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2 years ago

Charging method .Constant current method

Engineering

1 answer:

zaharov [31]2 years ago

6 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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An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a rate of 1.5 ft3/s th

photoshop1234 [79]

Answer:

mechanical power used to overcome frictional effects in piping is 2.37 hp

Explanation:

given data

efficient pump = 80%

power input = 20 hp

rate = 1.5 ft³/s

free surface = 80 ft

solution

we use mechanical pumping power delivered to water is

${W_{u}}= \eta {W_{pump}}$ .............1

put here value

${W_{u}}$ = (0.80)(20)

${W_{u}}$ = 16 hp

and

now we get change in the total mechanical energy of water is equal to the change in its potential energy

$\Delta{E_{mech}} = {m} \Delta pe$ ..............2

$\Delta {E_{mech}} = {m} g \Delta z$

and that can be express as

$\Delta {E_{mech}} = \rho Q g \Delta z$ ..................3

$\Delta {E_{mech}} = (62.4lbm/ft^3)(1.5ft^3/s)(32.2ft/s^2)(80ft)[\frac{1lbf}{32.2lbm\cdot ft/s^2}][\frac{1hp}{550lbf \cdot ft/s}]$ ......4

solve it we get

$\Delta {E_{mech}} = 13.614$ hp

so here

due to frictional effects, mechanical power lost in piping

we get here

${W_{frict}} = {W_{u}}-\Delta {E_{mech}}$

put here value

${W_{frict}}$ = 16 -13.614

${W_{frict}}$ = 2.37 hp

so mechanical power used to overcome frictional effects in piping is 2.37 hp

4 0

3 years ago

A 360 kg/min stream of steam enters a turbine at 40 bar pressure and 100 degrees of superheat. The steam exits the turbine as a

Brrunno [24]

Answer:
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Explanation:
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2 years ago

Explain the LWD process why is it important in drilling operations?

barxatty [35]

Answer:

Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible.

Explanation:

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6 0

2 years ago

To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105

GenaCL600 [577]

Answer:

$62.14\ \text{miles}$

$6213727.37\ \text{miles}$

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = $10^5\ \text{mm}^{-2}$

Volume of the metal = $1000\ \text{mm}^3$

$10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}$

$1\ \text{mile}=1609.34\ \text{m}$

$\dfrac{10^5}{1609.34}=62.14\ \text{miles}$

The chain would extend $62.14\ \text{miles}$

Dislocation density = $10^{10}\ \text{mm}^{-2}$

Volume of the metal = $1000\ \text{mm}^3$

$10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}$

$\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}$

The chain would extend $6213727.37\ \text{miles}$

3 0

3 years ago

The angle of twist can be computed using the material’s shear modulus if and only if: (a)- The shear stress is still in the elas

ollegr [7]

Answer:

The angle of twist can be computed using the material’s shear modulus if and only if the shear stress is still in the elastic region

Explanation:

The shear modulus (G) is the ratio of shear stress to shear strain. Like the modulus of elasticity, the shear modulus is governed by Hooke’s Law: the relationship between shear stress and shear strain is proportional up to the proportional limit of the material. The angle of twist can be computed using the material’s shear modulus if and only if the shear stress is still in the elastic region.

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3 years ago

Charging method .Constant current method​

Charging method .Constant current method