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3 years ago

Charging method .Constant current method

Engineering

1 answer:

mina [271]3 years ago

4 0

Answer:

There are three common methods of charging a battery; constant voltage, constant current and a combination of constant voltage/constant current with or without a smart charging circuit.

Constant voltage allows the full current of the charger to flow into the battery until the power supply reaches its pre-set voltage. The current will then taper down to a minimum value once that voltage level is reached. The battery can be left connected to the charger until ready for use and will remain at that “float voltage”, trickle charging to compensate for normal battery self-discharge.

Constant current is a simple form of charging batteries, with the current level set at approximately 10% of the maximum battery rating. Charge times are relatively long with the disadvantage that the battery may overheat if it is over-charged, leading to premature battery replacement. This method is suitable for Ni-MH type of batteries. The battery must be disconnected, or a timer function used once charged.

Constant voltage / constant current (CVCC) is a combination of the above two methods. The charger limits the amount of current to a pre-set level until the battery reaches a pre-set voltage level. The current then reduces as the battery becomes fully charged. The lead acid battery uses the constant current constant voltage (CC/CV) charge method. A regulated current raises the terminal voltage until the upper charge voltage limit is reached, at which point the current drops due to saturation.

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Vsevolod [243]

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<h2>The Invention of the Internal Combustion Engine (ICE)</h2>

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3 years ago

A certain part of the cast iron piping of a water distribution system involves a parallel section. Both parallel pipes have a di

Bezzdna [24]

Answer :

Explanation:

Given :

Length of pipe A $L_{A} = 1500$ m

Length of pipe B $L_{B} = 2500$ m

Flow rate through pipe A $Q_{A} = 0.4 \frac{m^{3} }{s}$

Diameter of pipe $D = 30 \times 10^{-2}$ m

Velocity from pipe A,

$V _{A} = \frac{Q_{A} }{A}$

$V _{A} = \frac{0.4 \times 4 }{\pi ( 30 \times 10^{-2} )^{2} }$

$V_{A} = 5.66$ $\frac{m}{s}$

Here, head loss is same because height is same.

$h_{a} = h_{b}$

$L_{A} V_{A} ^{2} = L_{B} V_{B} ^{2}$

$V_{B} = \sqrt{\frac{1500}{2500}} (5.66)$

$V_{B} = 4.38$ $\frac{m}{s}$

Now rate of flow from pipe B is,

$Q_{B} = V_{B} A$

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3 years ago

Using the degree day method calculate the annual kwh use in springfiled il with a heat loss of 12kwh an inddor twmperature of 7

vladimir2022 [97]

Answer:

The correct solution is "21024 KWh/degree day".

Explanation:

The given query is incomplete. Below is the attachment of complete query is provided.

The given values are:

Indoor design temperature:

= 70°F

Now,

According to the question,

The heat loss annually will be:

= $12\times 24\times 365$

= $105120 \ KW$

Degree days will be:

= $75-65$

= $5$

Hence,

Annual KWh use will be:

= $\frac{Heat \ loss \ annually}{degree \ days}$

On substituting the values, we get

= $\frac{105120}{5}$

= $21024 \ KWh/degree \ day$

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3 years ago

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vazorg [7]

Answer:

Hostile

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3 years ago

Charging method .Constant current method​

Charging method .Constant current method