
A pump is required to lift \[1000kg\] of water per minute from a well of depth \[10m\] and eject it with a speed of \[10m{s^{ - 1}}\]. The horsepower of the engine needed is: (assume \[g = 10m{s^{ - 2}}\])
A. $3.25$
B. $4.33$
C. $5.35$
D. $2.35$
Answer
493.8k+ views
Hint: The mechanical power of an electric motor is defined as the speed multiplied by the torque. Kilowatts \[\left( {kW} \right)\] or horsepower \[\left( {hp} \right)\] are common units of mechanical power, with one watt equals one joule per second or one Newton-Meter per second. The work done per unit of time is measured in horsepower.
Formula used:
\[P = \dfrac{{mgh + \dfrac{1}{2}m{v^2}}}{t}\]
Where,
\[P\] -power
\[m\] -mass of the object,
\[t\] -time
\[h\] -depth of the well
\[v\] -velocity,
\[g\] -acceleration due to gravity
Complete step by step solution:
We know the values from the question
\[m = 1000kg\]
\[t = 1\min = 60\sec \]
\[h = 10m\]
\[v = 10m/s\]
\[g = 10m/{s^2}\]
If we substitute these values into the above equation means we get,
\[P = \dfrac{{1000 \times 10 \times 10 + \dfrac{1}{2} \times 1000 \times 10 \times 10}}{{60}}\]
\[ \Rightarrow P = 2500W\]( $W = $ Watt)
We already know that,
\[1\] horsepower \[ = 746W\]
So, \[\dfrac{1}{{746}}\] horsepower \[ = 1W\]
\[ \Rightarrow 2500W = \dfrac{{2500}}{{746}}\] horsepower
\[ \Rightarrow 2500W = 3.25\] horsepower.
Additional information:
The horsepower is also represented as the ratio of the force and distance to the time. This ratio value also gives horsepower. Horsepower is classified as different types concerning the system. For example in the mechanical system, horsepower is also called mechanical horsepower. Similarly, electrical horsepower, hydraulic horsepower, boiler horsepower, etc., is there.
Note: The term "horsepower" refers to the amount of power produced by an engine. It's computed by multiplying the power required to move \[550\] pounds one foot in one second by \[33,000\] pounds one foot in one minute. The rate at which the work is completed is used to determine the power.
Formula used:
\[P = \dfrac{{mgh + \dfrac{1}{2}m{v^2}}}{t}\]
Where,
\[P\] -power
\[m\] -mass of the object,
\[t\] -time
\[h\] -depth of the well
\[v\] -velocity,
\[g\] -acceleration due to gravity
Complete step by step solution:
We know the values from the question
\[m = 1000kg\]
\[t = 1\min = 60\sec \]
\[h = 10m\]
\[v = 10m/s\]
\[g = 10m/{s^2}\]
If we substitute these values into the above equation means we get,
\[P = \dfrac{{1000 \times 10 \times 10 + \dfrac{1}{2} \times 1000 \times 10 \times 10}}{{60}}\]
\[ \Rightarrow P = 2500W\]( $W = $ Watt)
We already know that,
\[1\] horsepower \[ = 746W\]
So, \[\dfrac{1}{{746}}\] horsepower \[ = 1W\]
\[ \Rightarrow 2500W = \dfrac{{2500}}{{746}}\] horsepower
\[ \Rightarrow 2500W = 3.25\] horsepower.
Additional information:
The horsepower is also represented as the ratio of the force and distance to the time. This ratio value also gives horsepower. Horsepower is classified as different types concerning the system. For example in the mechanical system, horsepower is also called mechanical horsepower. Similarly, electrical horsepower, hydraulic horsepower, boiler horsepower, etc., is there.
Note: The term "horsepower" refers to the amount of power produced by an engine. It's computed by multiplying the power required to move \[550\] pounds one foot in one second by \[33,000\] pounds one foot in one minute. The rate at which the work is completed is used to determine the power.
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