Engr Help

Centrifugal Pumps

Example: (Cont.)

Step 2: Conservation of mass: mass flow rate at entrance equals mass flow rate at exit

dm/dt = ρ A₁ V_r1 = ρ A₂ V_r2 which yields: Q = 2π r₁ b₁ V_r1 = 2π r₂ b₂ V_r2

So the radial components of fluid velocity are: V_r1 = Q / 2π r₁ b₁ and V_r2 = Q / 2π r₂ b₂

V_r1 = Q / 2π r₁ b₁ = 28.6 ft/sec and V_r2 = Q / 2π r₂ b₂ = 19 ft/sec

Also r₁ ω = 66.66 ft/sec and r₂ω = 100.0 ft/sec

From the vector diagram on the left: tan β₁ = V_r1 / r₁ ω = 28.6 / 66.66 = 0.43

So β₁ = 23.3^o (result)

Step 3: Conservation of angular momentum T = ρQ [ r₂V_t2]

From the vector diagram on the left note that V_t2 = 0. From the vector diagram on the right:

V_t2 = r₂ ω ˗ W₂ cos β₂ = r₂ ω ˗ Wt₂

Note here V_r2 = W_r2 and tan β₂ = W_r2 / W_t2 So that W_t2 = W_r2 / tan β₂

V_t2 = r₂ ω ˗ W_r2 / tan β₂ = r₂ ω ˗ V_r2 / tan β₂ Since V_r2 = W_r2