Question

QUESTION:
An electric motor of 1hp is used to run water pump. The water pump takes 10 minutes to fill an overhead tank. The tank has a capacity of 800 liters and height of 15 m. Find the actual work done by the electric motor to fill the tank. Also find the efficiency of the system. (Density of water = 1000 kgm – 3 ) (Mass of 1 liter of water = 1 kg) (447600 J, 26.8 %)

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Answer 1

Given : An electric motor of 1hp is used to run water pump. The water pump takes 10 minutes to fill an overhead tank. The tank has a capacity of 800 liters and height of 15 m.

To Find : Find the work done by motor and Efficiency of the motor .

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SolutioN :

$\maltese$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ Power = \dfrac{Work \; Done}{Time} }}}}}$

• ${\underline{\boxed{\pmb{\sf{ Efficiency = \dfrac{W}{ W' } \times 100 }}}}}$

Where :

• W = Work Done
• W' = Required form of Output

$\maltese$ Calculating the Work Done :

• Converting the Values :

$\mapsto \; \; \sf{ 1 \; hp = \green{746 \; Watt} }$

$\mapsto \; \; \sf{ 10 \; min = \green{600 \; s} }$

• Let's Go :

${\dashrightarrow{\qquad{\sf{ Work \; Done = Power \times Time }}}} \\ \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Work \; Done = 746 \times 600 }}}} \\ \\ \\ \\ \ {\qquad \; \; {\dashrightarrow \; {\underline{\boxed{\pmb{\sf{ Work \; Done = 447600 \; Watt }}}}}}} \; {\red{\bigstar}}$

$\maltese$ Calculating the Efficiency :

• Converting the Values :

$\begin{gathered} \qquad \; \mapsto \; \sf { W' = mgh } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \mapsto \; \sf { W' = 800 \times 10 \times 15 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \mapsto \; \sf { W' = 800 \times 150 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \mapsto \; {\underline{\boxed{\pmb{\sf{ W' = 120000 \; Joules }}}}} \; {\orange{\bigstar}} \\ \\ \\ \end{gathered}$

• Let's Go :

${\dashrightarrow{\qquad{\sf{ Efficiency = \bigg\lgroup \dfrac{120000}{447600} \bigg\rgroup \times 100 }}}} \\ \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Efficiency = \bigg\lgroup \dfrac{120000}{4476\cancel{00}} \bigg\rgroup \times \cancel{100} }}}} \\ \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Efficiency = \bigg\lgroup \dfrac{120000}{4476} \bigg\rgroup }}}} \\ \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Efficiency = \bigg\lgroup \cancel\dfrac{120000}{4476} \bigg\rgroup }}}} \\ \\ \\ \\ \ {\qquad \; \; {\dashrightarrow \; {\underline{\boxed{\pmb{\sf{ Efficiency = 26.8 \; \% }}}}}}} \; {\purple{\bigstar}}$

$\therefore \;$ Work Done by the motor is 447600 Watt and Efficiency of the motor is 26.8 % .

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Answer 2

Power of pump, P = 1 hp = 746 W

time, t = 10 min = (10*60) s = 600 s

height, h = 15 m

mass, m = 800 kg (800 L; 1 L = 1 kg)

∴ Input Energy = P x t = 746 x 600 = 447600 J

Now, output energy = work done to lift water

∴ output energy or actual work done = mgh = 800 x 10 x 15 = 120000 J

So, now,
Efficiency = {Output Energy/Input Energy}*100

                = {120000/447600}*100

∴ Efficiency = 26.8 %