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2300_les12 - IV First Law of Thermodynamics D Applications...

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Unformatted text preview: IV. First Law of Thermodynamics D. Applications to steady flow devices 1. Heat exchangers - example: Clinker cooler for cement kiln Secondary air 520 C, 157,000 lbm/h Clinker ? C, 151 ton/h Cooler Clinker 1400 C, 151 ton/h Secondary air 20 C, 157,000 lbm/h Lesson 12, Geof Silcox, Chemical Engineering, University of Utah IV. First Law of Thermodynamics a. Assumptions i. changes in kinetic and potential energy are negligible ii. steady state iii. constant mass flow rates iv. negligible heat loss to surroundings v. negligible nonflow work vi. Evaluate heat capacities at average temperatures b. Energy balance over clinker cooler ⎛ ⎞ ⎛ ⎞ V2 V2 + gz ⎟ − ∑ me ⎜ h + + gz ⎟ 0 = Q − W + ∑ mi ⎜ h + ⎜ ⎟ ⎜ ⎟ 2 2 in ⎝ ⎠ i out ⎝ ⎠e 0 = ∑ mi hi − ∑ me he in out maCa (Tain − Taout ) = mcCc (Tcout − Tcin ) Lesson 12, Geof Silcox, Chemical Engineering, University of Utah 1 IV. First Law of Thermodynamics c. Analysis Solving for Tcout gives Tcout = maCa (Tain − Taout ) + Tcin mcCc Tcout = 157 ,000 lb / h(1040 J / kgK ) ( 20 C − 520 C ) + 1400 C = 1160 C 302 ,000 lb / h(1120 J / kgK ) Comment: since the heat capacities were evaluated at the average temperatures and the final temperature of the clinker was not known, the value of 1160 C can be used to re-evaluate Cc. Iterate until Tcout reaches a constant value. Lesson 12, Geof Silcox, Chemical Engineering, University of Utah IV. First Law of Thermodynamics 2. Nozzles and diffusers (subsonic flows) a. Nozzles increase velocity at the expense of pressure drop in direction of flow b. Diffusers increase pressure at the expense of decrease in velocity 0= QCV ⎛ V2 ⎞ ⎛ V2 ⎞ + ⎜h + ⎟ − ⎜h + ⎟ m⎝ 2 ⎠1 ⎝ 2 ⎠2 1 Nozzle 2 Units: 1 kJ/kg = 103 m2/s2 2 1 Diffuser A common example is the nozzle on a garden or fire hose. Lesson 12, Geof Silcox, Chemical Engineering, University of Utah 2 IV. First Law of Thermodynamics 3. Turbines and compressors a. In a turbine, a fluid does work on the turbine blades to rotate a shaft b. In a compressor, shaft work from the surroundings does work on the fluid to increase its pressure c. A fan slightly increases pressure in order to move a fluid from one location to another d. Neglect changes in KE and PE. Heat transfer may be negligible as well. Qin 2 Compressor 1 Win 0 = Win + Qin + m1h1 − m2 h2 Lesson 12, Geof Silcox, Chemical Engineering, University of Utah IV. First Law of Thermodynamics 4. Throttling Devices a. Used when a decrease in pressure is needed without any work effects. b. Accomplished by a partially open valve, a porous plug, or a long capillary tube. c. Commonly used in refrigeration where the pressure drop is accompanied by a drop in temperature. d. No work. Negligible heat transfer, ∆KE, ∆PE. h1 = h2 (5-41) P1 T1 P2 T2 Lesson 12, Geof Silcox, Chemical Engineering, University of Utah 3 IV. First Law of Thermodynamics e. Example of throttle - refrigeration cycle Refrigerant 134a is throttled from the saturated liquid state at 700 kPa to a pressure of 120 kPa. What is the drop in temperature? 4 Qout Condenser 1 Compressor Wcomp Boundary of system Throttle 3 (from cold Qin region) Evaporator 2 Lesson 12, Geof Silcox, Chemical Engineering, University of Utah IV. First Law of Thermodynamics e. Example of throttle - refrigeration cycle Recall (5-41), h1 = h2. From Table A-12 at 700 kPa, T1 = Tsat = 26.69 ºC and h1 = hf = 88.82 kJ/kg. From Table A-12 at 120 kPa, hf = 22.49 kJ/kg and hg = 236.97 kJ/kg. Because h1 = h2 = 88.82 kJ/kg, our final condition is a saturated mixture and T2 = Tsat = -22.32ºC Lesson 12, Geof Silcox, Chemical Engineering, University of Utah 4 IV. First Law of Thermodynamics 5. Pipe Flow a. May have significant heat transfer. b. Work term needed if there is a fan or pump in CV. c. With liquids ∆KE usually small. May be important with gases. d. ∆PE can be large for liquid flow, for example, neglecting friction and heat transfer, z2 Win Win + m1gz1 − m2 gz2 = 0 z1 Lesson 12, Geof Silcox, Chemical Engineering, University of Utah IV. First Law of Thermodynamics E. Unsteady Flow Processes - Example: rapid charging of an evacuated tank 1. Supply line is r-134a at 40 C and 0.14 MPa (absolute). If tank is initially evacuated, find the temperature in the tank when the flow stops (0.14 MPa). 2. Assumptions: (1) no heat transfer to the surroundings during filling, (2) negligible changes in KE and PE. r-134a 0.14 MPa 40 C Lesson 12, Geof Silcox, Chemical Engineering, University of Utah 5 IV. First Law of Thermodynamics 3. Energy and material balance on contents of tank ⎛ ⎞ ⎛ ⎞ dECV V2 V2 = Qin + Win + mi ⎜ h + + gz ⎟ − me ⎜ h + + gz ⎟ dt 2 2 ⎝ ⎠i ⎝ ⎠e dUCV = mi hi dt dmcv = mi dt a. Integrate energy balance with respect to time ∫ 2 1 2 2 dm dUCV cv dt = ∫ mi hi dt = hi ∫ dt 1 1 dt dt m2 u2 − m1u1 = ( m2 − m1 )hi b. Because tank is initially empty, m1 = 0 and u2 = hi Lesson 12, Geof Silcox, Chemical Engineering, University of Utah IV. First Law of Thermodynamics c. At 40ºC, 0.14 MPa, from Table A-13, hi = 288.70 kJ/kg d. Interpolate in Table A-13 to find T2 to give u2 = hi: T2 = 69.6ºC. e. Why is the final temperature higher than the inlet temperature, Ti = 40ºC? Lesson 12, Geof Silcox, Chemical Engineering, University of Utah 6 ...
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