Answer :
The velocity at point 2 is (last two digits of student ID +30) / 303.78 m/s.
Diameter of pipe at end point 1= 1.2m Diameter of pipe at end point 2= 1.1m Velocity at end point 1= (x+30) mm/h = (last two digits of student ID +30) mm/h Now we need to find the velocity at point 2. We can use the principle of conservation of mass to solve the problem. According to this principle, the mass of fluid entering a section of the pipe should be equal to the mass leaving the section of the pipe. Assuming the flow of fluid is steady, the volume of fluid entering the pipe in unit time should be equal to the volume of fluid leaving the pipe in unit time. Mass flow rate at point 1= Mass flow rate at point 2∴ ρ1A1V1 = ρ2A2V2 where, ρ is the density of the fluid, A is the cross-sectional area of the pipe and V is the velocity of the fluid. Substituting the values in the above equation, ρ1= ρ2 (density of fluid is constant)A1V1= A2V2V2= A1V1 / A2We know that, Diameter of pipe = 2× Radius of pipe A = πr²Thus, A ∝ d²∴ A1 / A2 = (d1/d2)²So,V2= (A1V1) / A2 = (πr1²× V1) / (πr2²)= (r1²× V1) / r2²= [(d1/2)² × V1] / [(d2/2)²]= (V1d1²) / (V2d2²)On substituting the given values,V2 = (V1 × 1.44) / 1.21= (last two digits of student ID +30) mm/h × 1.19 / 3600= [last two digits of student ID +30) / 303.78] m/s Therefore, the velocity at point 2 is (last two digits of student ID +30) / 303.78 m/s: Given, Diameter of pipe at end point 1= 1.2m Diameter of pipe at end point 2= 1.1mVelocity at end point 1= (x+30) mm/h = (last two digits of student ID +30) mm/h We need to find the velocity at point 2 using the principle of conservation of mass that states that the mass of fluid entering a section of the pipe should be equal to the mass leaving the section of the pipe. Substituting the values in the equation,ρ1A1V1 = ρ2A2V2 where ρ is the density of the fluid, A is the cross-sectional area of the pipe and V is the velocity of the fluid, we getV2 = (V1 × 1.44) / 1.21= (last two digits of student ID +30) mm/h × 1.19 / 3600= [last two digits of student ID +30) / 303.78] m/s. To find the velocity at point 2, we can use the principle of conservation of mass. This principle states that the mass of fluid entering a section of the pipe should be equal to the mass leaving the section of the pipe. Assuming the flow of fluid is steady, the volume of fluid entering the pipe in unit time should be equal to the volume of fluid leaving the pipe in unit time. Hence, we can use the following equation to find the velocity at point 2:ρ1A1V1 = ρ2A2V2where, ρ is the density of the fluid, A is the cross-sectional area of the pipe and V is the velocity of the fluid. Substituting the given values in the above equation, we get:V2 = (V1 × 1.44) / 1.21= (last two digits of student ID +30) mm/h × 1.19 / 3600= [last two digits of student ID +30) / 303.78] m/s. Therefore, the velocity at point 2 is (last two digits of student ID +30) / 303.78 m/s.
The velocity of the fluid at point 2 can be found using the principle of conservation of mass. By substituting the given values in the equation, we can get the value of the velocity at point 2.
To know more about velocity visit:
brainly.com/question/18084516
#SPJ11
