#### Transcript Control Volumes and Mass Balance

EGR 334 Thermodynamics Chapter 4: Section 1-3 Lecture 12: Control Volumes and Conservation of Mass Quiz Today? Today’s main concepts: • Be able to explain what a control volume is • Be able to write mass balance and mass rate balance equations for a control volume. • Be able to explain the continuity of mass flow equation. • Explain the difference between mass flow rate and volumetric flow rate. • Be able to set up problems involving mass balance Reading Assignment: • Read Chap 4: Sections 4-5 Homework Assignment: From Chap 4: 1, 6, 11, 22 Sec 4.1: Conservation of Mass 3 So far we have looked at closed systems Only Q and W pass the boundary Energy Balance Q and W Now move to open system, mass can also pass the boundary. add the mass balance. Mass From Chapter 2: Energy Balance [ E within the system ][ ] [ ] Mass Balance [ m within the system = net Q input + net W output ][ ] [ ] = net m input + net m output Sec 4.1: Conservation of Mass Mass Balance [ m within the system 4 ][ ] [ ] = net m input + net m dmCV i m e m dt output CV = control volume i = inlet, e = exit mi General form for multiple inlets/exits dmCV i m e m dt dmCV dt me Sec 4.1.2: Mass Flow Rate 5 Continuity Principle: mass flow is steady and continuous m V (Vn tA) where ρ = density Vn = normal velocity component A = cross sectional area V = volume Mass flow rate: dm m Vn A dt m Vn dA A Mass Flux Vn for constant Area for variable Area Sec 4.2 Mass Rate Balance 6 One-Dimensional Flow (Continuity Equation) • Flow is normal to the boundary • The fluid is homogeneous (intensive properties are uniform with position). AV m V AV v where: m = mass flow rate ρ = density V = velocity A = area V = volumetric flow rate v = specific volume General form for multiple inlets/exits dmCV Ai Vi Ae Ve dt vi ve At steady State, dmCV 0 dt and Ai Vi Ae Ve vi ve Sec 4.3 : Applications of the Mass Rate Balance Example: (4.16) Ammonia enters a control volume operating at steady state at pA = 14 bar, TA = 28oC, with a mass flow rate of 0.5 kg/s. Saturated vapor at pB = 4 bar leaves through one exit, with a volumetric flow rate of 1.036 m3/min and saturated liquid at pC=4 bar leaves through a second exit. Determine (a) the minimum diameter of the inlet pipe, in cm, so the ammonia velocity does not exceed 20 m/s (b) the volumetric flow rate of the second exit stream in m3/min. pA = 14 bar, TA = 28oC 0.5 kg/s Saturated vapor pB = 4 bar, 1.036 m3/min pC= 4 bar Saturated liquid 7 Sec 4.3 : Applications of the Mass Rate Balance pA= 14 bar TA= 28oC mA=0.5 kg/s Saturated vapor pB = 4 bar, 1.036 m3/min pC= 4 bar Saturated liquid Ammonia state Inlet A Exit B Exit C P (bar) 14 4 4 T (°C) 28 v (m3/kg) 8 What else can you determine about the states? State 1: from table A-14 (state A is compressed liquid… let vA≈ vf @28oC = 1.6714x10-3 m3/kg) State 2: from table A-14 ( vB = vg @4bar) = 0.3094 m3/kg and TB=Tsat = -1.9oC) State 3: from table A-14 ( vC = vf @4bar)= 0.0015597 m3/kg and TC=Tsat = -1.9oC) Sec 4.3 : Applications of the Mass Rate Balance Ammonia 9 Saturated vapor VB 1.036m3 / min state Liquid mA 0.5 kg / s Saturated liquid Inlet A Exit B Exit C P (bar) 14 4 4 T (°C) 28 -1.90 -1.90 1.6714x10-3 0.3094 1.5597x10-3 v (m3/kg) Consider mass flows: State A: already known… mA 0.5 kg / s State B: can be found from volumetric flow rate VB 1.036 m3 / min 1min mB B VB 0.0559 kg / s 3 vB 0.0016714 m / kg 60 s State C: can be found from mass rate balance mA mB mC mC mA mB 0.5 0.0559 0.4441 kg / s Sec 4.3 : Applications of the Mass Rate Balance Ammonia 10 Saturated vapor state VB 1.036m / min 3 Liquid mA 0.5 kg / s Saturated liquid Inlet A Exit B Exit C P (bar) 14 4 4 T (°C) 28 -1.90 -1.90 v (m3/kg) 1.6714x10-3 0.3094 1.5597x10-3 mA (kg/s) 0.5 0.0559 0.4441 What can we learn from the continuity equation: State A: mA VA vA m VA State B: (already known) VA v A mA (1.6714 10-3 m3 / kg )(0.5kg / s ) 0.0008357 m / s 0.05014 m / min 3 State C: VA V v v 3 VB 1.036m3 / min VC vC VC vC mC (1.5597 10-3 m3 / kg )(0.4441kg / s ) mC 0.0006927 m3 / s 0.04156 m3 / min Sec 4.3 : Applications of the Mass Rate Balance Ammonia state Saturated vapor Liquid Saturated liquid 11 Inlet A Exit B Exit C P (bar) 14 4 4 T (°C) 28 -1.90 -1.90 v (m3/kg) 1.6714x10-3 0.3094 1.5597x10-3 m (kg/s) 0.5 0.0559 0.4441 0.05014 1.036 0.04156 V (m3/min) Determine the size of the inlet pipe so that the velocity does not exceed VA = 20 m/s from the continuity equation: Area of a circular cross section: VA VA A therefore: d 4 VA VA VA VA d2 A d2 4 4 4(0.0514m3 / min) 1min 0.00738m 0.738cm (20m / s ) 60 s Sec 4.3 : Applications of the Mass Rate Balance 12 Example 2: ( from Prob 4.16) Liquid water at 70 oF enters a pump through an inlet pipe having a diameter of 6 in. The pump operates at steady state and supplies water to two exit pipes having diameters of 3 in and 4 in. The velocity of the 3 in pipe is 1.31 ft/s. At the exit of the 4 in pipe the velocity is 0.74 ft/s. The temperature of the water in each exit is 72 deg F. Determine a) the mass flow rate in lb/s in the inlet and Exit B each of the exit pipes. b) the volumetric flow rate at the inlet in ft3/min. Exit C Inlet A Identify what you know: state Inlet A V [ft/s] Exit B Exit C 1.31 0.74 T [°F] 70 72 72 d [in] 6 3 4 dm/dt [lbm/s] dV/dt [ft3/min] Sec 4.3 : Applications of the Mass Rate Balance Use continuity to find volumetric flow rates VA V m VA v v state 13 Inlet A V [ft/s] Exit B Exit C 1.31 0.74 T [°F] 70 72 72 d [in] 6 3 4 Volumetric Flow Rate: Area: V V A A d 2 4 Exit B VB VB dB2 4 (1.31 ft / s ) (3in) 2 1 ft 4 2 60 s 3.858 ft 3 / min 12in 1min Exit C VC VC dC 2 4 (0.74 ft / s ) (4in) 2 1 ft 4 2 60 s 3.874 ft 3 / min 12in 1min Sec 4.3 : Applications of the Mass Rate Balance Next find the mass flow rates state Specific Volumes may be found on Table A-2E: Exit C Inlet A: Inlet A V [ft/s] V m v Exit B 14 Exit B Exit C 1.31 0.74 T [°F] 70 72 72 d [in] 6 3 4 vA = vf @ 70 oF =0.01605 ft3/lbm vB = vC = vf @ 72 oF = 0.01606 ft3/lbm VB 3.858 ft 3 / min 1min mB 4.004 lbm / s 3 vB 0.01606 ft / lbm 60 s VB 3.874 ft 3 / min 1min mB 4.020 lbm / s 3 vB 0.01606 ft / lbm 60 s (apply mass balance) mA mB mC 4.004 4.020 8.024 lbm / s Sec 4.3 : Applications of the Mass Rate Balance 15 Finally, the volumetric flow rate of the inlet may be found: 60 s VA v A mA (0.01605 ft / lbm )(8.024lbm / s ) 7.727 ft 3 / min 1min 3 Summary: state Inlet A Exit B Exit C V [ft/s] 7.727 1.31 0.74 T [°F] 70 72 72 d [in] 6 3 4 v [ft3/lbm] 0.01605 0.01606 0.01606 dm/dt [lbm/s] 8.024 4.004 4.020 dV/dt [ft3/min] 7.727 3.858 3.874 16 End of slides for Lecture 12