MECH 314 Problems 6, Spring 2014

MECH 411 Introduction to Fluid Mechanics November 18, 2004 American University of Beirut, Fall 2004-2005 Handout # PS3 Problem Set 3 Solution Textbook: Conservation of mass: Problems 3.19, 3.23, 3.25, 3.29 Conservation of momentum: Problems 3.39, 3.53, 3.80, 3.82 Conservation of energy: Problems 3.132, 3.137, 3.143, 3.144 1 Problem 3.132 Step 1: The control volume is as shown in Figure. Step 2: The fluid is incompressible, isothermal and steady. Assume uniform velocity and properties at inlet and exit. Additionally assume that the change in kinetic energy be- tween 1 and 2 is small compared to change in potential energy. Step 3: Conservation of mass leads to uin = uout and ˙min = ˙mout = ρ Q. Apply the conservation of energy: d dt � V ρ � ˆu + |u2| 2 + gz � dV + � S ρ � ˆh + |u2| 2 + gz � (u · ˆn) dS = ˙Qgained − ˙Wviscous + � ˙W ext on fluid − � ˙W ext by fluid Then for the given conditions � S ρ � ˆh + |u2| 2 + gz � (u · ˆn) dS = − ˙Wviscous − � ˙W ext by fluid ⇒ ρ Q g (zout − zin) = −ρ Q g hf − ˙Wturbine ⇒ ˙Wturbine = −ρ Q g � CQ2� + ρ Q g H The turbine work is maximum when d( ˙Wturbine)/dQ = 0 leading to an optimal Qoptimal = � H/(3C). 11 Problem 3.137 Step 1: The control volume is as shown in Figure. Step 2: The fluid is incompressible, isothermal and steady. Assume uniform velocity and properties at inlet and exit. Step 3: Conservation of mass leads to Vin = VoutAout/Ain if Ain/Aout >> 1, then Vin ≃ 0. Note that “in” refers to the surface and not that inlet of the suction pipe at which the pressure is unknown. Apply the conservation of energy: d dt � V ρ � ˆu + |u2| 2 + gz � dV + � S ρ � ˆh + |u2| 2 + gz � (u · ˆn) dS = ˙Qgained − ˙Wviscous + � ˙W ext on fluid − � ˙W ext by fluid Then for the given conditions � S ρ � ˆh + |u2| 2 + gz � (u · ˆn) dS = − ˙Wviscous − � ˙W ext by fluid ⇒ ˙m 2 � V 2 out − V 2 in � + ˙m g (zout − zin) = − ˙m g hf + ˙Wpump ⇒ ˙Wpump = ˙m �V 2 out 2 + g (zout + hf − zin) � With ˙m = ρ Vout Aout = 167.7 lbm/s, Vout = 120 ft/s, g = 32.2 ft/s2, ρ = 64 lb/ft3. Then ˙Wpump = 40274 lbf.ft/s so that that motor driving the pump must be ˙Wpump/ηpump = 97 hp. 12 Problem 3.143 Step 1: The control volume is as shown in Figure. Step 2: The fluid is an ideal gas. The flow is compressible and unsteady. Assume spatially (but not temporally) uniform properties in the control volume. Neglect potential energy changes. Assume the velocity of the fluid inside the tank to be zero. Assume uniform properties at inlet and that the inlet kinetic energy is very small compared to inlet en- thalpy, i.e. V 2 in/2 << ˆhin . Step 3: Conservation of mass leads to d dt � V ρ dV = ˙min ⇒ d dt(ρV) = ˙min ⇒ V dρ dt = ˙min Apply the conservation of energy: d dt � V ρ � ˆu + |u2| 2 + gz � dV + � S ρ � ˆh + |u2| 2 + gz � (u · ˆn) dS = ˙Qgained − ˙Wviscous + � ˙W ext on fluid − � ˙W ext by fluid Under the stated assumptions, the energy equation reduces to d dt � V ρ ˆu dV − ˙min ˆhin = 0 ⇒ d dt � V ρ CvT dV − ˙min ˆhin = 0 ⇒ Cv V d dt (ρ T) − ˙min ˆhin = 0 ⇒ Cv ρ V dT dt + Cv T V dρ dt − ˙min Cp Tin = 0 Noting that from conservation of mass V dρ dt = ˙min, then ⇒ Cv ρ V dT dt = ˙min Cp Tin − ˙min Cv T The initial rate of increase of temperature in the tank is dT dt ����� t=0+ = ˙min (Cp − Cv) Tin ρ0 V Cv = 3.2 C/s where ρ0 = p0/RT0 = 200/(0.287×293) = 2.3783 kg/m3 is the initial density in the tank. 13 Problem 3.144 Step 1: The control volume is as shown in Figure. Step 2: The fluid is incompressible and steady. Assume uniform properties at sections 1, 2 and 3. The flow area A1 is much larger than A2 so that V1 ≃ 0. The friction head losses are between 1 and 2. The losses between 2 and 3 are negligible. Step 3: Conservation of mass between 1 and 2 leads to V1 ≃ 0 Conservation of energy: d dt � V ρ � ˆu + |u2| 2 + gz � dV + � S ρ � ˆh + |u2| 2 + gz � (u · ˆn) dS = ˙Qgained − ˙Wviscous + � ˙W ext on fluid − � ˙W ext by fluid In order to determine the exit angle θ that maximizes the horizontal distance so that z3 − z2 is constant is obtained by applying Newton’s second law for a small fluid element of volume δV and fixed mass ρ δV. Noting that the only force acting on this element is gravity, then d dt (ρ δV u) = ρ δV g du dt = g The horizontal and vertical components are dw dt = −g ⇒ w − w0 = −gt ⇒ w − V2 sin θ = −gt du dt = 0 ⇒ u = u0 = V2 cos θ ⇒ x = V2 cos θ t At 3 w = 0, x = R/2 where R is the horizontal range to be maximized. At 3, eliminating t from the above two equations leads to R = 2 V2 cos θ V2 sin θ g = V 2 2 sin 2θ g It is clear that R is maximum when θ = π/4. To determine V2, we first apply the conservation of energy between 2 and 3: ˙m �V 2 3 2 + gz3 � − ˙m �V 2 2 2 + gz2 � = 0 ⇒ V 2 2 − V 2 3 = 2 g (z3 − z2) To relate V3 to V2, we apply conservation of momentum in the x direction for the control volume consisting of the jet 2-3, we get ˙m V2 cos θ = ˙mV3 14 Substituting in the energy equation, we get ˙m �V 2 3 2 + gz3 � − ˙m �V 2 2 2 + gz2 � = 0 ⇒ V 2 2 � 1 − cos2 θ � = 2 g (z3 − z2) ⇒ V 2 2 sin2 θ = 2 g (z3 − z2) With θ = π/4, we get V 2 2 = 4 g (z3 − z2) With z3 − z2 = 25 m, we get V2 = 31.32 m/s. To determine the power delivered to the pump, we apply conservation of energy between 1 and 2 ˙m � ˆh2 + V 2 2 2 + gz2 � − ˙m � ˆh1 + V 2 1 2 + gz1 � = ˙Wpump − ˙m g hf ⇒ ˙m � CvT2 + p2 ρ2 + V 2 2 2 + gz2 � − ˙m � CvT1 + p1 ρ1 + V 2 1 2 + gz1 � = ˙Wpump − ˙m g hf The flow is assumed to be isothermal T1 = T2, also p1 ≃ p2 = pa, V1 ≃ 0, then ˙Wpump = ρ V2 πD2 2 2 �V 2 2 2 + g (z2 − z1 + hf) � = 26.245 kW 15