00:01
We're giving we have to start up the multiple so we're given h1 h1 is 3 ,455 kilojoues per kilogram and we have a v1 is 70 meters per second and h2 is 2 ,850 kilograms per mole kilograms for kg i mean kilojoules per kg so according according to the steady flow energy equation, our 8 of 1 is equal to h2 for kinetic energy k -e -2.
01:48
So now, if we rearrange this equation, you could say h -1 minus 8 -22, the form of kinetic energy, 1 -5b squared 2.
02:04
So v -2 squared, and this is per kilogram mass.
02:28
Now we can input of values, right? if we do that, we have 345 minus 2 ,850 times 10 to the third because it's converting to kilograms as equal to one -half v -sub -2 squared.
03:07
We could actually multiply this both sides by two to get rid of this fraction right here.
03:25
So that would be multiply both sides of the equation by two.
03:32
So this would be 390 ,000.
03:43
It's equal to v .2 squared.
03:48
Now, if we take the square width of both sides, we could find v.
03:53
So v sub 2 becomes 624 .49 meters to second.
04:16
For the second part of this problem, the reaction for the mass flow rate of the liquid.
04:29
So the area a sub 1, our initial area is 0 .1.
04:48
And the specific volume, we're told, is 0 .19.
04:57
So v .1 is 0 .19 meters per kilogram...