Thermodynamics - matambudziko nemhinduro
Mutemo wekutanga we thermodynamics
1. Zvichibva pagirafu yePV iri pazasi, chiyero chei che basa zvakaitwa negasi mukuita kwekutanga I, kubasa rakaitwa negasi mukuita kwekutanga II?
Zvinozivikanwa:
Maitiro 1:
Pressure (P) = 20 N/m2
Vhoriyamu yekutanga (V)1) = marita gumi = 10 dm3 = 10 x gumi-3 m3
Vhoriyamu yekupedzisira (V)2) = marita gumi = 40 dm3 = 40 x gumi-3 m3
Maitiro 2:
Maitiro (P) = 15 N/m2
Vhoriyamu yekutanga (V)1) = marita gumi = 20 dm3 = 20 x gumi-3 m3
Vhoriyamu yekupedzisira (V)2) = marita gumi = 60 dm3 = 60 x gumi-3 m3
Zvaidiwa: Chiyero chebasa rinoitwa negasi
Solution:
Basa rinoitwa negasi mukuita kwekutanga:
W = P ΔV = P (V)2–V1) = (20)(40-10)(10)-3 m3) = (20)(30)(10)-3 m3) = (600)(10-3 m3) = 0.6 m3
Basa rinoitwa negasi mukuita uku II:
W = P ΔV = P (V)2–V1) = (15)(60-20)(10)-3 m3) = (15)(40)(10)-3 m3) = (600)(10-3 m3) = 0.6 m3
Chiyero chebasa rinoitwa negasi mukuita kwekutanga kwekutanga uye maitiro ekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekutanga kwekupedzisira ...
0.6 M3 : 0.6 m3
1: 1
2.
Zvichibva pagirafu riri pazasi, ibasa ripi rinoitwa negasi rehelium mukuita kweAB?
Zvinozivikanwa:
Kumanikidzwa (P) = 2 x 105 N / m2 = 2 x gumi5 Pascal
Vhoriyamu yekutanga (V)1) = 5 cm3 = 5 x gumi-6 m3
Vhoriyamu yekupedzisira (V)2) = 15 cm3 = 15 x gumi-6 m3
Zvaidiwa: Basa rakaitwa negasi riri muprocess AB
Solution:
W = ∆P ∆V
W = P (V)2 - V1)
W = (2 x 10)5)(15 x 10-6 - 5 x 10-6)
W = (2 x 10)5)(10 x 10-6) = (2 x 105)(1 x 10-5)
W = 2 Joule
3.
Zvichibva pagirafu iri pazasi, ibasa ripi rinoitwa mukuita AB?
Zvinozivikanwa:
Kumanikidzwa kwekutanga (P)1) = 4 Pa = 4 N/m2
Kumanikidzwa kwekupedzisira (P)2) = 6 Pa = 6 N/m2
Vhoriyamu yekutanga (V)1) = 2 m3
Vhoriyamu yekupedzisira (V)2) = 4 m3
Zvaidiwa: basa rapera ndinoita basa iri
Solution:
Basa rinoitwa negasi = nzvimbo iri pasi pekona ab
W = nzvimbo yetriangle + nzvimbo yerectangle
W = ½ (6-4) (4-2) + 4 (4-2)
W = ½ (2)(2) + 4(2)
W = 2 + 8
W = 10 Joule
4. Zvichibva pagirafu iri pazasi, ibasa ripi rakaitwa panguva yekuongorora ABCA?
Solution:
Basa (W) = Nzvimbo yekatatu 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 inotora maJoules maviri ekupisa pakupisa kwakanyanya uye inoburitsa maJoules zana nembiri ekupisa pakupisa kwakaderera. Injini inoshanda sei?
Zvinozivikanwa:
Kupinza kupisa (Q)H) = 2000 Joule
Kupisa kunobuda (Q)L) = 1200 Joule
Basa rinoitwa neinjini (W) = 2000 – 1200 = 800 Joules
Zvaidiwa: kushanda zvakanaka (e)
Solution:
e = W / QH
e = 800/2000
e = 0.4 x 100%
e = 40%
6. Injini inotora kupisa pa960 Kelvin uye injini inoburitsa kupisa pa576 Kelvin. Injini inoshanda sei?
Zvinozivikanwa:
Tembiricha yepamusoro (T)H) = 960 K
Tembiricha yakaderera (T)L) = 576 K
Kuda: kushanda zvakanaka (e)
Solution:
Kushanda zvakanaka kweinjini yeCarnot = 0.4 x 100% = 40%
7. Zvichibva pagirafu iri pazasi, basa rinoitwa neinjini i6000 Joules. Chii chinopiswa neinjini mudenderedzwa rega rega?
Zvinozivikanwa:
Basa (W) = 6000 Joule
Tembiricha yepamusoro (T)H) = 800 Kelvin
Tembiricha yakaderera (T)L) = 300 Kelvin
Kuda: kupisa kunoburitswa neinjini
mhinduro :
Kushanda zvakanaka kweCarnot (yakanaka):
Kupisa kunotorwa neinjini yeCarnot:
W = e Q1
6000 = (0.625) Q1
Q1 = 6000/0.625
Q1 = 9600
Kupisa kunoburitswa neinjini yeCarnot:
Q2 =Q1 - W
Q2 = 9600 - 6000
Q2 = 3600 Joule
8. Kushanda zvakanaka kweinjini yeCarnot ndeye 40%. Kana kupisa kukanyudzwa pa727°C saka tembiricha yakaderera chii.
Zvinozivikanwa:
Kushanda zvakanaka (e) = 40% = 40/100 = 0.4
Tembiricha yepamusoro (T)H= = 727oC + 273 = 1000 K
Zvaidiwa: Kupisa kwakadzika
Solution:
TL = 600 Kelvin – 273 = 327oC
9. Zvichibva pagirafu iri pazasi, kana injini ikatora 800 J yekupisa, ibasa ripi rinoitwa neinjini.
Zvinozivikanwa:
Tembiricha yepamusoro (T)H) = 600 Kelvin
Tembiricha yakaderera (T)L) = 250 Kelvin
Kupinza kupisa (Q)1) = 800 Joule
Kuda: Basa (W)
Solution:
Kushanda zvakanaka kweinjini yeCarnot:
Injini yakaita basa rayo:
W = e Q1
W = (7/12)(800 maJoules)
W = 466.7 Joule
10. Tembiricha yepamusoro yeinjini yeCarnot i600 K. Kana injini ikatora 600 J yekupisa uye tembiricha yakaderera iri 400 K, ibasa rei rinoitwa neinjini.
Zvinozivikanwa:
Tembiricha yakaderera (T)L) = 400 K
Tembiricha yepamusoro (T)H) = 600 K
Kupinza kupisa (Q)1) = 600 Joule
Kuda: Basa rakaitwa neCarnot engine (W)
Solution:
Kushanda zvakanaka kweinjini yeCarnot:
Basa rakaitwa neCarnot engine:
W = e Q1
W = (1/3)(600) = 200 Joule
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Chii chinonzi Carnot cycle, uye nei ichikosha mu thermodynamics? mhinduro: Kutenderera kweCarnot idenderedzwa rethermodynamic rakarongwa rinopa muganho wepamusoro pakushanda zvakanaka kunogona kuitwa nechero injini yekare yethermodynamic panguva yekushandurwa kwekupisa kuita basa (kana zvinopesana). Zvakakosha nekuti inoisa muganho wekushanda zvakanaka zvichibva pakupisa kwematangi ekupisa ayo injini inoshanda pakati pawo.
