I-Thermodynamics - izinkinga nezixazululo
Umthetho wokuqala we-thermodynamics
1. Ngokusekelwe kugrafu ye-PV engezansi, iyini isilinganiso somsebenzi owenziwe yigesi enqubweni I, nomsebenzi owenziwe yigesi enqubweni II?
Kwaziwa:
Inqubo 1:
Ingcindezi (P) = 20 N/m 2
Umthamo wokuqala (V 1 ) = 10 amalitha = 10 dm 3 = 10 x 10 -3 m 3
Umthamo wokugcina (V 2 ) = amalitha angu-40 = 40 dm 3 = 40 x 10 -3 m 3
Inqubo 2:
Inqubo (P) = 15 N/m 2
Umthamo wokuqala (V 1 ) = 20 amalitha = 20 dm 3 = 20 x 10 -3 m 3
Umthamo wokugcina (V 2 ) = amalitha angu-60 = 60 dm 3 = 60 x 10 -3 m 3
Okufunwayo: Isilinganiso somsebenzi owenziwe yigesi
Isixazululo:
Umsebenzi owenziwe yigesi kule nqubo I:
W = P ΔV = P (V 2 –V 1 ) = (20)(40-10)(10 -3 m 3 ) = (20)(30)(10 -3 m 3 ) = (600)(10 -3 m 3 ) = 0.6 m 3
Umsebenzi owenziwe yigesi kule nqubo II:
W = P ΔV = P (V 2 –V 1 ) = (15)(60-20)(10 -3 m 3 ) = (15)(40)(10 -3 m 3 ) = (600)(10 -3 m 3 ) = 0.6 m 3
Isilinganiso somsebenzi owenziwe yigesi enqubweni I kanye nenqubo II:
0.6 m 3 : 0.6 m 3
1: 1
2.
Ngokusekelwe kugrafu engezansi, yimuphi umsebenzi owenziwe yigesi ye-helium enqubweni ye-AB?
Kwaziwa:
Ingcindezi (P) = 2 x 10 5 N/m 2 = 2 x 10 5 Ama-Pascal
Umthamo wokuqala (V 1 ) = 5 cm 3 = 5 x 10 -6 m 3
Umthamo wokugcina (V 2 ) = 15 cm 3 = 15 x 10 -6 m 3
Okufunwayo: Umsebenzi owenziwe yigesi enqubweni AB
Isixazululo:
W = ∆P ∆V
W = P (V 2 – V 1 )
W = (2 x 10 5 )(15 x 10 -6 – 5 x 10 -6 )
W = (2 x 10 5 )(10 x 10 -6 ) = (2 x 10 5 )(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/m 2
Ingcindezi yokugcina (P 2 ) = 6 Pa = 6 N/m 2
Umthamo wokuqala (V 1 ) = 2 m 3
Umthamo wokugcina (V 2 ) = 4 m 3
Kufunwa: umsebenzi usuqedile ngiyawucubungula
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 10 5 – 2 x 10 5 )
W = ½ (10)(4 x 10 5 )
W = (5)(4 x 10 5 )
W = 20 x 10 5
W = 2 x 10 6 amaJoule
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 ) = 2000 Joules
Ukukhishwa kokushisa (Q L ) = 1200 Joules
Umsebenzi owenziwe yinjini (W) = 2000 – 1200 = 800 Joules
Okufunwayo: ukusebenza kahle (e)
Isixazululo:
e = W / Q H
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 eliphakeme (TH ) = 960 K
Izinga lokushisa eliphansi (T L ) = 576 K
Okufunwayo: 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 (TH ) = 800 Kelvin
Izinga lokushisa eliphansi (T L ) = 300 Kelvin
Kufunwa: ukushisa okukhishwe yinjini
Isixazululo :
Ukusebenza kahle kwe-Carnot (okufanelekile):
Ukushisa okumuncwa yinjini yeCarnot:
W = e Q 1
6000 = (0.625) Q 1
Umbuzo 1 = 6000 / 0.625
Umbuzo 1 = 9600
Ukushisa okukhishwe yinjini yeCarnot:
Umbuzo 2 = Umbuzo 1 – U
Umbuzo 2 = 9600 – 6000
Q 2 = 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 eliphakeme (TH ) = 727 o C + 273 = 1000 K
Okufunwayo: Izinga lokushisa eliphansi
Isixazululo:
T L = 600 Kelvin – 273 = 327 o C
9. Ngokusekelwe kugrafu engezansi, uma injini imunca ama-J angu-800 okushisa, yimuphi umsebenzi owenziwa yinjini.
Kwaziwa:
Izinga lokushisa eliphezulu (TH ) = 600 Kelvin
Izinga lokushisa eliphansi (T L ) = 250 Kelvin
Okokufaka ukushisa (Q 1 ) = 800 Joules
Okufunwayo: Umsebenzi (W)
Isixazululo:
Ukusebenza kahle kwenjini yeCarnot:
Umsebenzi wenziwe yinjini:
W = e Q 1
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 eliphakeme (TH ) = 600 K
Okokufaka ukushisa (Q 1 ) = 600 Joules
Okufunwayo: Umsebenzi wenziwe yinjini yeCarnot (W)
Isixazululo:
Ukusebenza kahle kwenjini yeCarnot:
Umsebenzi wenziwe yinjini yeCarnot:
W = e Q 1
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.
