2 Continuity EquationContinuity equation represents the law mass conservation.Unsteady flow the continuity equationFor steady state condition(Mass flow rate into the system)(Mass flow rate out of the system)-=Rate of change of storage.(Mass flow rate into the system)(Mass flow rate out of the system)-=
3 Example No. 1A jet of water discharges into an open tank, and water leaves the tank through an orifice in the bottom at a rate of m3/s. If the cross-sectional area of the jet is m2 where the velocity of water is 7 m/s, at what rate is water accumulating in (or evacuating from) the tank?
4 Example No. 1Situation: Jet of water (7 m/s at m2) entering tank and water leaving at m3/s through orifice.Find: Rate of accumulation (or evacuation) in tank (kg/s).Water density is 1000 kg/m3.
5 Example No. 1 Plan There is one inlet and one outlet. 1. Develop an equation for accumulation rate by applying the continuity equation.2. Analyze equation term by term.3. Calculate the accumulation rate.
6 mass through the control Example No. 11. Continuity equationThe accumulation rateof mass in the controlvolumeThe net outflow rate ofmass through the controlSurface+= 0Because there is only one inlet and outlet, the equation reduces to:
7 Example No. 12. Term-by-term analysis: The inlet mass flow rate is calculated using:
8 Example No. 12. Term-by-term analysis: The outlet mass flow rate is calculated using:
9 Example No. 1 3. Accumulation rate: Note that the result is positive so water is accumulating in the tank.
10 Example No. 2A river discharges into a reservoir at a rate of 400,000 ft3/s (cfs), and the outflow rate from the reservoir through the flow passages in the dam is 250,000 cfs. If the reservoir surface area is 40 mi2, what is the rate of rise of water in the reservoir?
11 Example No. 2Situation: Reservoir with 400,000 cfs entering and 250,000 cfs leaving. Area is 40 mi2.Find: Rate of water rise (ft/hr) in reservoirAssumptions:Water density is constantThe mass in the control volume is constant.Plan1. Apply the continuity equation.2. Analyze term by term.3. Evaluate rise rate.
12 Example No. 2SolutionApply Continuity equation:
13 Example No. 2 Solution 2. Term-by-term analysis: Mass in the control volume is constant, so dmcv/dt = 0.Inlet port 1 is river flow rateOutlets are reservoir surface and dam outlet,Substitution of terms back into continuity equation:
15 Example No. 3A tank has a hole in the bottom with a cross-sectional area of m2. The cross-sectional area of the tank is 0.1 m2. The velocity of the liquid flowing out the bottom hole is V=(2gh)0.5, where h is the height of the water surface in the tank above the outlet. At a certain time the surface level in the tank is 1 m and rising at the rate of 0.1 cm/s. The liquid is incompressible. Find the velocity of the liquid through the inlet.
17 Example No. 4A 10 cm jet of water issues from a 1 m diameter tank. Assume that the velocity in the jet is m/s where h is the elevation of the water surface above the outlet jet. How long will it take for the water surface in the tank to drop from h0 = 2 m to hf = 0.50 m?17
18 Example No. 4Situation: Water draining by a 10 cm jet from 1 m diameter tank.Find: Time (in seconds) to drain from depth of 2 m to 0.5 m.Plan1. Apply the continuity equation.2. Analyze term by term.3. Solve the equation for elapsed time.4. Calculate time to change levels.18