00:01
Right here we're told that the mass first of all we have to convert the mass flow rate from kilograms per minute to kilograms per second so we know the mass is equal to 80 so we have 80 kilograms per minute and we could convert that to seconds so we know one there's 60 seconds of one minute so you multiply this by one over 60 so now the mass would be equal to 1 .33 kilograms per second.
01:26
So now if we calculate the area of the cross section of the pipe, so we know mass is equal to the density, the density of water times the area, times v1, which v1 is the inlet's velocity.
02:08
So to calculate the area, we could solve this equation and isolate for the area, by dividing both sides by the density and the volume so pv over 1 so area is equal to our mass divided by the density times of volume so density over the volume over the velocity so now we can substitute our values into this equation and we have our areas equal to 1 .33 divided by 1000 times 1 .5 so the area is equal to 8 .89 times 10 to negative 4 so now if we apply bernercle's equations at point one and point two respectively we have p1 1 divided by pg so p1 of pg plus v1 squared divided by 2g plus z 1 that is equal to p 2 over g 2 p 2 p 2 over p 2 over p times g plus v sub 2 squared plus v sub 2 squared over 2 over 2 plus z sub 2.
05:35
A quick thing to remember, p .1 and p sub 2 are the pressures at point 1 and 0 .2, while v sub 1 and ov sub 2 are the inlet and exit velocities.
05:52
And z sub 1 and z sub 2 are the dantam hats.
05:58
Yg is the acceleration due to gravity.
06:05
So now we know what they are.
06:08
We can substitute our values into the respective equation.
06:16
So now we have piece of one divided by 1000 times 9 .81 plus 1 .5 squared divided by 2 times 9 .81.
07:07
Plus 0 that is equal to 0 plus 16 .5 squared divided by 2 times 9 .81 plus 0.
07:54
So now piece of 1 you can isolate piece of 1 right so piece of 1 will be equal to 1 ,000 times 9 .81 times 16 .5 squared minus 1 .5 squared divided by 2 times 9 .81...