MARINE ELECTRO-TECHNIQUE 1

 Less prone to memory than NiCd  Shallow discharge better than deep Degrades after 200-300 deep cycles Need regular full discharge to avoid crystals  Self discharge 1.5-2.0 more than NiCd  Longer charge time than for NiCd To avoid overheating

 Features ◦ 50 cycles at 50% discharge ◦ No memory effect ◦ Shallow discharge better than deeper

 Chemistry Lead Sulfuric acid electrolyte  Features +Least expensive +Durable ◦ Low energy density ◦ Toxic

 Low self-discharge ◦ 40% in one year (three months for NiCd)  No memory  Cannot be stored when discharged  Limited number of full discharges  Danger of overheating during charging

 Ratings CCA: cold cranking amps (0F for 30 sec) RC: reserve capacity (minutes at 10.5v, 25amp)  Deep discharge batteries Used in golf carts, solar power systems 2-3x RC, 0.5-0.75 CCA of car batteries Several hundred cycles

 Chemistry Graphite (-), cobalt or manganese (+) Nonaqueous electrolyte  Features +40% more capacity than NiCd +Flat discharge (like NiCd) +Self-discharge 50% less than NiCd ◦ Expensive

 300 cycles  50% capacity at 500 cycles

 Chemistry Graphite (-), cobalt or manganese (+) Nonaqueous electrolyte  Features +Slim geometry, flexible shape, light weight +Potentially lower cost (but currently expensive) ◦ Lower energy density, fewer cycles than Li-ion

Type Capacity Density (mAh) (Wh/kg) Alkaline AA 2850 Rechargeable 1600 124 750 80 NiCd AA 41 NiMH AA 1100 Lithium ion 51 Lead acid 1200 100 2000 30

Type Voltage Peak Optimal Drain Drain Alkaline NiCd 1.5 0.5C < 0.2C 1.25 20C 1C Nickel metal 1.25 5C < 0.5C Lead acid 2 5C 0.2C Lithium ion 3.6 2C < 1C

Type Cycles Charge Discharge Cost per Alkaline (to 80%) time per month kWh 50 (50%) 3-10h 0.3% $95.00 NiCd 1500 1h 20% $7.50 NiMH 300-500 2-4h 30% $18.50 Li-ion 500-1000 2-4h 10% $24.00 Polymer 300-500 2-4h 10% Lead acid 200-2000 8-16h 5% $8.50

 The lead-acid battery is one of the most common batteries in use today. [Common examples of Lead acid batteries are car batteries, alarm system backup batteries]  The nickel-cadmium battery is being used with increasing (type of rechargeable battery [Nickel cadmium is the most commonly used battery for Low Earth Orbit (LEO) missions] 46

Lead-acid battery plate arrangement. 47

Nickel-cadmium cell. 48

 Lead-acid cell charge can be assessed with an instrument called a hydrometer, which measures the density of the electrolyte liquid or to check the specific gravity of the battery 49

(A) Connecting batteries in series: (i) increases voltage,(ii) does not increase overall amp-hour capacity.(iii) must have the same amp-hour rating (iv) increase resistance 50

(B) Connecting batteries in parallel: (i) increases total current capacity by decreasing total resistance, and it also (ii) increases overall amp-hour capacity.(iv) must have the same voltage rating. 51

 Batteries can be damaged by excessive cycling and overcharging.  Water-based electrolyte batteries are capable of generating explosive hydrogen gas, which must not be allowed to accumulate in an area. 52

CHAPTER 5 MAGNETISM AND ELECTROMAGNETIC

II. MAGNETISM AND ELECTROMAGNETIC The knowledge that students can obtain from this topic are: a. The concept of magnetism and electromagnetism with simple explanation. b. Laws and rule that related to magnetic and electric field. c. Understanding of magnetic hysteresis and hysteresis loop

2.1. Magnetic Field. The pointer of a compass is - from steel - call a permanent magnet The earth’s North Pole (or N) and the other end is called the S N south (or S). Fig. 2.1 Suspended permanent magnet

• Magnetism is a force field that acts on some materials (magnetic materials) but not on other materials (non magnetic materials). • Magnetic field around a bar magnet • Two “poles” dictated by direction of the field • Opposite poles attract (aligned magnetic field) • Same poles repel (opposing magnetic field)

• By convention, flux lines leave the north- seeking end (N) of a magnet and enter its south-seeking end (S).

The idea of lines of magnetic flux. Fig. 2.2. Use of steel filings for determining distribution of magnetic field

2.2. Magnetic field due to an electric current. IN S The needle was deflected clockwise or anticlockwise, depending upon the direction of the current. A phenomenon discovered by Oersted at Copenhagen in 1820. Fig. 2.3 Oersted’s experiment

The lines of magnetic flux Conductor Fig. 2.4 Magnetic flux due to current in a straight conductor

Approaching current Note the interesting convention for showing the direction of current flow in a conductor. Departing current Fig. 2.5. Current convention.

2.3. Faraday’s Law of Electromagnetic Induction. •First Law It states:- Whenever the magnetic flux linked with a circuit changes, an e.m.f. is always induced in it. Or Whenever a conductor cuts magnetic flux, an e.m.f. is induced in that conductor. •Second Law It states : - The magnitude the induced e.m.f. is equal to the rate of change of flux linkages.

Explanation : Suppose a coil has N turns and flux trough it changes from an initial value of Ф1 webers to the final value of Ф2 webers in time t second. Initial flux linkages = NФ1 Final flux linkages = NФ2 N 2 − N1 The induced e.m.f = e= t Or e = N d volt dt

2.4 Fleming’s Right-hand Rulec M O T I O N SN FIELD

I.e. dynamically induced e.m.f. A v A B B θv (a) (b) e = Blv volt e = Blv sin θ volt B= flux density l = length of conductor v = velocity θ = angle

2.5 Lenz's Law CAD B N NS (a) G CA D B S NS (b) G

i.e. Statically Induced E.M.F L = N or ; eL = L dI Volt I dt M = N 21 or M = N1 N 2 ; eM = M dI1 Volt I1 dt I / or A Where : L = Coefficient of self- inductance M= Coefficient of mutual inductance

2.7 Magnetic Hysterisis H= NI/l produced by solenoid- called magnetizing force The value of H can be increased or decreased by increasing or decreasing current through the coil. Fig. 2.11 The hysterisis curve

2.8 Hysteresis Loop

2.9 Variations in Hysteresis Curves

AREA OF HYSTERISIS LOOP l = mean of iron bar. A = its area of cross section N= No. of turns of wire of the solenoid. B = flux density

The power of rate of expenditure of energy = eI watt = H.l xN.A dB = A.l.H dB Watt N dt dt Energy spent in time ‘dt’ = A.l.H dB xdt = A.l.H.dB dt Where; e = N d Volt Ф = BA Volt dt H = NI/l e = NA dB dt

Total network done for one cycle of magnetization is W = A.l. H.dB Joule Hence H.dB = area of the loop i.e. the area between B/H curve and the B-axis. So, work done/cycle = A.l x (area of the loop) joule. Now A.l. = volume of material. So net work done/cycle/m3 = (loop area ) joule. Or Wh = (area of B/H loop) joule/m3/cycle.

Example 2.1 The field coils of a 6-pole dc generator each have 500 turns, are connected in series. When the field is excited, there is a magnetic flux of 0.02 Wb/pole. If the field circuit is opened in 0.02 second and the residual magnetism is 0.002 Wb/pole. Calculate the average voltage which is induced across the field terminals. In which direction is this voltage directed relative to the direction of the current.

Solution •Total number of turn, N = 6 x 500 = 3000 •Total initial flux = 6 x 0.02 = 0.12 Wb •Total residual flux = 6 x 0.002 = 0.012 Wb •Change in flux, d̀Φ = 0.12 – 0.012 = 0.108 Wb •Time of opening the circuit, dt = 0.02 second •Induced e.m.f =N d = 3000 x 0.108 dt 0.02 = 16,200 V

Example 2.2 A coil of resistance 100 Ω is placed in a magnetic field of 1 mWb. The coil has 100 turns and galvanometer of 400 Ω resistances is connected in series with it. Find the average e.m.f and the current if the coil is moved in 1/10th second from the given field to a field of 0.2 mWb.



Solution. Here dф = 1-0.2 = 0.8 mWb = 0.8 x 10-3 Wb dt = 1/10 = 0.1 second ; N = 100 e = 100 x 0.8 x 10-3 /0.1 = 0.8 V Total circuit resistance = 100 + 400 = 500 Ω Current induced = 0.8 / 500 = 1.6 x 10-3 A = 1.6 mA

Example 2.3 A conductor of length 1 meter moves at right angles to a uniform magnetic field of flux density 1.5 Wb/m2 with a velocity of 50 meters /second . Calculate the e.m.f induced in it. Find also the value of induced e.m.f when the conductor moves at an angle of 30o to the direction of the field.

1 meter 50 m /s 30o 1.5 Wb/m2

Solution Here B = 1.5 Wb/m2 ; l = 100 cm = 1 m ; v = 50 m/s Now e = B.l.v = 1.5 x 1 x 50 = 75 Volt In the second case θ = 30o sin 30o = 0.5  e = 75 x 0.5 = 37.5 Volt

Example 2.4 A conducting rod AB makes contact with metal rails AD and BC which are 50 cm apart in uniform magnetic field B = 1.0 Wb/m2 perpendicular to the plane ABCD. The total resistance (assumed constant) of the circuit ABCD is 0.4 Ω a).What is the direction and magnitude of the e.m.f induced in the rod when it is moved to the left with velocity of 8 m/s ? b). What force is required to keep the rod in motion? c). Compare the rate at which mechanical work is done by the force F with the rate of development of electric power in the circuit.

Solution a) Since AB moves to the left, direction of the induced current as found by applying Fleming’s left hand rule, is from A to B. Magnitude of the induced e.m.f is given by. e = B.l.v Volt = 1 x 0.5 x 8 = 4 Volt. b) Current through AB = 4 / 0.4 10 A. Force in AB i.e. F = B.I.l = 1 x 10 x 0.5 = 5 N. The direction of this force, as found by applying Fleming left hand rule, is to the right. c) Rate of doing mechanical work = F x v = 5 x 8 = 40 j/s or W Electric power produced = e. I = 4 x 10 = 40 W From the above, it is obvious that the mechanical work done in moving the conductor against force F is converted into electric energy.

Example 2.5 In a 4-pole dynamo, the flux per pole is 15 mWb. Calculate the average e.m.f induced in one of the armature conductors if armature is driven at 600 rpm.