00:01
Hello students, in this portion first draw this distance let's say this distance with s.
00:10
Now first write what we have given in this portion.
00:14
We have given water flows through a horizontal pipe that is given as one gallon per second and the pipe diameter is given as four inch and second let's say b2 so the diameter for this direction will be given as 2 inch.
00:35
Now the density of the mercury is 847 per cube and density for water will be 62 .4 cube.
00:53
Now from bernoulli's equation from between the point 1 suppose this is point 1 and point 2 so we have the equation that is p1 upon rho g plus v1 square upon 2g where is the volume plus z1 will be equal to p2 upon rho g plus v2 square up 2g plus z2 so it is note that z1 is equal to z2 so for this we can write p1 minus p2 will be equal to rho w v2 square minus v1 square upon 2.
01:45
Now taking this equation 1 now we let the differential height of mercury manometer vh so differential height of mercury manometer vh and the distance between the center line and the mercury level that is shown in the figure mercury level in the tube will where mercury is raised by distance of s.
02:21
Now we can express the pressure difference so we get p1 plus rho gs plus h is equal to p2 plus rho gs plus rho for mercury gh.
02:43
Now for further solving this we get p1 minus p2 that is the difference in pressure will be equal to density for mercury minus density for water gh...