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 Mass Control entering leaving volume (inlet) (exit) CV Q W lesson 11 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) ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝in lake ⎠ ⎝lake ⎠ ⎝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 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 =−∑mm∑ (kg/s) (5-9) dt in out (5-9) can be integrated with respect to time to give mmm ∆ CV =−∑ ∑ (kg) (5-8) in out lesson 11 IV. First Law of Thermodynamics 3. Simplified cases a. Steady flow dm cv ==0(/)∑∑m − m kg s (5-18) dt in out b. One-dimensional flow (velocity and density are assumed constant or averaged over cross-sectional area of inlet or exit). VAV mA ===ρV (kg/s)(5-19) vv where V is the volumetric flow rate (m3/s) of the fluid. lesson 11 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 Given: Area = πr2 r1 = 0.0075 m 10 L/min r2 = 0.00375 m ρ = 1000kg/m3 Find:V1 and V2 Model: one-dimensional, steady flow lesson 11 IV. First Law of Thermodynamics c. Example Analysis Because we are at steady state, dm cv = 0 = m − m or m = m = m dt 1 2 1 2 From the inlet volumetric flow rate, L 1 min 1 kg kg m = ρV A = ρV =10 = 0.1667 1 1 min 60 s 1 L s lesson 11 3 IV. First Law of Thermodynamics c. Example Analysis The inlet velocity is kg 0.1667 m s m V1 = = = 0.943 ρA kg 2 s 1 1000 π ()0.0075 m m3 and the exit velocity is 2 A1 m ⎛ 1.5 ⎞ m V2 = V1 = 0.943 ⎜ ⎟ = 3.77 A2 s ⎝ 0.75 ⎠ s lesson 11 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 Mass Control entering leaving volume (inlet) (exit) or system Q W ⎛Time rate of change⎞ ⎛net rate of energy ⎞ ⎛total rate of ⎞ ⎛total rate of ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜of energy within ⎟ = ⎜cros sin g boundary ⎟ + ⎜energy entering ⎟ − ⎜energy exiting ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝control volume ⎠ ⎝as work and heat ⎠ ⎝CV with mass ⎠ ⎝CV with mass ⎠ lesson 11 4 IV. First Law of Thermodynamics 2. Mathematical statement of conservation of energy for a control volume (single inlet, single outlet). 1 2 ⎛ kJ ⎞ Recall that e = u + V + gz ⎜ ⎟ 2 ⎝ kg ⎠ dE ⎛⎞⎛⎞VV22 CV =−QWin,net out ,net +++−++ mugzmugz i⎜⎟⎜⎟ e (A) dt⎝⎠⎝⎠ 2ie 2 lesson 11 IV. First Law of Thermodynamics 3. The work term a. Flow work WF = iVel (kW) WPA = iiVel flow 1 mA = iVel (kg/s) WmPvflow = () v b. Rewrite work term in (A) WWWout ,net=+ nonflow flow =−W m Pv + m Pv =−+ W m Pv m Pv nonflow i()ie e () CV i () ie e () P P i CV e m i m e lesson 11 5 IV. First Law of Thermodynamics c. The control volume energy equation (A) becomes dE ⎛⎞⎛⎞VV22 CV =QCV ,in − W CV ,out + m i⎜⎟⎜⎟ u +++ Pv gz − m e u +++ Pv gz dt⎝⎠⎝⎠ 2ie 2 dE ⎛⎞⎛⎞VV22 CV =−QCV ,in W CV ,out +++−++ m i⎜⎟⎜⎟ h gz m e h gz dt⎝⎠⎝⎠ 2ie 2 Single inlet - single outlet dE ⎛⎞⎛⎞VV22 CV =−QWCV ,in CV ,out +∑∑ mhgzmhgz i⎜⎟⎜⎟ ++− e ++ dtin⎝⎠⎝⎠ 2ie out 2 Multiple inlet - multiple outlet (see 5-59, p. 255) lesson 11 IV. First Law of Thermodynamics 4. Steady state control volume equations a. Conservation of energy ⎛⎞⎛⎞22 VV 0QWmhgzmhgz=−+CV CV∑∑ i⎜⎟⎜⎟ ++− e ++ (5-37) in⎝⎠⎝⎠22ie out b. Conservation of mass 0 = ∑mi − ∑me (5-18) in out lesson 11 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⎟ − m2 ⎜h + + gz⎟ ⎝ 2 ⎠1 ⎝ 2 ⎠2 Recall 0 = m1 − m2 ⎡ V 2 − V 2 ⎤ 2 1 (5-38) Then Qin,net −Wout ,net = m⎢()h2 − h1 + + g()z2 − z1 ⎥ ⎣ 2 ⎦ 22 VV21− or qw−=() h21 − h + + gz() 21 − z (5-39) 2 lesson 11 7.