2300_les06 - II. Energy and the First Law of Thermodynamics...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: II. Energy and the First Law of Thermodynamics A. Generic Statement of the First Law for a Closed System ⎛Time rate of change ⎞ ⎜ ⎟ ⎛ rate energy ⎞ ⎛ rate energy ⎞ ⎟−⎜ ⎟ ⎜ of energy within ⎟=⎜ ⎜ ⎟⎜ ⎟ ⎜ system ⎟ ⎝ enters system ⎠ ⎝ leaves system ⎠ ⎝ ⎠ No mass enters or leaves. Heat Rate form of energy balance: System Work dEsystem dt = Ein − Eout (4-12) Integrated form of energy balance: ∆Esystem = E in − Eout (4-11) Note: ∆Esystem + ∆Esurroundings = 0 Lesson 6, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics B. Specific Statement of First Law for a Closed System 1. Energy is conserved. 2. Energy can cross the boundary of a closed system by only two mechanisms: heat transfer and work transfer. 3. The change in energy of a closed system is equal to the net heat transferred to the system minus the net work performed by the system (4-17). Total energy: Energy per unit mass: ∆E = Qin ,net − Wout ,net ( kJ ) E = mu + 1 mV 2 + mgz (kJ ) 2 ∆e = qin,net − w out ,net (kJ/kg) ⎛ kJ ⎞ 12 V + gz ⎜ ⎟ 2 ⎝ kg ⎠ w = W / m and q = Q / m e=u+ de = δ qin,net − δ w out ,net (kJ/kg) The differential form of (4-17): dE = δ Qin,net − δ Wout ,net ( kJ ) Lesson 6, Geof Silcox, Chemical Engineering, University of Utah 1 II. Energy and the First Law of Thermodynamics 4. (Rate form) The rate of change in energy of a closed system is equal to the rate of heat transfer to the system, minus the rate of work performed by the system plus. dE = Qin ,net − Wout ,net (kW ) dt Integration gives the previous form, (4-17): ∆E = ∫ t2 t1 dE dt dt Q = ∫ Qdt t1 t2 W = ∫ Wdt t1 t2 ∆E = Qin ,net − Wout ,net ( kJ ) Lesson 6, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics 5. Example 1 (ideal gas). We have two rigid, insulated chambers connected by a valve. Chamber A is filled with air at 10 bar (gage) and 300 K and B is empty. At these conditions, the air behaves as an ideal gas. Now open the valve and allow the entire system (chambers A and B) to reach equilibrium. Find the change in U and T. The system, A + B, is isolated. Air, p1, T1 A B 0 0 Ideal gas Vacuum X ∆U = Q + W = 0 ∴U = 5 mRT = constant 2 PV = mRT = constant Because U is constant and a function of only T (because the air behaves as an ideal gas) this is a constant temperature process. Joule verified this experimentally. ∴ At equilibrium, T1 = T2 and P1V1 = P2V2. This is an irreversible process. Lesson 6, Geof Silcox, Chemical Engineering, University of Utah 2 II. Energy and the First Law of Thermodynamics 6. Example 2 (ideal gas). A cylinder, containing V1 = 2.0 L of air at 11 bar (gage) and 300 K, is fitted with a frictionless piston. The constant atmospheric pressure outside the cylinder is 1 bar. The air expands reversibly and isothermally to a final pressure of 0 bar (gage). Treating the air as an ideal gas, calculate Qin, Wout, and V2. Patm = 1 bar 2L air Wnet,out 2 L air (system) T1 = 300 K T2 = 300 K P1 = 11 bar (gage) P2 = 0 bar (gage) Qin First Law: ∆U = Qin − Wout = 0 ⇒ Qin = Wout Reversible expansion: Wout = ∫ 2 1 PdV Lesson 6, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics 6. Example 2. Isothermal expansion of ideal gas. Ideal gas: PV = mRT or P = Calculate work: mRT V Wout = ∫ PdV = mRT ∫ 1 2 2 1 V P dV = mRT ln 2 = PV1 ln 1 1 V V1 P2 Wout = Qin = PV1 ln 1 P1 12 = (1200 kPa ) 0.002 m 3 ln = 5.964 kJ 1 P2 Note: in using the ideal gas law, always use absolute pressure rather than gage. ( ) Calculate final volume: V2 = V1 P1 12 = 0.002 = 0.024 m 3 1 P2 Lesson 6, Geof Silcox, Chemical Engineering, University of Utah 3 II. Energy and the First Law of Thermodynamics 7. Example 3 (ideal gas). A cylinder containing 2.0 L of air at 11 bar (gage) and 300 K is fitted with a frictionless piston. The constant atmospheric pressure outside the cylinder is 1 bar. The air is expanded reversibly and adiabatically to a final pressure of 0 bar (gage). Treating the air as an ideal gas, calculate Qin, Wout, V2 and T2. Patm = 1 bar Qin = 0 2L air Wnet,out 0 First Law: 2 2 L air (system) T1 = 300 K V1 = 0.002 m3 P1 = 11 bar (gage) P2 = 0 bar (gage) ∆U = Qin − Wout Reversible expansion: Wout = ∫ PdV or δ W = PdV ( kJ ) or δ w = Pdv ( kJ / kg ) 1 Lesson 6, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics 7. Example 3. Reversible, adiabatic expansion of ideal gas. Write first law in differential form: Qin = 0 2L air ∆ ( mu ) = Q − W ( kJ) Wnet,out du = δ q − δ w ( kJ/kg) du = −Pdv 0 5 5 RT or du = RdT or du = cv dT 2 2 5 RT cv = R Ideal gas: Pv = RT or P = 2 v 2 dT 2 dv T v Integrate 1st law: cv ∫ or cv ln 2 = −R ln 2 = −R ∫ 1T 1v T1 v1 Recall for air that u= Lesson 6, Geof Silcox, Chemical Engineering, University of Utah 4 II. Energy and the First Law of Thermodynamics 7. Example 3. Reversible, adiabatic expansion of ideal gas. Rearrange: cv ln ⎛V ⎞ T2 v T = −R ln 2 to give 2 = ⎜ 1 ⎟ T1 v1 T1 ⎝ V2 ⎠ R cv By convention, this is written T2 ⎛ v1 ⎞ =⎜ ⎟ T1 ⎝ v 2 ⎠ k −1 or Tv k −1 = constant where k = R +1 cv For air at room temperature, k = 7/5 = 1.4. Lesson 6, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics 7. Example 3. Reversible, adiabatic expansion of ideal gas. Use the ideal gas law to rearrange previous equation: T2 v 2P2 = T1 v1P1 P2 ⎛ v1 ⎞ = ⎜ ⎟ or Pv k = constant P1 ⎝ v 2 ⎠ k T2 ⎛ P2 ⎞ =⎜ ⎟ T1 ⎝ P1 ⎠ k −1 k or TP 1− k k = constant Lesson 6, Geof Silcox, Chemical Engineering, University of Utah 5 II. Energy and the First Law of Thermodynamics 7. Example 3. Reversible, adiabatic expansion of ideal gas. We are finally ready to calculate Qin, Wout, V2 and T2. Qin = 0 because the process is adiabatic. Wout = −∆U = mcv (T1 − T2 ) k −1 ⎡ ⎤ k ⎛ T2 ⎞ ⎢1 − ⎛ P2 ⎞ ⎥ = mcvT1 ⎜ 1 − ⎟ = mcvT1 ⎢ ⎜ P1 ⎟ ⎥ ⎝ T1 ⎠ ⎢⎝⎠⎥ ⎣ ⎦ 2 ⎡ ⎤ 7 5 ⎢1 − ⎛ P2 ⎞ ⎥ = 5 (1200 kPa ) 0.002 m 3 = PV1 ⎜⎟ 1 ⎢ ⎝ P1 ⎠ ⎥ 2 2 ⎢ ⎥ ⎣ ⎦ ( ) 2 ⎡ ⎤ ⎛ 1 ⎞7 ⎥ ⎢1 − = 3.050 kJ ⎢ ⎜ 12 ⎟ ⎥ ⎝⎠ ⎣ ⎦ Lesson 6, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics 7. Example 3. Reversible, adiabatic expansion of ideal gas. Calculate T2 and V2. ⎛P ⎞ T2 = T1 ⎜ 2 ⎟ ⎝ P1 ⎠ R k −1 k ⎛ 1 ⎞7 = 300 K ⎜ ⎟ = 147.5 K ⎝ 12 ⎠ cv 2 T2 ⎛ V1 ⎞ cv V ⎛T ⎞R = ⎜ ⎟ or 1 = ⎜ 2 ⎟ T1 ⎝ V2 ⎠ V2 ⎝ T1 ⎠ V1 0.002 m3 V2 = = = 0.01180 m3 5/2 cv ⎛ T2 ⎞ R ⎛ 147.5 ⎞ ⎜ 300 ⎟ ⎜⎟ ⎝ ⎠ T1 ⎠ ⎝ Lesson 6, Geof Silcox, Chemical Engineering, University of Utah 6 II. Energy and the First Law of Thermodynamics 8. Example 4. Work done against the atmosphere and net work available in Examples 2, 3. In Examples 2 and 3, we have not considered the work done against the atmosphere. In both cases that will be Patm = 1 bar 2L air Watm = ∫ Patm dV = Patm (V2 − V1 ) 2 1 Qin Wnet,out In Example 2, Watm = 100 kPa 0.0240 − 0.002 m 3 = 2.20 kJ The net work available in Ex. 2 is Wnet,out = 5.96 - 2.20 = 3.76 kJ. In Example 3, Watm = 100 kPa 0.0118 − 0.002 m 3 = 0.980 kJ The net work available in Ex. 3 is Wnet,out = 3.05 - 0.98 = 2.07 kJ. Lesson 6, Geof Silcox, Chemical Engineering, University of Utah ( ) ( ) II. Energy and the First Law of Thermodynamics 9. Example 5. Estimating the energy released in an explosion. A 2-L, plastic soda bottle is filled with air at 11 bar (gage) and 300 K. The atmospheric pressure is 1 atm. The bottle suddenly explodes. Estimate the energy released. Key assumptions: (1) ideal gas, (2) adiabatic. This process is highly irreversible. 0 First Law: ∆U = Qin − Wout Wout = Watm = Patm (V2 − V1 ) = P2 (V2 − V1 ) Work performed: Ideal gas: mRT2 P2 5 5 Change in U: u = RT or ∆U = mcv (T2 − T1 ) where cv = R 2 2 PV = mRT or V2 = Lesson 6, Geof Silcox, Chemical Engineering, University of Utah 7 II. Energy and the First Law of Thermodynamics 9. Example 5. Estimating the energy released in an explosion of a 2-L soda bottle. From first law: ∆U = −Wout or mcv T2 − T1 = Patm V1 − V2 ( ) ( ) P2 cv 15 + P1 R 12 + 2 Solving for T2: T2 = T1 = = 221.4 K cv 5 1+ 1+ P2 cv ⎛ 2 R + ⎜ cv P1 R ⎜1− Energy released: Wout = −∆U = mcv (T1 − T2 ) = PV1 1 c R⎜ 1+ v ⎜ R 1 5⎞ ⎝ ⎛ +⎟ 5⎜ Wout = (1200 kPa ) 0.002 m3 ⎜ 1 − 12 2 ⎟ = 1.57 kJ 5⎟ 2⎜ 1+ ⎝ 2⎠ ( ) ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Lesson 6, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics 10. Summary of relations used to calculate work in ideal gas, closed system (batch), compression and expansion processes. Isobaric (constant pressure): P = constant Isothermal: Pv = constant R Adiabatic and reversible: Pv k = constant where k = + 1 cv Polytropic: Pv = constant , where 1 < n < k (most of the time) n Wout ,poly = P2V2 − PV1 mR (T2 − T1 ) 1 = , n ≠1 1− n 1− n Lesson 6, Geof Silcox, Chemical Engineering, University of Utah 8 II. Energy and the First Law of Thermodynamics 11. Final remarks on reversible and actual processes a. Examples 2 and 3 involved batch reversible expansion processes. Actual processes deliver less work than reversible processes. Define an efficiency to account for this: Wactual = ηWreversible , 0 < η < 1, expansion b. Actual compression processes require more work than reversible compression processes. Define an efficiency to account for this: Wactual = Wreversible η , 0 < η < 1, compression For batch compression and expansion processes, η ≅ 0.75. The efficiency must be determined experimentally. Lesson 6, Geof Silcox, Chemical Engineering, University of Utah 9 ...
View Full Document

This note was uploaded on 09/02/2010 for the course PHYS 2300 taught by Professor Silcox during the Fall '09 term at University of Utah.

Ask a homework question - tutors are online