Magnetohydrodynamic Propulsion for Sea Vehicles 



Suppose we now use Eq. (16) to estimate the magnetic field necessary for 

 the described propulsion system. Some of the parameters will be taken from 

 the paper "Prospects for the Electromagnetic Submarine," by S. Way and C. 

 Devlin, presented at the AIAA 3rd Propulsion Joint Specialist Conference in 

 Washington, D.C., in July, 1967 (AIAA paper 67-432). The system they de- 

 signed and tested is reported there and may be referred to for actual design 

 details. Their measured value of o- for sea water was a - 4.5 mho/m or 

 a = .045 mho/cm. A typical drag coefficient will be taken as C-q = .004. In 

 Eq. (16) if we wish B in gauss, then pV^ must be expressed in dynes/cm^ and 

 rms = lO'^crWh^ if cr is in mhos/cm, V in cm/sec, and i> in cm. Putting 

 knots and S in meters for convenience (1 knot = 51.4 cm/sec), 



V in 



B = 1.6 



104 



1 - 77 



Se 



Ae 



2Cr.S, 



1 + 



D^W 



Ae 



- 1 



1 + 



1 + 



Ae 



(17) 



where B^ gauss, v-^ knots, cr-^ mho/cm, and h^ meters. 



Consider a typical case. For a submarine shape, the wetted area, if the 

 length is L and its maximum diameter d, is about S^ = ttDL . Suppose Ae , the 

 exit area of the propulsion unit, is about 1/5 {nD^/^). Then C^s^^/Ae = 



CpT^DL / 1/57TD2/4 =(20)CnL/D. If L/D = 10 and Cj. = .004, then CnS^./Ae 



0.8. 



In this case, y l + 2CpS^/Ae 

 efficiency 77 p = 86.5%, and 



'2.6= 1.613, so Ue/v= 1.31 and the propulsive 



B ^ 1.18 



10^ 



ah 1 



(18) 



In Eq. (18) the term h^/Ae is the area of the propulsive duct divided by the 

 exit area. If this term is small, B is small, because the velocity in the duct 

 gets larger, increasing the fluid coupling with the electromagnetic fields. The 

 fluid attains its maximum value of velocity there, however, and care must be 

 taken to avoid cavitation. For our purposes here, assume s^ = 1/2 Ae. Also 

 assume the propulsive duct length x is half the length of the vessel, so the elec- 

 trode area might be bk = b(L/2). Then, if a = .045 mho/cm. 



B = 5.56 



10^ 



1 - 77 



(19) 



where B-^- gauss, L~ meters, and v-^ knots. 



For a submarine tanker with l about 200 meters and V = 20 knots and 

 with 77 = 0.5, a field of 1.76 x 10'' or 17,600 gauss is required. This is not a 

 difficult field to produce in small volumes, but generally does require heavy 

 equipment. Of course, with superconducting magnets the prospects are much 

 better, but the engineering details are indeed challenging. 



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