Answer:
Outlet Velocity = 192 m/s
Outlet Pressure = 510 kPa
Explanation:
Givens:
Inlet Velocity, V₁ = 12 m/s
Inlet Pressure, P₁ = 600 kPa = 600,000 Pa
Inlet Radius r₁ = 0.5 cm
Outlet Velocity, V₂ = not given (we are asked to find this)
Outlet Pressure, P₂ = not given (we are asked to find this)
Outlet Radius, r₂ = 0.5 cm
From these, we can find the following:
Inlet Area, A₁ = π (r₁)² = π(2)² = 4π cm²
Outlet Area, A₂ = π (r₂)² = π(0.5)² = 0.25π cm²
<u>Part A :</u>
Assuming that water is incompressible, we can reason that within the same given time, the amount of volume of water entering the inlet must equal the volume of water exiting the outlet. Hence by the continuity equation (i.e. conservation of mass)
Inlet Volume flow rate = Outlet Volume flow rate
(recall that Volume flow rate in a pipe is given by Velocity x Cross Section Area), Hence the equation becomes
V₁ x A₁ = V₂ x A₂ (substituting the values that we know from above)
12 x 4π = V₂ x 0.25π (we don't have to change all to SI units because the conversion factors on the left will cancel out the conversion factors on the right).
V₂ = (12 x 4π) / (0.25π)
V₂ = 192 m/s (Answer)
<u>Part B:</u>
For Part B, if we assume a closed ideal system (control volume method), we can simply apply the energy equation (i.e Bernoulli's equation)
P₁ + (1/2)ρV₁ + ρgh₁ = P₂ + (1/2)ρV₂ + ρgh₂
Because the pipe is horizontal, there is no difference between h₁ and h₂, hence we can neglet this term:
P₁ + (1/2)ρV₁ = P₂ + (1/2)ρV₂ (rearranging)
P₂ = P₁ + (1/2)ρV₁ - (1/2)ρV₂
= P₁ + (1/2)ρ (V₁-V₂)
Assuming that the density of water is approx, ρ = 1000 kg/m³
P₂ = 600,000 + (1/2)(1000) (12-192)
= 600,000 + ( -90,000)
= 510,000 Pa
= 510 kPa (Answer)
