Report

EGR 334 Thermodynamics Chapter 3: Section 6-8 Lecture 07: Enthalpy and Internal Energy Quiz Today? Today’s main concepts: • Define enthalpy, h • Recognize that enthalpy and internal energy are both intensive state properties that can also be determined on the steam tables. • Learn to pull values for enthalpy and internal energy from the steam tables. • Learn to use computer tools to find property values. • Use steam table values to solve 1st Law balance problems. Reading Assignment: • Read Chap 3: Sections 9-10 Homework Assignment: From Chap 3: 35, 41, 42, 45 3 Last time we learned that if you know two of the three intensive properties: Temperature, T Specific Volume, v and Pressure, p ...then you could find the other properties at that state. Other intensive properties include: Enthalpy, h Internal energy, u and Entropy, s (which are also found on the steam tables. h h(T , v) or u u ( p, v ) or s s (T , p ) Sec 3.6.1 : Introducing Enthalpy 4 Internal Energy (U) Specific Internal Energy (u) : energy / mass Kinetic Energy translational rotational vibrational Potential Energy electric energy of atoms, molecules, or crystals Chemical Bonds Enthalpy (H) Specific Enthalpy (h) : energy / mass Internal Energy + Work H U PV h u Pv 5 Saturated Steam Table Enthalpy, h Internal Energy, u Notice there are additional columns on the steam table that we ignored last time. For the time being, we’re going to still ignore this last quantity. 6 What is enthalpy? Enthalpy is defined in terms of internal energy and the work of expansion as Extensive Form H U pV Intensive Form h u pv Your author indicates that this term is defined, simply because it’s convenient. This combination of terms will show up very frequently, especially when we begin working with control volumes. It’s a way to conveniently deal with the internal energy in combination with the energy associated with expansion or compression of the substance. Enthalpy is another formulation of energy and is commonly expressed in units of kJ or ft-lbf. 7 Example 1: Determine property values Determine specific volume (v), internal energy (u), and enthalpy (h) of the following state properties of H20. a) p = 10 bar T = 179.9 deg C. b) p = 10 bar T = 320 deg. C. c) p = 10 bar T = 450 deg C. x = 30% 8 Example 1: Determine property values Determine specific volume, internal energy, and enthalpy of the following state properties of H20. a) p = 10 bar T = 179.9 deg C. x = 30% from table A.3: at 10 bar the saturated temperature is 179.9 which means the substance is a mixture of liquid and vapor. 9 From table A-3 the following values may be directly read: quality x = 30% p=10 bar spec. vol. int. energy enthalpy [m3/kg] [kJ/kg] [kJ/kg] vf=0.001127 uf=761.68 hf=762.81 vg=0.1944 ug=2583.6 hg=2778.1 therefore: v v f x (v g v f ) v 0.0011 0.3(0.1944 0.0011) v 0.0591 m3 / kg u u f x(u g u f ) u 761.7 0.3(2583.6 761.7) u 1308.3 kJ / kg h 762.8 0.3(2778.1 762.8) h 1367.4 kJ / kg h h f x(hg h f ) 10 Example 1: Determine property values Determine specific volume, internal energy, and enthalpy of the following state properties of H20. b) p = 10 bar T = 320 deg. C. vapor therefore: v = 0.2678 m3/kg u = 2826.1 kJ/kg h = 3093.9 kJ/kg 11 Example 1: Determine property values c) p = 10 bar T = 450 deg C. This also is a gas or superheated vapor and will need to be found by interpolating from table A-4. v v1 (T T1 ) (v2 v1 ) (T2 T1 ) 0.3257 (450 440) 0.3304 m3 / kg u u1 (T T1 ) (0.3541 0.3257) (500 440) (u2 u1 ) (T2 T1 ) 3023.6 (450 440) (3124.4 3023.6) (500 440) 3040.4 kJ / kg h h1 (T T1 ) (h2 h1 ) (T2 T1 ) 3349.3 (450 440) 3370.8 kJ / kg (3478.5 3349.3) (500 440) 12 Linear Interpolation: yunknown y1 xknown x1 y2 y1 x2 x1 yunknown y2 y1 y1 ( xknown x1 ) x2 x1 * y2 yunknown y1 * x1 xknown x2 Your turn: Find the value for enthalpy at T = 400 deg. C and p = 50 bar at 400°C and p1=40 bar h1 = 3213.6 kJ/kg at 400°C and p2=60 bar h2 = 3177.2 kJ/kg hunknown h2 h1 h1 ( pknown p1 ) p2 p1 = 3195.4 kJ/kg On Tests and Quizzes, you will be expected to pull out data from the printed steam tables. That’s not your only option when working homework. Computer Method 1: Web based Steam Table Calculator at: http://www.dofmaster.com/steam.html For: T = 400 deg C. p = 50 bar = 500 kPa Enthalpy per unit mass: h = 3195.4 kJ/kg Computer Method 2: Interactive Thermodynamics (IT) Software you can download to your computer from: http://www.wiley.com/college/moran IT is more than just a look up table. It is also an equation solver that can automatically look up properties values and attempt to find a solution using an iterative solver. h = 3195 kJ/kg Sec 3.5.2 : Saturation Tables Example: Use IT to solve this problem. A cylinder-piston assembly initially contains water at 3 MPa and 300oC. The water is cooled at constant volume to 200 oC, then compressed isothermally to a final pressure of 2.5 MPa. Find the specific volume and enthalpy at each of these three states. 17 State 1: p1 = 3 MPa= 30 bar and T1 = 300 deg C. i) Start with superheated vapor table A-4 ii) Interpolate between 280 and 320 deg C. vb va v1 va (T1 Ta ) Tb Ta v1 0.0771 (0.0850 0.0771) (300 280) (320 280) v1 0.0811 m3 / kg 18 State 2: v2 = v1 = 0.0811 m3/kg and T2 = 200 deg C. Recognize that this is in the liquid-vapor mixture range. Refer to Table A-2 i) Locate vg and vf at T = 200 deg C. v f 1.1565 103 vg 0.1274 v2 0.0811 ii) Determine quality, x2 v v f x (v g v f ) x v vf vg v f 0.0811 0.0011565 0.633 63.3% 0.1274 0.0011565 19 State 3: p3 = 2.5 MPa = 25 bar and T2 = T3= 200 deg C. Recognize that this is in the compressed liquid. Refer to Table A-5 i) Locate p3=25 bar and T3=200 deg C ii) Read the values directly. v3 1.1555 103 m3 / kg u3 849 kJ / kg h3 852.8 kJ / kg 20 Repeat this work, but use IT instead. Results are the same as shown yesterday. Sec 3.6.2: Retrieving u and h Data Workout Problem Given Ammonia at 12°C and v = 0.1217 m3/kg determine P, u and h. a) Using Tables in Appendix: b) Using IT: Sec 3.6.2: Retrieving u and h Data Workout Problem Given Ammonia at 12°C and v = 0.1217 m-3/kg determine P, u and h. a) Using Tables in Appendix: Sec 3.6.2: Retrieving u and h Data Example: Given Ammonia at 12°C and v=0.1217 m-3/kg determine P, u and h. Notice that vf = 0.001608 Since: and vg = 0.1923 vf < v < vg the substance is a liquid-gas mixture Saturation Properties are: Tsat = 12 deg C. Quality if found using: v v f x vg v f Psat= 658.9 kPa x v vf vg v f Sec 3.6.2: Retrieving u and h Data Example: Given Ammonia at 12°C and v=0.1217 m-3/kg determine P, u and h. x v vf vg v f with vf = 1.608x10-3, vg = 0.1923 m3/kg x x 0.1217 0.001608 0.1923 0.001608 0.63 63% Sec 3.6.2: Retrieving u and h Data Example: Given Ammonia at 12°C and v=0.1217 m-3/kg determine P, u and h. Internal Energy: kJ u 1 x u f xug 1 0.63255.16 kJ 0 . 63 1346 . 8 kg kg u 942.9 kJ kg Enthalpy: kJ h 1 x h f xhg 1 0.63256.2 kJ kg 0.63 1473.6 kg h 1023.2 kJ kg Check: h u Pv 942.9 kJ kg 658.9kPa0.1217 103 N m 2 m3 kJ kg 103 Nm kPa b) Using IT: Sec 3.8: Appling the Energy Balance using the Property Tables Workout Problem: (3.60) A rigid, insulated tank fitted with a paddle wheel is filled with water, initially a two phase liquid-vapor mixture at 20 psi, consisting of 0.07 lb of saturated liquid and 0.07 lb of saturated vapor. The tank contents are stirred by the paddle wheel until all of the water is saturated vapor at a pressure greater than 20 psi. Kinetic and potential energy effects are negligible. For the water, determine (a) Volume occupied, in ft3, Pi = 20 psi (b) Initial temperature, in °F, mf = 0.07 lb (c) Final pressure, in psi, ml = 0.07 lb (d) Work, in BTU x mvapor mliquid mvapor 0.07 0.5 0.07 0.07 27 28 Sec 3.8: Appling the Energy Balance using the Property Tables 29 Example: (3.77) A system consisting of 1 kg or water undergoes a power cycle composed of the following processes. Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor. Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C Process 3-4: Isothermal compression with Q34 = -815.8 kJ Process 4-1: Constant-volume heating Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and potential energy effects, determine the thermal efficiency. 30 31 End of Slides for Lecture 07 Sec 3.8: Appling the Energy Balance using the Property Tables Workout Problem: (3.60) A rigid, insulated tank fitted with a paddle wheel is filled with water, initially a two phase liquid-vapor mixture at 20 psi, consisting of 0.07 lb of saturated liquid and 0.07 lb of saturated vapor. The tank contents are stirred by the paddle wheel until all of the water is saturated vapor at a pressure greater than 20 psi. Kinetic and potential energy effects are negligible. For the water, determine (a) Volume occupied, in ft3, Pi = 20 psi (b) Initial temperature, in °F, mf = 0.07 lb (c) Final pressure, in psi, ml = 0.07 lb (d) Work, in BTU x mvapor mliquid mvapor 0.07 0.5 0.07 0.07 32 Sec 3.8: Appling the Energy Balance using the Property Tables (a) (b) (c) (d) Volume occupied, in ft3, Initial temperature, in °F, Final pressure, in psi, Work, in BTU x 33 Pi = 20 psi mf = 0.07 lb ml = 0.07 lb mvapor mliquid mvapor 0.07 0.5 0.07 0.07 Sec 3.8: Appling the Energy Balance using the Property Tables (a) (b) (c) (d) a) Volume occupied, in ft3, Initial temperature, in °F, Final pressure, in psi, Work, in BTU 34 Pi = 20 psi mf = 0.07 lb ml = 0.07 lb xi = 0.5 V 1 x v f xvg 0.07lb0.01683 1.407 ft 3 b) T = Tsat = 227.96 °F ft 3 lb 0.07lb20.094 ft 3 lb Sec 3.8: Appling the Energy Balance using the Property Tables (a) (b) (c) (d) Volume occupied, in ft3, Initial temperature, in °F, Final pressure, in psi, Work, in BTU 35 Pi = 20 psi mf = 0.07 lb ml = 0.07 lb xi = 0.5 V 1.407 ft 3 ft 3 10.05 c) v m 1.4lb lb Interpolate values to find P2 = 42 psi & u2 = 1093 BTU/lb U K PE Q W W U mu2 u1 ui u f xug u f 196.19 0.51082196.19 639.1 BTU lb U 0.14lb1093 639.1 BTU lb 63.6BTU W Sec 3.8: Appling the Energy Balance using the Property Tables 36 Example: (3.77) A system consisting of 1 kg or water undergoes a power cycle composed of the following processes. Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor. Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C Process 3-4: Isothermal compression with Q34 = -815.8 kJ Process 4-1: Constant-volume heating Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and potential energy effects, determine the thermal efficiency. Sec 3.8: Appling the Energy Balance using the Property Tables 37 Example: (3.77) A system consisting of 1 kg or water undergoes a power cycle composed of the following processes. Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor. Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C Process 3-4: Isothermal compression with Q34 = -815.8 kJ Process 4-1: Constant-volume heating Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and potential energy effects, determine the thermal efficiency. Sec 3.8: Appling the Energy Balance using the Property Tables 38 Example: (3.77) A system consisting of 1 kg or water undergoes a power cycle composed of the following processes. Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor. Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C Process 3-4: Isothermal compression with Q34 = -815.8 kJ Process 4-1: Constant-volume heating Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and potential energy effects, determine the thermal efficiency. state 1 2 3 P (bar) 10 10 5 T (°C) Tsat=179.9 v (m3/kg) v1=0.1944 u (kJ/kg) 2583.6 160 v2= v3=0.3835 3231.8 2575.2 4 160 v4=v1 Sec 3.8: Appling the Energy Balance using the Property Tables 39 Example: (3.77) A system consisting of 1 kg or water undergoes a power cycle composed of the following processes. Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor. W12 PdV Pm dv P m v2 v1 V2 v2 V1 v1 105 N m 2 m3 kJ 1kg 0.3835 0.1944 W12 10bar bar kg 103 Nm W12 189.1kJ U12 Q12 W12 Q12 U12 W12 mu2 u1 W12 Q12 1kg 3231.8 2583.6 kJ kg 189.1kJ 837.3kJ Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C Q23 U 23 W23 mu3 u2 0 Q23 1kg 2575.2 3231.8 kJ kg 656.6kJ Sec 3.8: Appling the Energy Balance using the Property Tables 40 Example: (3.77) A system consisting of 1 kg or water undergoes a power cycle composed of the following processes. Process 3-4: Isothermal compression with Q34 = -815.8 kJ W34 Q34 U34 Q34 mu3 u2 v4 v f 0.1944 1.102103 x4 0.6317 3 vg v f 0.3071 1.10210 u4 u f x4 ug u f 674.86 0.63172568.4 674.86 1871kJ kg W34 Q34 mu3 u4 815.8kJ 1kg1871 2575.2 kJ kg 111.6kJ Process 4-1: Constant-volume heating Q41 U 41 W41 mu1 u4 0 Q23 1kg 2583.6 1871 kJ kg 712.6kJ Sec 3.8: Appling the Energy Balance using the Property Tables 41 Example: (3.77) A system consisting of 1 kg or water undergoes a power cycle composed of the following processes. Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor. Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C Process 3-4: Isothermal compression with Q34 = -815.8 kJ Process 4-1: Constant-volume heating Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and potential energy effects, determine the thermal efficiency. Q (kJ) W (kJ) Process 1-2 837.3 189.1 Process 2-3 -656.6 0 Process 3-4 -815.8 1871 Process 4-1 712.6 0 Total 77.5 77.5 Qin Q12 Q41 837.3 712.6 1549.9kJ Wcycle Qin 77.5kJ 0.05 5% 1549.9kJ