1. Reynolds Transport Theorem We need to relate time derivative of a property of a system to rate of change of that property within a certain region (C.V.) If B is any property of the fluid like; mass, energy, momentum. Intensive value for a small portion of the fluid is The total amount of B in the control volume is where; Reynolds’ Transport Theorem wants to relate Now let B = m We get conservation of mass principle [Continuity Equation]
Prof. Dr. MOHSEN OSMAN

2. If the control volume has only a number of one dimensional inlets and outlets, we can write For steady flow within the control volume and conservation of mass law reduces to: → → Or for one-dimensional steady flow Special Case The conservation of mass law for incompressible flows, wether steady or unsteady is
Prof. Dr. MOHSEN OSMAN

3. Example # 1 Water is being added to a storage tank at a rate of 200 liters/min. At the same time, water flows out the bottom through a 5 cm in- side diameter pipe with an average velocity of 18 m/s. The storage tank has an inside diameter of 3 cm. Find the rate at which the water level rises or falls. Solution Applying continuity equation (conservation of mass principle for a fixed control volume) Since the fluid is water, then it is an incompressible then then (I) since
Prof. Dr. MOHSEN OSMAN

4. Differentiate with respect to (w. r. t
Differentiate with respect to (w.r.t.) time (II) Substitute into equation (I) Example # 2 An upright cylindrical tank with a base area of 60 m2 is being filled at the rate of 10 m 3. Fluid is flowing out of the tank through a hole near the tank bottom. The cross–sectional area of this exit is 0.25 m2, and the velocity of the fluid at the exit is , where h is the elevation in meters of the free surface of the liquid above the exit. Find h for steady state conditions and develop an expression for h(t).
Prof. Dr. MOHSEN OSMAN

5. Time is the only independent variable, and so the conservation of mass for the control volume is given by can be interpreted as follows Fluid crosses the control volume at the inlet and the exit, hence Substitute both equations (I) and (II) into continuity equation: Divide by ρ Steady-state conditions will be reached when or when →
Prof. Dr. MOHSEN OSMAN

6. Integrating Arbitrary Fixed Control Volume & The Reynolds transport theorem for an arbitrary fixed control volume generalizes to
Prof. Dr. MOHSEN OSMAN

7. Since the control volume is fixed in space We can write Flux terms = Flux terms = = Control Volume Moving at Constant Velocity For a control volume moving uniformly at velocity Vs as shown in figure, the Reynolds’ transport theorem of this case is where
Prof. Dr. MOHSEN OSMAN

8. If flow crosses the boundaries of the control surface only at certain simplified inlets and exits which are approximately one dimensional, the flux terms reduce to a simple sum Conservation of Linear Momentum In Newton’s law = the property being differentiated is the linear momentum Therefore our dummy variable is and , and application of the Reynolds’ transport theorem gives the linear – momentum relation for fixed control volume (C.V.)
Prof. Dr. MOHSEN OSMAN

9. 1 – The term is the fluid velocity relative to non accelerating coordinate system. 2 – The term is the vector sum of all forces acting on the material considered as a free body (surface forces on all fluids & solids cut by the C.S. plus all body forces acting on the masses). 3 – Entire equation is a vector relation. The equation thus has three components. If we want only, say, the x – component, the equation reduces to: Similarly,
Prof. Dr. MOHSEN OSMAN

10. One-Dimensional Momentum Flux Momentum Flux Term = For one-dimensional cross section ve for outlet flux and –ve; i.e For inlet flux Thus if the control volume has only one-dimensional inlets and outlets, the linear momentum conservation law reduces to: Net vector = Rate of change of + Vector sum of outlet – Vector sum of inlet force on vector momentum momentum fluxes momentum fluxes fixed C.V. within the C.V. Net Pressure Force on a Closed Control Surface Generally speaking, the surface forces on a control volume are due to: (1) Forces exposed by cutting through solid bodies which protrude through the surface and, (2) Forces due to pressure and viscous stresses of the surrounding fluid.
Prof. Dr. MOHSEN OSMAN

11. Pressure–force computation by subtracting out a uniform distribution: (a) Uniform pressure, (b) Non-uniform pressure, For the shown convergent nozzle Net Pressure Force acting on direction of fluid flow is
Prof. Dr. MOHSEN OSMAN

12. Conservation of Linear Momentum For one dimensional inlets and outlets the linear momentum conservation law reduces to
Prof. Dr. MOHSEN OSMAN