Need help on a physics problem please.?
A horizontal pipe of diameter 0.831 m has a smooth constriction to a section of diameter 0.4986 m. The density of oil flowing in the pipe is 821 kg/m3. If the pressure in the pipe is 8040 N/m2 and in the constricted section is 6030 N/m2, what is the rate at which oil is flowing? Answer in units of m3 /s.
- ?Lv 71 month ago
Bernoulli: p₀ + ρgh₀ + ½ρv₀² = p₁ + ρgh₁ + ½ρv₁²
and for a horizontal pipe, the h terms cancel.
For constant volume rate of flow,
v₀*A₀ = v₁*A₁
and so taking the wider section to be ₀,
v₀*(π/4)(0.831m)² = v₁*(π/4)(0.4986m)²
v₀ = v₁ * 0.36
8040Pa + ½*821kg/m³*(0.36v₁)² = 6030Pa + ½*821kg/m³*v₁²
solves to v₁ = 2.37 m/s
volume flow rate
Q = 2.37m/s * (π/4)(0.4986m)² = 0.863 m³/s
If you find this helpful, please select Favorite Answer. You get points too!