2300_les11 - IV. First Law of Thermodynamics A....

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Unformatted text preview: IV. First Law of Thermodynamics A. Introduction to Open Systems 1. Analysis of flow processes begins with the selection of an open system. 2. An open system is a region of space called a control volume (CV). Mass entering (inlet) Q Control volume CV W Mass leaving (exit) Lesson 11, Geof Silcox, Chemical Engineering, University of Utah IV. First Law of Thermodynamics 3. Example We can write a balance equation for the conservation of any extensive property of the open system. As an example, consider the mass of water in the Great Salt Lake. We might be interested in (1) the instantaneous rate of change of mass or (2) the change that occurs over a period of time. ⎛ rate of change ⎞ ⎛ rate at which ⎞ ⎛ rate at which ⎞ ⎟ ⎟⎜ ⎟⎜ ⎜ ⎜ of mass of water ⎟ = ⎜ water enters ⎟ − ⎜ water leaves ⎟ (1) ⎟ ⎟ ⎜ lake ⎟ ⎜ lake ⎜ in lake ⎠ ⎠⎝ ⎠⎝ ⎝ ⎛ change in mass ⎞ ⎛ amount of water ⎞ ⎛ amount of water ⎞ ⎟ ⎟⎜ ⎟⎜ ⎜ ⎜ of water in lake ⎟ = ⎜ which enters lake ⎟ − ⎜ which leaves lake ⎟ (2) ⎜ during February ⎟ ⎜ during February ⎟ ⎜ during February ⎟ ⎠ ⎠⎝ ⎠⎝ ⎝ We obtain Form (2) by integrating Form (1). We use both forms for mass, energy, and entropy. Lesson 11, Geof Silcox, Chemical Engineering, University of Utah 1 IV. First Law of Thermodynamics B. Conservation of Mass for a Control Volume 1. For a closed system, msys = constant and dmsys/dt = 0 2. For an open system ⎛ Time rate of change of ⎞ ⎛ total rate of mass ⎞ ⎛ total rate of mass ⎞ ⎟ ⎟⎜ ⎟⎜ ⎜ ⎜ mass within a control ⎟ = ⎜ entering a control ⎟ − ⎜ leaving a control ⎟ ⎟ ⎜ volume at time t ⎟ ⎜ volume at time t ⎟ ⎜ volume at time t ⎠ ⎠⎝ ⎠⎝ ⎝ dmCV = ∑m − ∑m dt in out (kg/s) (5-9) (5-9) can be integrated with respect to time to give ∆mCV = ∑ m − ∑ m in out (kg) (5-8) Lesson 11, Geof Silcox, Chemical Engineering, University of Utah IV. First Law of Thermodynamics 3. Simplified cases a. Steady flow dmcv = 0 = ∑ m − ∑ m (kg / s ) dt in out (5-18) b. One-dimensional flow (velocity and density are assumed constant or averaged over cross-sectional area of inlet or exit). m = ρ VA = VA V = v v ( kg / s ) (5-7, 19) where V is the volumetric flow rate (m3/s) of the fluid. Lesson 11, Geof Silcox, Chemical Engineering, University of Utah 2 IV. First Law of Thermodynamics c. Example - liquid water with a constant density of 1000 kg/m3 enters a nozzle at the rate 10 L/min. The inlet of the nozzle has a diameter of 1.50 cm, the diameter of the exit is 0.75 cm. Find the velocities of the water at the inlet and exit. 1 2 10 L/min Given: Area = πr2 r1 = 0.0075 m r2 = 0.00375 m ρ = 1000kg/m3 Find: V1 and V2 Model: one-dimensional, steady flow Lesson 11, Geof Silcox, Chemical Engineering, University of Utah IV. First Law of Thermodynamics c. Example Analysis Because we are at steady state, dmcv = 0 = m1 − m2 or dt m1 = m2 = m From the inlet volumetric flow rate, m = ρV1 A1 = ρV = 10 L 1 min 1 kg kg = 0.1667 min 60 s 1 L s Lesson 11, Geof Silcox, Chemical Engineering, University of Utah 3 IV. First Law of Thermodynamics c. Example Analysis The inlet velocity is kg 0.1667 m m s = = 0.943 V1 = kg s ρA1 1000 π (0.0075 m )2 m3 and the exit velocity is V2 = V1 m A1 m ⎛ 1.5 ⎞ = 0.943 ⎜ ⎟ = 3.77 A2 s ⎝ 0.75 ⎠ s 2 Lesson 11, Geof Silcox, Chemical Engineering, University of Utah IV. First Law of Thermodynamics C. Conservation of Energy for an Open System 1. The law of conservation of energy can be applied to a control volume Mass entering (inlet) Q Control volume or system W Mass leaving (exit) ⎞ ⎛ total rate of ⎞ ⎛Time rate of change ⎞ ⎛ net rate of energy ⎞ ⎛ total rate of ⎟ ⎟⎜ ⎟⎜ ⎟⎜ ⎜ ⎟ = ⎜ cros sin g boundary ⎟ + ⎜ energy entering ⎟ − ⎜ energy exiting ⎟ ⎜ of energy within ⎟ ⎜ as work and heat ⎟ ⎜ CV with mass ⎟ ⎜ CV with mass ⎟ ⎜ control volume ⎠ ⎠⎝ ⎠⎝ ⎠⎝ ⎝ Lesson 11, Geof Silcox, Chemical Engineering, University of Utah 4 IV. First Law of Thermodynamics 2. Mathematical statement of conservation of energy for a control volume (single inlet, single outlet). Recall that e = u + 12 V + gz 2 ⎛ kJ ⎞ ⎜⎟ ⎜ kg ⎟ ⎝⎠ ⎛ ⎞ ⎛ ⎞ dECV V2 V2 = Qin,net − Wout ,net + mi ⎜ u + + gz ⎟ − me ⎜ u + + gz ⎟ dt 2 2 ⎝ ⎠i ⎝ ⎠e (A) Lesson 11, Geof Silcox, Chemical Engineering, University of Utah IV. First Law of Thermodynamics 3. The work term a. Flow work W = F i Vel ( kW ) Wflow = P i Ai Vel Wflow = m ( Pv ) b. Rewrite work term in (A) m= 1 Ai Vel ( kg / s ) v Wout ,net = Wnonflow + Wflow = Wnonflow − mi ( Pv )i + me ( Pv )e = WCV − mi ( Pv )i + me ( Pv )e Pi CV mi me Pe Lesson 11, Geof Silcox, Chemical Engineering, University of Utah 5 IV. First Law of Thermodynamics c. The control volume energy equation (A) becomes ⎛ ⎞ ⎛ ⎞ dECV V2 V2 = QCV ,in − WCV ,out + mi ⎜ u + Pv + + gz ⎟ − me ⎜ u + Pv + + gz ⎟ dt 2 2 ⎝ ⎠i ⎝ ⎠e ⎛ ⎞ ⎛ ⎞ dECV V2 V2 = QCV ,in − WCV ,out + mi ⎜ h + + gz ⎟ − me ⎜ h + + gz ⎟ dt 2 2 ⎝ ⎠i ⎝ ⎠e Single inlet - single outlet ⎛ ⎞ ⎛ ⎞ dECV V2 V2 = QCV ,in − WCV ,out + ∑ mi ⎜ h + + gz ⎟ − ∑ me ⎜ h + + gz ⎟ dt 2 2 in ⎝ ⎠i out ⎝ ⎠e Multiple inlet - multiple outlet (see 5-59) Lesson 11, Geof Silcox, Chemical Engineering, University of Utah IV. First Law of Thermodynamics 4. Steady state control volume equations a. Conservation of energy ⎛ ⎞ ⎛ ⎞ V2 V2 0 = QCV − WCV + ∑ mi ⎜ h + + gz ⎟ − ∑ me ⎜ h + + gz ⎟ 2 2 in ⎝ ⎠i out ⎝ ⎠e b. Conservation of mass (5-37) 0 = ∑ mi − ∑ me in out (5-18) Lesson 11, Geof Silcox, Chemical Engineering, University of Utah 6 IV. First Law of Thermodynamics 5. Steady state control volume equations c. single inlet - single outlet ⎛ ⎞ ⎛ ⎞ V2 V2 0 = Qin ,net − Wout ,net + m1 ⎜ h + + gz ⎟ + gz ⎟ − m2 ⎜ h + ⎜ ⎟ ⎜ ⎟ 2 2 ⎝ ⎠2 ⎝ ⎠1 Recall Then 0 = m1 − m2 ⎡ ⎤ V 2 − V12 + g (z2 − z1 )⎥ (5-38) Qin ,net − Wout ,net = m ⎢(h2 − h1 ) + 2 2 ⎣ ⎦ or q − w = ( h2 − h1 ) + V22 − V12 + g ( z2 − z1 ) 2 (5-39) Lesson 11, Geof Silcox, Chemical Engineering, University of Utah 7 ...
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