I-Thermodynamics - izinkinga nezixazululo
Umthetho wokuqala we-thermodynamics
1. Ngokusekelwe kugrafu ye-PV engezansi, iyini isilinganiso se- umsebenzi kwenziwe yigesi enqubweni I, emsebenzini owenziwe yigesi enqubweni II?
Kwaziwa:
Inqubo 1:
ingcindezi (P) = 20 N/m2
Ivolumu yokuqala (V)1) = amalitha ama-10 = ama-dm ama-103 = 10 x10-3 m3
Umqulu wokugcina (V)2) = amalitha ama-40 = ama-dm ama-403 = 40 x10-3 m3
Inqubo 2:
Inqubo (P) = 15 N/m2
Ivolumu yokuqala (V)1) = amalitha ama-20 = ama-dm ama-203 = 20 x10-3 m3
Umqulu wokugcina (V)2) = amalitha ama-60 = ama-dm ama-603 = 60 x10-3 m3
Kufunwa: Isilinganiso somsebenzi owenziwe yigesi
Isixazululo:
Umsebenzi owenziwe yigesi kule nqubo I:
W = P ΔV = P (V)2-V1) = (20)(40-10)(10-3 m3) = (20)(30)(10-3 m3) = (600)(10-3 m3= 0.6m3
Umsebenzi owenziwe yigesi kule nqubo II:
W = P ΔV = P (V)2-V1) = (15)(60-20)(10-3 m3) = (15)(40)(10-3 m3) = (600)(10-3 m3= 0.6m3
Isilinganiso somsebenzi owenziwe yigesi enqubweni I kanye nenqubo II:
0.6 m3 : 0.6 m3
1: 1
2.
Ngokusekelwe kugrafu engezansi, yimuphi umsebenzi owenziwe yigesi ye-helium enqubweni ye-AB?
Kwaziwa:
Ingcindezi (P) = 2 x 105 N / m2 = 2 x105 Easter
Ivolumu yokuqala (V)1= 5cm3 = 5 x10-6 m3
Umqulu wokugcina (V)2= 15cm3 = 15 x10-6 m3
Kufunwa: Umsebenzi owenziwe yigesi enqubweni ye-AB
Isixazululo:
W = ∆P ∆V
W = P (V)2 - V1)
W = (2 x 105)(15 x 10-6 - 5 x 10-6)
W = (2 x 105)(10 x 10-6) = (2 x 105)(1 x 10-5)
W = 2 Joule
3.
Ngokusekelwe kugrafu engezansi, yimuphi umsebenzi owenziwayo enqubweni ye-AB?
Kwaziwa:
Ingcindezi yokuqala (P)1) = 4 Pa = 4 N/m2
Ingcindezi yokugcina (P)2) = 6 Pa = 6 N/m2
Ivolumu yokuqala (V)1= 2m3
Umqulu wokugcina (V)2= 4m3
Kufunwa: umsebenzi osuqediwe ngiwucubungula
Isixazululo:
Umsebenzi owenziwe ngegesi = indawo engaphansi kwejika ab
W = indawo kanxantathu + indawo yesikwele
W = ½ (6-4)(4-2) + 4(4-2)
W = ½ (2)(2) + 4(2)
W = 2 + 8
W = 10 Joule
4. Ngokusekelwe kugrafu engezansi, yimuphi umsebenzi owenziwe ngenqubo ye-ABCA.
Isixazululo:
Umsebenzi (W) = Indawo kanxantathu i-ABC
W = ½ (20-10)(6 x 105 - 2 x 105)
W = ½ (10)(4 x 105)
W = (5)(4 x 105)
W = 20 x 105
W = 2 x 106 Joule
5. Injini imunca amaJoules angu-2000 okushisa ekushiseni okuphezulu futhi ikhipha amaJoules angu-1200 okushisa ekushiseni okuphansi. Iyini ukusebenza kahle kwenjini?
Kwaziwa:
Okokufaka ukushisa (Q)H) = amaJoules ayi-2000
Umphumela wokushisa (Q)L) = amaJoules ayi-1200
Umsebenzi owenziwe yinjini (W) = 2000 – 1200 = 800 Joules
Kufunwa: ukusebenza kahle (e)
Isixazululo:
e = W / QH
e = 800/2000
e = 0.4 x 100%
e = 40%
6. Injini imunca ukushisa ku-960 Kelvin kanti injini ikhipha ukushisa ku-576 Kelvin. Iyini ukusebenza kahle kwenjini.
Kwaziwa:
Izinga lokushisa eliphezulu (T)H) = 960 K
Izinga lokushisa eliphansi (T)L) = 576 K
Okufunayo: ukusebenza kahle (e)
Isixazululo:
Ukusebenza kahle kwenjini yeCarnot = 0.4 x 100% = 40%
7. Ngokusekelwe kugrafu engezansi, umsebenzi owenziwe yinjini ungama-Joules angu-6000. Kungakanani ukushisa okukhishwa yinjini kumjikelezo ngamunye?
Kwaziwa:
Umsebenzi (W) = 6000 Joule
Izinga lokushisa eliphezulu (T)H) = 800 Kelvin
Izinga lokushisa eliphansi (T)L) = 300 Kelvin
Okufunayo: ukushisa okukhishwe yinjini
Isixazululo :
Ukusebenza kahle kwe-Carnot (okufanelekile):
Ukushisa okumuncwa yinjini yeCarnot:
W = e Q1
6000 = (0.625) Q1
Q1 = 6000/0.625
Q1 = 9600
Ukushisa okukhishwe yinjini yeCarnot:
Q2 =Q1 - W
Q2 = 9600 - 6000
Q2 = 3600 amaJoule
8. Ukusebenza kahle kwenjini yeCarnot kungu-40%. Uma ukushisa kumuncwa ku-727°C, pho izinga lokushisa eliphansi lingakanani.
Kwaziwa:
Ukusebenza kahle (e) = 40% = 40/100 = 0.4
Izinga lokushisa eliphezulu (T)H= = 727oC + 273 = 1000 K
Kufunwa: Ukushisa okuphansi
Isixazululo:
TL = 600 Kelvin – 273 = 327oC
9. Ngokusekelwe kugrafu engezansi, uma injini imunca ama-J angu-800 okushisa, yimuphi umsebenzi owenziwa yinjini.
Kwaziwa:
Izinga lokushisa eliphezulu (T)H) = 600 Kelvin
Izinga lokushisa eliphansi (T)L) = 250 Kelvin
Okokufaka ukushisa (Q)1) = amaJoules ayi-800
Okufunayo: Umsebenzi (W)
Isixazululo:
Ukusebenza kahle kwenjini yeCarnot:
Umsebenzi wenziwe yinjini:
W = e Q1
W = (7/12)(800 amaJoules)
W = 466.7 Joule
10. Izinga lokushisa eliphezulu lenjini yeCarnot lingu-600 K. Uma injini imunca ukushisa okungu-600 J kanti izinga lokushisa eliphansi lingu-400 K, yimuphi umsebenzi owenziwa yinjini.
Kwaziwa:
Izinga lokushisa eliphansi (T)L) = 400 K
Izinga lokushisa eliphezulu (T)H) = 600 K
Okokufaka ukushisa (Q)1) = amaJoules ayi-600
Okufunayo: Umsebenzi wenziwe yinjini yeCarnot (W)
Isixazululo:
Ukusebenza kahle kwenjini yeCarnot:
Umsebenzi wenziwe yinjini yeCarnot:
W = e Q1
W = (1/3)(600) = 200 amaJoules
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Uyini umjikelezo weCarnot, futhi kungani ubalulekile ku-thermodynamics? Impendulo: Umjikelezo weCarnot uwumjikelezo we-thermodynamic olungiselelwe kahle ohlinzeka ngomkhawulo ophezulu ekusebenzeni kahle okungafinyelelwa yinoma iyiphi injini ye-thermodynamic yakudala ngesikhathi sokuguqulwa kokushisa kube ukusebenza (noma okuphambene nalokho). Kubalulekile ngoba ubeka umkhawulo oyisisekelo wokusebenza kahle ngokusekelwe emazingeni okushisa ezindawo zokushisa lapho injini isebenza khona.
