A Faraday MHD generator (see Figure 13.9) is operating under the following conditions:

• Electrical resistivity of the ionized gas is ρ = 0.07 Ω m
• Magnetic field strength is B = 3.7 T
• Plasma velocity is u = 1000 m/s
• Electrode area is A = 1.4 m²
• Interelectrode gap in the channel is s = 0.9 m

Calculate the maximum power output and energy conversion efficiency of the generator

Question

A Faraday MHD generator (see Figure 13.9) is operating under the following conditions:

• Electrical resistivity of the ionized gas is ρ = 0.07 Ω m
• Magnetic field strength is B = 3.7 T
• Plasma velocity is u = 1000 m/s
• Electrode area is A = 1.4 m²
• Interelectrode gap in the channel is s = 0.9 m

Calculate the maximum power output and energy conversion efficiency of the generator

Accepted Answer

1. Open-circuit voltage

[latex]V_{\mathrm{oc}}=u\;B\;s=1000\;\mathrm{m/s} imes3.7\,T imes0.9\;\mathrm{m}=3330\,\mathrm{V}[/latex]

2. Short-circuit current

[latex]I_{\mathrm{sc}}\,=\,A\,u\,B/{\bf r}\,=\,1.4\,{ m}^{2} imes1000\,{ m}/{\bf s} imes3.7\,\mathrm{T}/0.07\,{\bf \Omega}\,{ m} )=74,000\,{ A}[/latex]

3. Current corresponding to the maximum power

[latex]I_{\mathrm{m}}={u B A/(2{\bf r})}={ 1000~\mathrm{m/s} imes3.7\,T imes1.4~\mathrm{m^{2}}/(2 imes0.07~\Omega~\mathrm{m})}=37,000~\mathrm{A}[/latex]

4. Voltage corresponding to the maximum power

[latex]V_{\mathrm{m}}={ u}\,B s-I_{\mathrm{m}}{\bf r}\,s/{ A}=1000\,{ m}/{ s} imes3.7\,{{T}} imes37,000\,{ A} imes0.07\,{\Omega}\,{ m}\,[/latex]

[latex] imes \,0.9\,{\mathrm{m/1.4}}\,{\mathrm{m^{2}}}\,=\,1665\,{\mathrm{V}}[/latex]

5. Maximum power output

[latex]P_{\mathrm{m}}=V_{\mathrm{m}}I_{\mathrm{m}}=1665 imes37000=61,600,000\,\mathrm{W}=61.6~\mathrm{MW}[/latex]

6. MHD generator conversion efficiency

[latex]{\bf h}_{\mathrm{MHD}}\,=\,V_{\mathrm{m}}/(u B)\,=\,1 665\,V(\,1000\,\mathrm{m}/{ \mathrm{s}} imes3.7\,\mathrm{T})\,=\,0.45[/latex]

Question 13.6: A Faraday MHD generator (see Figure 13.9) is operating under......

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