A pipe of diameter 80mm is fitted with a branch section into which fit two pipes of equal diameter. If it is desired that the velocity remain constant throughout, determine the required diameter of each of the branch pipes.

Alice/Pam/mat -- pick a name and keep it please.

Pam

the area of the outlet pipes have to be 1/2 of the original area (law of continuity).

1/2 PI (40^2)=PI R^2
R= 40 sqrt1/2=40*.707 mm

24

To determine the required diameter of each of the branch pipes, we need to use the principle of conservation of mass. According to this principle, the mass flow rate of fluid remains constant throughout the pipe system.

Since the velocity needs to remain constant, we can assume that the mass flow rate is the same in the main pipe and the two branch pipes.

Let's denote the diameter of the branch pipes as D. The cross-sectional area of the main pipe is given by A_main = π * (80/2)^2, and the cross-sectional area of the branch pipes is given by A_branch = 2 * π * (D/2)^2.

Since the mass flow rate is conserved, we can equate the mass flow rate in the main pipe (ρ * V_main * A_main) with the sum of the mass flow rates in the branch pipes (2 * ρ * V_branch * A_branch). Here, ρ is the density of the fluid, V_main is the velocity in the main pipe, and V_branch is the velocity in the branch pipes.

Setting up the equation and canceling out common terms, we get:
V_main * A_main = 2 * V_branch * A_branch

Substituting the values of A_main and A_branch, we get:
V_main * π * (80/2)^2 = 2 * V_branch * 2 * π * (D/2)^2

Simplifying the equation further, we get:
V_main * (80/2)^2 = V_branch * D^2

Now, since the velocity needs to be constant throughout, we can assume that V_main = V_branch.

Therefore, the equation becomes:
(80/2)^2 = D^2

Solving for D, we get:
D^2 = (80/2)^2
D^2 = 40^2
D = 40 mm

Therefore, the required diameter of each of the branch pipes is 40 mm.