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# Part 2: Discharging

### Internal Resistance

Batteries transfer energy to electrons so that they ‘flow’ around a circuit, the Electro Motive Force (EMF) is the total amount of energy per coulomb of charge that a battery can supply and is measured in volts. The EMF of a lead-acid cell is provided by that chemical reactions described above (figures 1 and 2) and can be seen as the maximum possible voltage across the cell’s terminals (the open circuit voltage). The path taken when current passes through the lead-acid cell will have resistance. This internal resistance depends on the cell’s design, construction, age and condition. On discharge this internal resistance (Rc) causes the voltage measured across the cell’s terminals to be less than the EMF (E) of the cell (the voltage drop = I x Rc, figure 3a). Thus when a current (I) flows the terminal voltage (U) is given by: $U=E-IR_C$ Figure 3: The internal resistance of a battery and the voltage measure across the terminals: (a) current flowing through a load; (b) no current flowing.

Example
A cell has an internal resistance of 0.02&omega and an EMF of 2.2V. what is its terminal potential difference if it delivers (a) 1A, (b) 10A and (c) 50A? $\text{(a)}~U=E-IR_C=2.2-\left (1\times 0.02\right )=2.18V$ $\text{(b)}~U=E-IR_C=2.2-\left (10\times 0.02\right )=2V$ $\text{(c)}~U=E-IR_C=2.2-\left (50\times 0.02\right )=1.2V$

Note that if a high resistance voltage meter is used to measure the voltage across a battery’s terminals it will register the batteries EMF; as long as there is no current flowing through a load from the battery (figure 3b). If the terminal voltage is measure when a current is flowing through a load from the battery, the meter will register the EMF minus the voltage drop across the internal resistance (figure 3a).

When charging a cell the voltage applied across the terminals must be great enough to push the desired current against the cell’s EMF. Therefore the effective voltage across the internal resistance is the difference between the terminal voltage (in this case applied to the cell) and the the cells EMF. Therefore the current that flows is given by: $I=\dfrac{U-E}{R_C}$

and: $U=E+IR_C$

Example
A cell with an EMF of 2V and an internal resistance of 0.08Ω is to be charged at 5A. What terminal voltage must be applied? $U=E+IR_C=2+\left (5\times 0.08\right )$

### Cell and Battery Voltage

A well maintained cell should have a cell EMF of about 2.2V falling to about 2V when fully discharged. Once the internal resistance has been taken into account the terminal voltage (the potential difference across the cell terminals) of each cell will be about 2.1V, but this value will drop depending on how much current is being drawn. Six cells in series make up a twelve volt battery which when fully charged will have a terminal voltage of 12.6 to 12.8V. The EMF of lead-acid cells is dependent on chemistry although the actual terminal voltage differs depending on the battery design, this must be taken into account when using a voltmeter to determining the batteries state of charge.

### Battery Capacity

The capacity of a battery is usually expressed as a number of ampere-hours (Ah). One ampere-hour is the amount charge delivered when a current of one ampere is delivered for one hour. Since the capacity of lead-acid batteries depend on the rate at which they are discharged a discharge rate is also quoted. For example a battery with a 300Ah capacity when discharged over 10 hours (10 hour rate) can give (300÷10) 30A continuously however, it may only have a 250Ah capacity when discharged at the 5 hour rate which will occur if (250÷5) 50A are continually draw from it (figure 4). In short the more slowly you discharge a battery the greater its capacity, deep-cycle battery capacities are normally quoted for the 20 hour rate. Figure 4 shows a typical battery capacity versus discharge rate graph. Figure 4: The battery capacity vrs. discharge rate graph for a Surrette series 500 battery. The continuous current at various rates are also shown.

Capacities are sometimes expressed in terms of kilowatt-hours (kWh) which can be calculated from the ampere-hour rate using the following equation: $kWh=\dfrac{Ah\times \text{battery voltage}}{1000}$

Therefore a 12V battery with a capacity of 300Ah at the 10 hour rate will have a capacity of (12×300/1000) 3.6kWh.

The Ampere-hour is a measure of the amount of charge that a battery can deliver: one ampere is a flow of charge at the rate of one coulomb per second, therefore a number of amperes multiplied by a time (i.e. hours) gives us a quantity of charge. Similarly, the watt-hour is a measure of the amount of energy that a battery can deliver: one watt is the supply of energy at the rate of one joule per second, therefore a number of watts multiplied by a time gives us a quantity of energy.

The capacities of lead-acid batteries are very dependent on the temperature at which the battery is operating. The Capacity is normally quoted for a temperature of 25°C however, the capacity will reduce by about 50% at -25°C and will increase to about 100% at 45°C (figure 5).

### Battery Life

The life span of a deep-cycle battery is normally quoted in the number of cycles that it can be expected to perform, a cycle being a discharge followed by recharging. Deep cycle batteries should not be discharged by more than 60% of their capacity and the less you regularly discharge a battery the longer it will last. A battery in daily use and discharged by no more than 40% of its capacity should last for more than 3000 cycles and may not need replacing for up to 12 years. A battery that is frequently heavily discharged may last no longer than 2 years. Figure 6 shows the variation in battery life with the depth to which it is discharged.

### Battery Efficiency

Due to internal resistance and the fact that the charging voltage is greater than the discharge voltage, the energy returned by the battery upon discharge will be less than the energy used for recharging. Typically a lead-acid battery will be 80 to 90% efficient when considering ampere-hours (i.e. charge transferring efficiency). This figure assumes that the charging and discharging voltages are the same since: $\text{ampere-hour efficiency}=\dfrac{\text{discharged Ah}}{\text{charging Ah}}\times 100\%$

Capacities are quoted in terms of the number of ampere-hours that a full battery can discharge, but this will only be about 80% of the ampere-hours needed to completely recharge the same battery from empty. The charging voltage is the sum of the cell EMF and the internal voltage drop (due to internal resistance) whereas the discharge terminal voltage is their difference. A truer method is to calculate an energy efficiency for a battery using watt-hours: $\text{watt-hour efficiency}=\dfrac{\text{discharged watt-hours}}{\text{charging watt-hours}}\times 100\%$

The watt-hour efficiency is typically 65% for a lead-acid battery. Ampere-hour efficiencies are still useful for solar power sizing calculations since these often use ampere-hours when sizing the panel array needed to charge the battery bank but be careful.

Example
A discharged 12V battery is charged for 10 hours at 12A, the average charging terminal voltage being 15V. When connected to a load current of 10A for 9 hours at an average terminal voltage of 12V the battery is discharged. $\text{ampere-hour efficiency}=\dfrac{10\times 9}{10\times 12}\times 100\% =75\%$ $\text{watt-hour efficiency}=\dfrac{10\times 9\times 12}{10\times 12\times 15}\times 100\% =60v\%$

Note that because of the internal resistance of the battery the efficiency will depend on the charging and discharging rate.