Question

a pump on the ground floor of a building can pump up water to fill a tank of volume 30cube in 15mmts . if the tank is 40 m above the ground and efficiency of the pump is 30 percentage . how much power is consumed by the pump ? ( density of water 10 cube kg / m cube , g= 9.8 m/s square .​

Answer 1

Answer:

39.2*10^-2 w

Explanation:

given:

height=40m

volume of water(v)=30 cm³=30*10^-6 m³

time(t)=15 min=15*60 s

density of water(d)=10³ kg/m³

g=9.8 m/s²

so,mass of water(m)=dv

=10³*30*10^-6 kg

=30*10^-3 kg

power output=work output/time

=mgh/t

=30*10^-3*9.8*40/15*60

=39.2*10^-3/3 w

efficiency %=power output/power input*100

power input=efficiency %*power output

=30*39.2*10^-3/3 w

=39.2*10^-2 w

Answer 2

$\huge{\bold{\underline{\underline{Correct question:-}}}}$

a pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ in 15minutes . if the tank is 40 m above the ground and efficiency of the pump is 30 percentage . how much power is consumed by the pump ? ( density of water 10³ kg / m cube , g= 9.8 m/s square .

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

$\large{ \rho_{water} = 10^3 \: Kg/m^3}$

$\large{ g = 9.8 \: m/s^2}$

$\large{ Volume \: of \: Tank = 30m^3}$

$\large{ Time = 15 \: minutes = 15 \times 60 = 900 \: Seconds}$

$\large{ Height = 40 \: m}$

$\huge{\bold{\underline{Explanation:-}}}$

Mass of the water is unknown,

So, for that ,

$\large{\bold{Density = \frac{Mass}{Volume}}}$

Now,

$\large{ \rho = \frac{M}{V}}$

Substituting the values,

$\large{ 10^3 = \frac{M}{30}}$

$\large{\boxed{M = 30 \times 10^3 \: Kg}}$

Now, Work done

$\large{\bold{W = mgh}}$

$\large{W = 30000 \times 9.8 \times 40}$

$\large{W = 30000 \times 392}$

$\large{\boxed{W = 11760000 \: J}}$

As we know power is rate of doing work.

Let the power delivered is P.

$\large{\bold{P = \frac{W}{T}}}$

Substituting the values,

$\large{P = \frac{11760000}{900}}$

$\large{\boxed{P = 1306.67 \: Watts}}$

Power delivered is 30%.

Now,

$\large{30 \% P' = P}$

$\large{30 \% P' = 1306.67}$

$\large{P' = \frac{1306.67}{30 \%}}$

$\large{P' = \frac{1306.67 \times 100}{30}}$

$\large{P' = 43.5555 \times 100}$

$\large{P' = 4355.55}$

$\huge{\boxed{\boxed{P' = 4.3555 \: KW}}}$

So, Power consumed by the pump is 4.3555 Kilo Watts.