CEE 3310 – Control Volume Analysis 13.6 Review• Conservation of Linear MomentumddtZC.V.~vρ d∀ +ZC.S.~vρ(~v ·~n) dA =X~FC.V.3.7 Moment of Momentum Equation (Conservationof Angular Momentum)X~F = m~a =ddt(m~v)Now, taking the moment of each side about some point A with position vector ~rX~r ×~F =ddt(~r × m~v) =ddt(~r × ρ∀~v)AndX(~r ×~Fext) =ddtZsys(~r ×~v)ρ d∀Let H =Rsys(~r ×~v)ρd∀ be the moment-of-momentum of the system. Now we turn to theReynolds Transport Theorem to find the conservation law for the moment-of-momentumwith B = H and b = ~r ×~vdHsysdt=ddtZCV(~r ×~v)ρ d∀+ZCS(~r ×~v)ρ(~v ·~n) dAFor a non-deformable inertial control volume we haveX(~r ×~F )ext=∂∂tZCV(~r ×~v)ρ d∀+ZCS(~r ×~v)ρ(~v ·~n) dAIf we have 1-D inlets and outlets:X(~r ×~F )ext=∂∂tZCV(~r ×~v)ρ d∀+X(~r ×~v)out˙mout−X(~r ×~v)in˙minNote that the (~v ·~n) term leads to a positive or negative flux of angular momentum, as wefound with the conservation of mass and linear momentum. However, angular momentum2itself has a sign (as does linear momentum), which arises from the ~r ×~v term so we musttrack two signs, which may cancel each other out. We note that our sign convention forthe sign of angular momentum is the same as we used for torque, namely the right-hand-rule with a counter-clockwise sense yielding positive angular momentum and a clockwisesense yielding negative angular momentum. Hence the inward flux of clockwise angularmomentum is positive and the outward flux of clockwise angular momentum is negative.3.8 Moment-of-Momentum Examples3.8.1 Example 1 – Single-Arm Lawn SprinklerConsider the following single-arm lawn sprinklerWhat is the optimal coordinate system and control volume?A moving control volume ⇒ need absolute velocity, thereforeV = velocity relative to nozzle + velocity of nozzle=QA− ΩRCEE 3310 – Control Volume Analysis 3where A is the area of the pipe. Now, what is the solution?Ω =QAR−T0ρQR2Aha! Even if T0, the retarding torque from friction, is zero Ω is finite. What is Ω whenT0= 0? ⇒ Ω = V0/R, where V0is the nozzle exit velocity, why? ⇒ The absolutevelocity out of the nozzle is zero!3.8.2 Example 2 – Pipe Section and Bracket TorqueConsider the following pipe section and supporting bracket:What is the reaction torque at the bracket wall mounting point (A)? You may assume thefluid in the pipe is constant density and the surrounding environment is at atmosphericpressure.How do we draw the control volume?4What is the equation?What isP(~r ×~F )CV?What are the pressure forces?TA= h2(P2A2+ ˙mV2)−h1(P1A1+ ˙mV1) or since ˙m = ρQ = ρA1V1= ρA2V2TA= h2A2(P2+ρV22)−h1A1(P1+ρV21)CEE 3310 – Control Volume Analysis 53.9 Review• Moment of momentum equation example problems3.10 The First Law of Thermodynamics and the En-ergy EquationTime rate of change oftotal system energy=Time rate of change byheat transfer+Time rate of change bywork transferordEdtsys=˙Q +˙Wwhere E is energy, Q is heat, and W is work. Recall that ( ˙ ) indicates time rate ofchange.Sign conventions:˙Q → is the transfer by radiation, conduction, convection of heat. Transfer into thecontrol volume is positive.W > 0 is work done on the system by the surroundings and W < 0 is work done by thesystem on the surroundings.˙W =˙Wshaft+˙Wpressure+˙Wviscous stressNote that gravitational work will enter our energy budget through potential energy.˙Wshaft= TshaftΩ where Tshaftis the torque and Ω is the angular velocity.63.10.1 Stress Induced Work per Time (Power)Work occurs by applying a force over a distance. Therefore pressure (normal stress) andshear (tangential stress) can produce work. If we look at the rate of work, or work perunit time, we have power and we see that we can think of this as applying a force on asystem with a given velocity.δ˙W = δ~F ·~vTherefore for pressure we haveδ˙Wpres= −P~n δA ·~v = −P~v ·~n δATherefore˙Wpres= −ZCSP~v ·~n dAThis term is usually moved to the right-hand-side flux term of the energy equation as itis a flux, which is how we will treat it.For shear stress we haveδ˙Wshear= ~τδA ·~v = ~τ ·~v δATherefore˙Wshear=ZCS~τ ·~v dABut shear is internally self-canceling. On the control surface ~v = 0 at solid boundaries(no-slip boundary condition), if the control surface is normal to the flow then ~τ ⊥ ~v andhence ~τ ·~v = 0. Hence it is often reasonable to assume that the shear stress induced workis small.What is a good environmental example of when this assumption breaks down?3.11 The Energy EquationWith our definitions of work and energy we are now ready to apply the ReynoldsTransport Theorem to produce the conservation of energy equation. Let B = E andCEE 3310 – Control Volume Analysis 7b = e = E/m, the energy per unit mass. We can decompose e into the following compo-nentse = ˘u +v22+ gz + otherwhere ˘u is the internal energy per unit mass, v2/2 is the kinetic energy per unit mass,and gz is the potential energy per unit mass. We will ignore other sources of internalenergy but from the definition they are easily included.We can write the Reynolds Transport Theorem flux term (including the pressure workterm)ZCS˘u +v22+ gz +Pρρ(~v ·~n) dAThereforedEdtsys=˙Q+˙Wshaft=ddtZCV˘u +v22+ gzρ d∀+ZCS˘u +v22+ gz +Pρρ(~v·~n) dAThe 1-D form is˙Q +˙Wshaft=ddtZCV˘u +v22+ gzρ d∀+Xoutflows˘u +v22+ gz +Pρ˙m −Xinflows˘u +v22+ gz +Pρ˙mWe sometimes will choose to write (˘u + P/ρ) =˘h = enthalpy.3.11.1 Forms of the Energy EquationWe often find it convenient to cast the energy equation in alternate forms.Velocity Squared form:If the flow is at steady state then conservation of mass gives us that ˙mout= ˙min= ˙mand we can normalized all of our quantities by ˙m which leaves our homogeneous equationwith terms having units of velocity squared˙Q˙m+˙Wshaft˙m= ˘uout+v2out2+ gzout+Poutρ− ˘uin−v2in2− gzin−Pinρ8Definingq =˙Q˙m=heat transferunit massand ws=˙Wshaft˙m=shaft workunit masswe can write the above as˘uout+v2out2+ gzout+Poutρ= ˘uin+v2in2+ gzin+Pinρ+ q + wsHead form:Manometers have historically lead to a desire to think of energy in units of length, orhead. We see that we can arrive at units of length by dividing our velocity squared formof the energy equation by gravity. Hence we have˘uoutg+v2out2g+ zout+Poutγ=˘uing+v2in2g+ zin+Pinγ+ hq+ hswhere hqand hsare q/g and ws/g,