c) At the restored speed \(v_0\), the extra power required because of the magnetic field, above the original field-free power, is \[ \boxed{\Delta\mathcal P_{\mathrm{mag}} =\frac{LB^2hw}{\rho}\,v_0^2.} \] If the reference is instead the slowed operating state in (b), the total turbine-power increase is \[ \boxed{\Delta\mathcal P_{(b)\to v_0} =\frac{LB^2hw}{\rho}\,v_0^2+Pwh(v_0-v).} \]

  • d) Let \(\omega=2\pi f\) be the laboratory angular frequency and define the phase lag across the section as \(\Phi=kL\). To first order in \(v_0/c\), \[ \boxed{\delta\Phi \equiv\Phi(v_0)-\Phi(0) =-\frac{\omega L v_0}{c^2}(n^2-1) =-\frac{2\pi f L v_0}{c^2}(n^2-1).} \] Thus the flow reduces the phase lag. The outgoing wave’s corresponding phase advance has the same magnitude and positive sign.

Solution

a) Magnetic force on the fluid

The fluid velocity and applied magnetic field are \[ \mathbf v=v\hat{x}, \qquad \mathbf B=B\hat{z}. \] Consequently, \[ \mathbf v\times\mathbf B=-vB\hat{y}. \] The magnetic force on moving charge carriers therefore drives conventional current through the mercury in the \(-\hat{y}\) direction, between the copper walls.

The walls are separated by \(w\), so the motional electromotive force has magnitude \[ \mathcal E=vBw. \] The current travels a distance \(w\) through mercury with conducting cross-sectional area \(Lh\). Its resistance is therefore \[ R=\rho\frac{w}{Lh}. \] Neglecting the resistance of the copper walls and shorting wires, the current is \[ I=\frac{\mathcal E}{R} =\frac{vBLh}{\rho}. \]

The force on this current in the mercury is \[ \mathbf F_B =I(-w\hat{y})\times(B\hat{z}) =-IwB\hat{x}. \] Hence \[ \boxed{\mathbf F_B=-\frac{LB^2hw}{\rho}\,v\,\hat{x}.} \] Its direction is opposite to the fluid velocity, as expected for magnetic braking.

b) Speed after applying the magnetic field

Let the pipe cross-sectional area be \[ A=wh. \] The turbine’s pressure rise supplies the driving force \[ F_P=PA. \]

The assumed proportionality between speed and applied driving force is equivalent, for steady motion, to a linear viscous resistance \(\kappa v\). Before the magnetic field is applied, \[ PA=\kappa v_0, \] so \[ \kappa=\frac{PA}{v_0}. \]

With the field applied and the same pressure rise \(P\), the steady force balance is \[ PA-\kappa v-\frac{LB^2A}{\rho}v=0. \] Substitution for \(\kappa\) gives \[ PA=\left(\frac{PA}{v_0}+\frac{LB^2A}{\rho}\right)v. \] Canceling \(A\) and solving, \[ \boxed{v=\frac{v_0}{1+\dfrac{LB^2v_0}{\rho P}}.} \]

The result approaches \(v_0\) as \(B\to0\), and the field always reduces the speed.

c) Additional turbine power

At the restored speed \(v_0\), the viscous resistance returns to its original value, \[ F_{\mathrm{visc}}=\kappa v_0=PA. \] The turbine must also overcome the magnetic force \[ |F_B(v_0)|=\frac{LB^2A}{\rho}v_0. \] Thus the required pressure rise is \[ P_{\mathrm{restored}} =P+\frac{LB^2v_0}{\rho}. \]

For incompressible flow, mechanical power supplied by a pressure rise is pressure rise times volume flow rate: \[ \mathcal P=P_{\mathrm{rise}}Av. \] Therefore the total power at the restored speed is \[ \mathcal P_{\mathrm{restored}} =\left(P+\frac{LB^2v_0}{\rho}\right)Av_0 =PAv_0+\frac{LB^2A}{\rho}v_0^2. \]

The original field-free power was \(\mathcal P_0=PAv_0\). The additional power needed at the same speed because of magnetic braking is consequently \[ \boxed{\Delta\mathcal P_{\mathrm{mag}} =\mathcal P_{\mathrm{restored}}-\mathcal P_0 =\frac{LB^2hw}{\rho}v_0^2.} \]

This equals the Joule heating in the mercury: \[ I^2R =\left(\frac{v_0BLh}{\rho}\right)^2 \frac{\rho w}{Lh} =\frac{LB^2hw}{\rho}v_0^2, \] which checks the energy balance.

There is a distinction if “additional power” is measured from the slowed state in (b). In that state, the turbine supplies only \[ \mathcal P_{\mathrm{slowed}}=PAv. \] The increase from that operating state is \[ \boxed{ \begin{aligned} \mathcal P_{\mathrm{restored}}-\mathcal P_{\mathrm{slowed}} &=\frac{LB^2hw}{\rho}v_0^2+Pwh(v_0-v)\\ &=\frac{LB^2hw}{\rho}v_0^2 \left(1+\frac{1}{1+\dfrac{LB^2v_0}{\rho P}}\right). \end{aligned}} \] The second term accounts for restoring the original viscous dissipation as the speed increases.

d) Contribution of the water’s motion to the phase difference

Treat the supplied refractive index \(n\) as constant, neglecting dispersion. In the water’s rest frame, the wave’s phase velocity is \[ u'=\frac{c}{n}. \]

The water moves along the propagation direction at speed \(v_0\). Relativistic velocity addition gives the laboratory phase velocity \[ u=\frac{u'+v_0}{1+\dfrac{u'v_0}{c^2}} =\frac{\dfrac{c}{n}+v_0}{1+\dfrac{v_0}{nc}}. \] Expanding to first order in \(v_0/c\), \[ u\simeq \left(\frac{c}{n}+v_0\right) \left(1-\frac{v_0}{nc}\right) =\frac{c}{n}+v_0\left(1-\frac{1}{n^2}\right). \] The coefficient \(1-1/n^2\) is the Fresnel drag coefficient.

For a laboratory monochromatic wave written as \[ E(x,t)=E_0\cos(\omega t-kx), \] the entrance-to-exit phase lag is \[ \Phi=kL=\frac{\omega L}{u}. \] The source fixes the laboratory frequency \(\omega\).

Expanding the reciprocal velocity, \[ \frac1u\simeq \frac nc-\frac{(n^2-1)v_0}{c^2}. \] Thus \[ \Phi(v_0)\simeq \frac{n\omega L}{c} -\frac{\omega L(n^2-1)v_0}{c^2}. \] For stationary water, \[ \Phi(0)=\frac{n\omega L}{c}. \] Subtracting, \[ \boxed{\delta\Phi =-\frac{\omega L(n^2-1)v_0}{c^2}.} \]

Equivalently, with laboratory vacuum wavelength \(\lambda_0=c/f\), \[ \boxed{\delta\Phi =-\frac{2\pi L v_0}{\lambda_0c}(n^2-1).} \] The negative sign denotes a reduction of the phase lag for propagation along the flow. If phase difference is defined as exit phase minus entrance phase, its change has the opposite sign.

The solution was checked against the official solution.