Question

From a waterfall, water is pouring down at the rate of \[160{\rm{ kg}}\] per second on the blades of a turbine. If the height of the fall be \[50{\rm{ m}}\], the power delivered to the turbine is approximately equal to (Take \[g = 9.8{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}\])
A. \[7.84 \times {10^4}{\rm{ W}}\]
B. \[7.84 \times {10^7}{\rm{ W}}\]
C. \[8400{\rm{ W}}\]
D. \[784{\rm{ kW}}\]

Answer 1

Hint: We know that the given turbine's power can be expressed as the product of the mass flow of water through its vanes, height from which water is falling, and acceleration due to gravity. We will use this expression to find the power of the turbine.

Complete step by step answer:
Given:
The mass flow rate of water on the turbine blades is \[\dot m = 160{\rm{ }}{{{\rm{kg}}} {\left/
 {\vphantom {{{\rm{kg}}} {\rm{s}}}} \right.
 } {\rm{s}}}\].
The height of the fall is \[h = 50{\rm{ m}}\].
The value of acceleration due to gravity is \[g = 9.8{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}\].

We have to find the value of turbine power due to the flow of water through its blades.

Let us write the expression for the power of the given turbine.
\[P = \dot mgh\]
On substituting \[160{\rm{ }}{{{\rm{kg}}} {\left/
 {\vphantom {{{\rm{kg}}} {\rm{s}}}} \right.
 } {\rm{s}}}\] for \[\dot m\], \[9.8{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}\] for g and \[50{\rm{ m}}\] for h in the above expression, we get:

$ P = \left( {160{\rm{ }}{{{\rm{kg}}} {\left/
 {\vphantom {{{\rm{kg}}} {\rm{s}}}} \right.
 } {\rm{s}}}} \right)\left( {9.8{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}} \right)\left( {50{\rm{ m}}} \right)\\$
 $P= 78400{\rm{ }}{{{\rm{kg}} \cdot {{\rm{m}}^2}} {\left/
 {\vphantom {{{\rm{kg}} \cdot {{\rm{m}}^2}} {{{\rm{s}}^3}}}} \right.
 } {{{\rm{s}}^3}}} $

We can rewrite the unit expression in terms of Joule per second.
$P = 78400{\rm{ }}{{{\rm{kg}} \cdot {{\rm{m}}^2}} {\left/
 {\vphantom {{{\rm{kg}} \cdot {{\rm{m}}^2}} {{{\rm{s}}^3}}}} \right.
 } {{{\rm{s}}^3}}} \times \left( {\dfrac{{\rm{J}}}{{{{{\rm{kg}} \cdot {{\rm{m}}^2}} {\left/
 {\vphantom {{{\rm{kg}} \cdot {{\rm{m}}^2}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}}}} \right)\\ $
$P = 78400{\rm{ }}{{\rm{J}} {\left/
 {\vphantom {{\rm{J}} {\rm{s}}}} \right.
 } {\rm{s}}}
$

We know that Joule per second gives us the power in terms of Watt so we can write the above expression as below:
$P = 78400{\rm{ }}{{\rm{J}} {\left/
 {\vphantom {{\rm{J}} {\rm{s}}}} \right.
 } {\rm{s}}} \times \left( {\dfrac{{\rm{W}}}{{{{\rm{J}} {\left/
 {\vphantom {{\rm{J}} {\rm{s}}}} \right.
 } {\rm{s}}}}}} \right)\\ $
$P= 78400{\rm{ W}} $

We can again convert the above expression so that it can match with any of the given options.
$P = \dfrac{{78400}}{{10000}}{\rm{ }} \times {\rm{1}}{{\rm{0}}^4}{\rm{ W}}\\ $
$P= 7.84 \times {\rm{1}}{{\rm{0}}^4}{\rm{ W}} $

Therefore, we can say that the power obtained from the given turbine due to the flow of water over its vanes is \[7.84 \times {\rm{1}}{{\rm{0}}^4}{\rm{ W}}\]

So, the correct answer is “Option A”.

Note:
We can note that Newton-metre gives us the unit of power, Joule. It would be an added advantage if we can remember the relationship between various power units to solve similar kinds of problems.