fluid flowcontinuity equation

1204 Fluid Flow

The Continuity Equation

The mass of a moving fluid doesn't change as it flows. This leads to an important relationshipt called the continuity equation.
In steady flow the total mass in the tube is constant, so dm1=dm2dm_1 = dm_2

dm1=dm2ρA1v1dt=ρA2v2dt or A1v1=A2v2\begin{aligned} dm_1 &= dm_2\\ \rho A_1v_1 dt &= \rho A_2v_2 dt \quad \text{ or }\\ A_1 v_1 &= A_2 v_2 \end{aligned}

The product AvAv is the volume flow rate dVdt\dfrac{dV}{dt}, the rate at which volume crosses a section of the tube:

dVdt=Av\begin{aligned} \frac{dV}{dt} &= Av \end{aligned}

AA - Cross-sectional area of flow tube
vv - Speed of flow

Examples

(12.6) Incompressible oil of density 850kg/m3850 kg/m^3 is pumped through a cylindrical pipe at a rate of 9.5 liters per second.
(a) The first section of the pipe has a diameter of 8.0 cm. What is the flow speed of the oil? What is the mass flow rate?
(b) The second section of the pipe has a diameter of 4.0 cm. What are the flow speed and mass flow rate in that section?

Solution

Let the volume flow rate be dVdt=0.0095m3/s\dfrac{dV}{dt} = 0.0095 m^3/s, and the mass flow rate be dmdt\dfrac{dm}{dt}. (Part a and part b have the same mass flow rate)

dmdt=ρ⋅dVdt=850⋅0.0095=8.075kg\begin{aligned} \frac{dm}{dt} &= \rho \cdot \frac{dV}{dt}\\ &= 850 \cdot 0.0095 = 8.075kg \end{aligned}

According to the continuity equation, we have

dVdt=Av=π⋅(d2)2v⇒v=dVdt⋅4πd2va=0.0095⋅4π⋅0.082=1.89m/svb=0.0095⋅4π⋅0.042=7.56m/s\begin{aligned} \frac{dV}{dt} &= Av = \pi \cdot (\frac{d}{2})^2 v\\ \To v &= \frac{dV}{dt} \cdot \frac{4}{\pi d^2}\\ v_a &= 0.0095 \cdot \frac{4}{\pi \cdot 0.08^2} = 1.89 m/s\\ v_b &= 0.0095 \cdot \frac{4}{\pi \cdot 0.04^2} = 7.56 m/s \end{aligned}

Exercises

38

(12.38) Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.070m20.070 m^2, and the magnitude of the fluid velocity is 3.503.50 m/s. (a) What is the fluid speed at points in the pipe where the cross-sectional area is (a) 0.105m20.105 m^2 and (b) 0.047m2.047 m^2? (c) Calculate the volume of water discharged from the open end of the pipe in 1.00 hour.

Solution

We calculate the volume flow rate dVdt\dfrac{dV}{dt} first

dVdt=Av=0.07⋅3.5=0.245m3/sva=dVdt0.105=2.3m/svb=dVdt0.047=5.2m/sVtotal=dVdt⋅3600=882m3\begin{aligned} \frac{dV}{dt} &= Av = 0.07 \cdot 3.5 = 0.245 m^3/s\\ v_a &= \frac{\frac{dV}{dt}}{0.105} = 2.3 m/s\\ v_b &= \frac{\frac{dV}{dt}}{0.047} = 5.2 m/s\\ V_{total} &= \frac{dV}{dt} \cdot 3600 = 882 m^3 \end{aligned}