Two kilowatt motor  pump means that 2kJ wotk is done in $1$ sec 

$\text{Power}=\dfrac{\text{workdone}}{\text{time}}=\dfrac{\text{pressure }\times \text{ change in volume}}{\text{time}}=\dfrac {2000\times 10^{-6}}{1}= 0.02 \ W$

The power of pump = $\dfrac{d(mgh)}{dt}$

Electrical Power is measured in watts (W), kilo-watts (kW), mega-watts (MW), and giga-watts (GW). The power of a load can change quickly, especially if the load is turning on and off.

Electrical Energy is measured in kilowatt-hours (kWh), megawatt-hours, and sometimes (less commonly) in watt-hours or watt-seconds. Energy is the primary measurement used by the utility company to determine your bill each month (although demand and power factor are also sometimes used). Unlike power, energy does not change quickly, but instead accumulates gradually.

Electrical Energy is measured in kilowatt-hours (kWh). A unit is the primary measurement used by the utility company to determine your bill each month (although demand and power factor are also sometimes used). Unlike power, energy does not change quickly, but instead accumulates gradually.

Power, $P = \dfrac{w}{t} = \dfrac{mgh}{t} =\dfrac{mgh}{t\times 746} HP = 8.35HP$
$\therefore Power$  $= 9HP$ 

Using   $E = Pt$       $\implies$ Energy can be expressed in $Wh$

The kilowatt hour (symbolized kW⋅h as per SI) is a composite unit of energy equivalent to one kilowatt (1 kW) of power sustained for one hour. One watt is equal to 1 J/s. One kilowatt hour is 3.6 MJoules, which is the amount of energy converted if work is done at an average rate of one thousand watts for one hour.

1 kilowatt-hour is the amount of energy energy consumed by a 1000 watt electric appliance when it is operated for one hour.

$1\ kWh$ $ = 1000 \times 3600$ $ = 3.6\times 10^6$ $J$

Smallest commercial unit of energy is called Watt hour. It is defined as the amount of electric energy, which flows through a conductor in one hour, at a power of one watt.

$E = \dfrac{28.8 \times {10}^{3}}{3.6 \times {10}^{6}}$
   $= 8 \times {10}^{-3}  kwh$

The biggest commercial unit of energy is kilowatt hour $\left(kwh\right)$. It is also known as kilo-volt-ampere hour $\left(kVAh\right)$ and board of trade unit (BOT) or simply electric unit.

Commercial unit of electrical energy is  Kilowatt-hour  i.e  $kWh$ and 1 kWh is equal to 1 unit of energy.

As watt hour is the unit of energy, thus watt hour meter measures the amount of energy consumed for a given time period.

$1  kwh = 1  kw \times 1  h$
     $= 1000  w \times 1 \times 60 \times 60$
     $= 3.6 \times {10}^{6}  J$

Energy consumed by lamp in 2 hour     $E _l = 0.02\times 2 = 0.04$ kWh per day

$1$ kW  $ = 1000$ $\dfrac{J}{s}$            

We know  $1$ kWh $ = 3.6\times 10^6$ $J$

$1$ kW $ = 1000$ $\dfrac{J}{s}$

A unit is defined as $kWh$, means a $1000 joule/ sec$ is used for $1 hour$ ,

The kilowatt-hour (symbolized kWh) is a unit of energy equivalent to one kilowatt (1 kW) of power expended for one hour. The kilowatt-hour is commercially used as a billing unit for energy delivered to consumers by electric utilities.

KiloWatt is the unit of power and hour is the unit of time. Product of power and time equals Energy.

Let's consider a body is moved along a straight line by a machine delivering a constant power $P$. The distance moved by the body is $S$. 

Given that ,

Given that,

acceleration due to gravity $g = 10m/sec^2$

Energy produced by $1.0\ cm^3$ of fuel = $34\ kJ$

$Just$ $before$ $hitting$ $the$ $earth$ $θ$ $=$ $0°.$ $Hence,$ $the$ $power$ $exerted$ $by$ $the$ $gravitational$ $force$ $is$ $greatest$ $at$ $the$ $instant$ $just$ $before$ $the$ $body$ $hits$ $the$ $earth.$

Kilowatt-hours is the power generated in one hour=$\dfrac{volt\times current\times time( hour)}{1000}$

Time of operation $t = 90$ min $ = 1.5 $ hr

Power =F.V

$1$ kilowatt-hour(kWh) is a unit of energy. Normally, we want energy to be expressed in joules(J) and time in seconds(s).
Energy(kWh)$=$Power(kW)$\times$(h)$=1000$W$\times 3600$s$=1000$J/s$\times 3600$s$=3600000$J$=3.6\times 10^6$J.