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Thermodynamics basics

Thermodynamics basics. Zeroth Law. C. A. B. If A and B and B and C are in thermal equil, then A and C are in thermal equil. [ie. At same T]. R Johnson/UAF/CEM/ME 2005. KE and PE. KE = [1/2]mv 2 ; PE = mgh. So, m = 1 kg, v = 100 m/s, h = 100 m, .

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Thermodynamics basics

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  1. Thermodynamics basics Zeroth Law C A B If A and B and B and C are in thermal equil, then A and C are in thermal equil. [ie. At same T] R Johnson/UAF/CEM/ME 2005

  2. KE and PE KE = [1/2]mv2 ; PE = mgh So, m = 1 kg, v = 100 m/s, h = 100 m,  KE = 0.5[104] kg[m2/s2 ] = 5000 J via J = N-m and N = kg m /s2 PE = 1[9.8][100] kg m2 /s2 = 980 J

  3. Thermal, chemical, nuclearenergy 1 kg water with T = 100 C changes internal energy by U = 420 kJ via mc T with c ~ 4.2 kJ/kg/K If evap at P ~ 1 atm, U ~ 2 MJ 1 kg liquid or gaseous fossil fuel has htg vl ~ 44MJ E = m c2 9 x 1016 J via c = 3 x 108 m /s RJ 4.02

  4. Energy amounts BMR ~ 70 W ~ 240 Btu/hr Home htg in Fbks ~ 30 K Btu/hr in winter Home electrical use ~ 1000 kWh/mo ~ 3.4 MBtu or rate of ~ 1.4 kW or ~ 5 K Btu/hr US energy use ~ 90 Q/yr with Q = 1015 Btu  10 kW/cap RJ 4.02

  5. Relative masses

  6. Properties h = const C P SH T liq saturated P = const. s or v Water: Crit Pt at P = 22 MPa, T = 374 oC RJ UAF

  7. SH steam RJ UAF

  8. Sat. steam

  9. 50 50 0.012 0.012 12.0 12.0 209.3 209.3 2234.2 2234.2 209.3 209.3 2382.7 2382.7 0.704 0.704 7.372 7.372 100 100 0.101 0.101 1.67 1.67 418.9 418.9 2087.6 2087.6 419.0 419.0 2257.0 2257.0 1.307 1.307 6.048 6.048 150 150 0.476 0.476 0.393 0.393 631.7 631.7 1927.9 1927.9 632.2 632.2 2114.3 2114.3 1.842 1.842 4.996 4.996 200 200 1.554 1.554 0.127 0.127 250 250 3.973 3.973 0.050 0.050 1080.4 1080.4 1522.0 1522.0 1085.4 1085.4 1716.2 1716.2 2.793 2.793 3.280 3.280 300 300 8.681 8.681 0.021 0.021 374 374 22.09 22.09 0.0031 0.0031 2029 2029 0 0 2099 2099 0 0 4.430 4.430 0 0 Properties of Saturated Water T P vg uf ufg hf hfg s sfg oC MPa m3/kg kJ/kg kJ/kg/K

  10. vg vs. P P = 0.02 0.05 0.10; vg = 7.65 3.24 1.694 P = 1 2 4 6; vg = 0.194 0.0996 0.0498 0.0324

  11. hg vs P P = 0.02 0.05 0.1 0.2 0.5 1 ; hg = 2610 2646 2676 2707 2749 2778

  12. First First Law Conservation of Energy E = Q - W Q [for system] W E eg. Q = 100 kJ W = 60 kJ E = 40 kJ mi me Each term > 0 if by system

  13. For a cycle E = 0 so Qnet = W which leads to W = QH - QL 2nd Law says all of QH can’t be converted into W or  < 100 %

  14. Closed transient T Q - W = m[u2 -u1] v eg. Let SH steam at P1 = 5 MPa and 900 oC cool at const v = 0.1076 m3 /kg to 450 oC STs u1 = 3841 kJ/kg STs  P2 = 3 MPa and u2 = 3020 kJ/kg So Q/ m = - 821 kJ/kg

  15. Heat Xfer Qdot = -k [dT/dx] k < 0.1 W/m/K for good insulators qdot

  16. Required insulation thickness keep resting adult warm at T = -10 oC. ka = 0.026 W/[m oK] and Edot = - Qdot Edot = 70 W = kA [T/x] with T= 40 oC. 70 W = 1.04 [A/x] W with A (m2 ) & x (m ) A ~ 2 m2 x ~ [2/70] m ~ 3 cm 30 oC Ta = - 10 oC 2m

  17. 1st Law for CV dEcv/dt = mdot [hin - hex ] + Qdot - Wdot Ecv = mcv [u + 0.5v2 + gz]cv We could also put KE & PE terms on RHS of 1st eqn.

  18. Open sys in SSSFQcv + mi hi = me he + Wcv Qcv= - 200 kW Q W i mi = me= 3 kg/s e hi= 3000 kJ/kg;he= 2600 kJ/kg Wcv= 3[3000 – 2600] – 200 =1000 kW Note: dot omitted over Q, W, m

  19. Throttling hi = he i e X Eg. Steam at 4 MPa and 700 C exits valve at 0.5 MPa h = 3906 kJ/kg so Te = 691oC via T = 600 + [(3906 – 3702)/(3926 – 3702)] 100 With [ ] = 0.91 T P s  from 7.62 to 8.47 because of irreversibilities s

  20. Steam Properties

  21. Boiler e i T Now take sat liq at 4 MPa entering boiler and heat to 800 C s hi = hf = 1087 and he = 4142 kJ/kg So q = he - hi = 3055 kJ/kg Note: Tsat = 250 C

  22. Heating of Liquid Q = m cp [T2 - T1 ] Above true whether P const or not as heating occurs For water, cp = c = 1 Btu/lbm/ oF = 4.2 kJ/kg/ K

  23. US Solar Insolation

  24. Solar water heater http://www.eren.doe.gov/erec/factsheets/solrwatr.pdf

  25. NREL photo

  26. Solar Hot Water Vancouver Int. Airport $ 375 K cost and annual savings of $ 67 K 100 panels heat 800 gph http://www.solaraccess.com/news/story?storyid=5271

  27. Solar water heating solarwaterFbks.m NREL Fbks data; vls are monthly averages [kWh/m^2/day] for March - Nov. for surface tilted at lat angle of 64 deg mo = 2:10; effic1 = [0.5 0.6 0.7*ones(1,4) 0.6 0.5 0.4]; Sinsoltilt9 = [2.4, 4.7, 5.6, 5.3, 5.2, 4.9, 4.2, 3.4, 2.0]; % avg = 3.3 for yr & if mult by 3600, we convert to kJ/m^2/day Tl = 10; Th = 40; rhow = 1; % kg/liter Cpw = 4.2; effic = 0.7; Toi = [Th Tl]; % kJ/kg/K Volw = effic*3600.*Sinsoltilt9/Cpw/rhow/(Th - Tl); Volw1 = 3600.*effic1.*Sinsoltilt9/Cpw/rhow/(Th - Tl); figure(2); plot(mo,Volw,mo,Volw1,'linewidth',2); grid on; title('Warm water prepared','fontsize',16); % etc

  28. Fbks NREL solar data

  29. Solar Air Heater 1 PV panel Solar air heater at UAF

  30. Data from Fieldpoint Panel output voltage fan current amps Pyranometer W/m^2 Outlet T [deg F] Inlet T [deg F] 11:11 4.254 0.04 0.0 -0.5 -1.4 12:11 4.234 0.076 153.9 19.5 1.9 12:41 4.122 0.116 38.5 10.2 2.2 12:56 12.036 0.776 307.8 18.6 4.6 13:26 3.776 0.136 384.7 32.6 3.2 13:41 10.596 0.656 38.5 11.7 2.5 14:11 2.53 0.152 384.7 49.0 5.3 14:56 4.216 0.04 38.5 14.3 2.3 16:41 1.068 0 0.0 2.9 0.1 1/12/03 UAF Energy Center http://pug.engr.uaf.edu/data_html\011203_ecenter.html

  31. Solar Air Heater ~ 8 hr period represented

  32. Solar Air Heater

  33. Heat Engines Receive heat at high T and reject to low T QL QH W  = W/ QH ~ 30 % RJ UAF

  34. Carnot Cycle = 0 QH TH 2 3 QL = TL [s4 – s1] QH /TH = QL /TL T QH = TH [s3 – s2] 4 1 QL TL W = QH - QL s  = [QH - QL] /QH  = 1 - TL /TH Is upper bound for heat engine

  35. Applications Diesel electric generators, gas turbines power plants http://www.gensetcentral.com/pdf/js170uc.pdf RJ UAF

  36. Example - DEG Wel 6 gph fuel 138 K Btu/gal QL QH = 818 K Btu/hr = 242 kW and  = 33 %  Wel = 81 kW with rest of output as heat flux up exhaust and rejected to jacket water and ambient [ 13.5 kWh/gal ] RJ UAF

  37. Second Law Processes only proceed in certain drcts: Clausius: Can't build cyclic device whose sole effect is heat xfer from cold to hotter body.

  38. Entropy or Heat xfer thru finite T is irreversible

  39. Clausius Inequality < = 0 Consider Rankine Cycle 3 P2 = P3 = 1 MPa T2 = 100 C T3 = 350 C SH vap T B 4 x = 1 2 1 C Sat liq P P1 = P4 = 100 kPa and sat so T = 100 C

  40. T 3 350 C boiler 2a 180 C turb 2 4 100 C 1 cond s3 = 7.30 s 1.30 7.36

  41. Now calc = q23 /T23 + q41 / T34 q23= h3 - h2 = 3051 – 417 = 2634 kJ/kg q41= h1 - h4 = - 2258 kJ/kg q41 / T34 = - 2258/373 = -6.05 kJ/kg/K Adiabatic from 1..2 and 3..4

  42. To calc. Split into three parts In CL regime, =[ h2a – h2 ]/Tavg = [762 – 418]/413 = 0.83 In sat regime, = hfg /Tsat = 2015/453 = 4.45 • [h3 - hg ]/Tavg = [3051 - 2778]/538 • 0.51 so In SH regime, = 5.79 = 5.79 - 6.05 = - 0.26 Note : wnet = q23 - q14 = 515 kJ/kg

  43. SSSF Open sys in SSSF Sgen = m [se - si] – Q/T I = To Sg Wrev = W + I Q e i

  44. Example - ST 8 kg/s at 3 MPa Q = - 300 kW 450 oC W hi = 3344; he = 2769 W = 8[hi - he ] - 300 = 4. 3 MW 0.2 MPa 150 oC Ex 7-15 C&B RJ UAF

  45. Ex. [cont.] si = 7.083 se = 7.280 kJ/kg/K so Sg = 8[se - si] + 300/298 = 2.56 kW/oK I = To Sg = 765 kW Wrev = W + I = 5.07 MW 2 = W/Wrev = 85 % RJ UAF

  46. If 1 kg water heated isothermally and reversibly at 100 oC from sat liquid to sat vapor, 1st Law Q = m[u2 – u1 ] + P[v2 - v1 ] = m[h2 – h1] = 1[h2 - h1 ] = 1[hfg ] = 2257 kJ S = Q/T = 2257/373 = 6.048 kJ/K = m sfg T 2 1 s or v

  47. Environmental Impacts > 1272 grams of fossil fuel and chemicals used to produce a 32-bit DRAM memory chip [mass of 2 grams] Another 400 grams of fossil fuel required to produce electricity to operate it over its lifetime. This doesn’t even account for ultimate disposal cf ratio of 2:1 for automobile Environmental Science & Technology / January 1, 2003

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