00:01
Hello students, to determine the mass flow rate of water through the pipe at position 1, we can use bernoulli's equation bernoulli's equation which states that which states that the sum of the pressure, kinetic energy and potential energy per unit volume of a liquid is constant along a streamline.
00:28
So at the position 1, the pressure is given as 3 atmosphere and the velocity and height are 0.
00:37
Velocity and height is equal to 0 and at the position 2, the pressure is 1 atmosphere and the height is given as 0 .75 m since there are no frictional loads, the kinetic energy per unit volume is same at both these positions.
00:54
Therefore, we can write p1 plus 1 by 2 rho v1 square plus rho into g into h1 is equal to p2 plus 1 by 2 rho into v2 square plus rho into g into h2 where p is the pressure, rho is the density of water, v is the velocity, g is the acceleration due to gravity and h1 and h2 are the height.
01:24
So since the diameter of the pipe changes from 8 cm to 3 cm from 8 cm to 3 cm at position 2, the velocity of water also changes.
01:38
So we can use the continuity equation which states that the mass flow rate of a fluid is constant along a streamline to relate the velocities at position 1 and 2.
01:50
That is a1v1 is equal to a2v2 where a is the cross -sectional area of the pipe.
02:02
The cross -sectional area of the pipe at position 1a1 is equal to pi into r square 0 .08 m by 2 square which is equal to 0 .005026 m square.
02:20
Now similarly we can find a2 is equal to pi into 0 .03 by 2 square which is equal to 0 .0007069 m square...