Izibonelo ezi-5 zoMthetho kaBernoulli
I-1 P1 dan v1 ukucindezela kanye nesivinini somoya ngaphezu kwephiko, P2 dan v2 ukucindezela kanye nesivinini somoya ngaphansi kwephiko. Ukuze indiza isuke, izimo...
A. P1 =P2 dan v1 = v2
B. P1 < P2 dan v1 > v2
C.P1 < P2 dan v1 < v2
D.P1 > P2 dan v1 > v2
E.P1 > P2 dan v1 < v2
Ingxoxo:
Kuyaziwa ukuthi:
P1 = Umfutho womoya emaphikweni
P2 = Umfutho womoya ngaphansi kwamaphiko
v1 = isivinini somoya emaphikweni
v2 = isivinini somoya ngaphansi kwamaphiko
Kubuziwe :
Yiziphi izimo lapho indiza iphakama khona noma indiza ikhuphukela phezulu?
Jawab :
Umthetho kaBernoulli noma okubizwa ngokunembile Isimiso sikaBernoulli ithi uma ijubane loketshezi liphezulu, ingcindezi yoketshezi iphansi, futhi okuphambene nalokho, uma ijubane loketshezi liphansi, ingcindezi yoketshezi iphakeme. Ama-fluid ayizinto ezingageleza. Izibonelo zama-fluid umoya namanzi.
Uma isivinini somoya siphezulu, umfutho womoya uphansi, ngokuphambene nalokho, uma isivinini somoya siphansi, umfutho womoya uphakeme.
Uma isivinini somoya esingaphezu kwephiko sikhulu kunesivinini somoya esingaphansi kwephiko, ingcindezi yomoya engenhla kwephiko ingaphansi kwengcindezi yomoya engaphansi kwephiko. Ngenxa yokuthi ingcindezi yomoya engaphansi kwephiko inkulu, ukusunduza komoya ongaphansi kwephiko nakho kukhulu. Ngokuphambene nalokho, ingcindezi yomoya engenhla kwephiko incane, ngakho ukusunduza komoya ngaphezu kwephiko nakho kuncane (P = F/A – lapho ingcindezi inkulu, P, amandla amakhulu, F).
Ukusunduza komoya ngaphansi kwephiko kukhulu, ukusunduza komoya ngaphezu kwephiko kuncane, ngakho-ke kukhona amandla avelayo noma amandla aphelele aqondiswa phezulu. La mandla aphelele aqondiswa phezulu okwenza indiza iphakame..
Jadi izimo zokuthi indiza isuke ngu-P1 < P2 kanye no-v1 > v2
Impendulo efanele ingu-B.
2.
Kulesi sibalo, amanzi adonswa yi-compressor ngokucindezela okungu-120 kPa angene epayipini elingezansi (1) bese egeleza phezulu ngesivinini esingu-1 m/s.-1 (g = 10 ms-2 kanti ubuningi bamanzi buyi-1000 kg.m-3). Umfutho wamanzi epayipini eliphezulu (II) u….
A. 52,5 kPa
B. 67,5 kPa
C. 80,0 kPa
D. 92,5 kPa
E. 107,5 kPa
Ingxoxo
Kuyaziwa ukuthi:
Irediyasi enkulu yepayipi (r)1= 12cm
Irediyasi encane yepayipi (r)2= 6cm
Umfutho wamanzi emapayipini amakhulu (p)1) = 120 kPa = 120.000 amaPascal
Ijubane lamanzi emapayipini amakhulu (v)1) = 1 ms-1
Ukuphakama kwepayipi elikhulu (h)1= 0m
Ukuphakama kwepayipi elincane (h)2= 2m
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 ms-2
Ubuningi bamanzi (ρ) = 1000 kg.m-3
Kubuziwe: Umfutho wamanzi epayipini 2 (p)2)
Impendulo:
Ijubane lamanzi epayipini 2 libalwa kusetshenziswa i-Continuity Equation:
A1 v1 =A2 v2
(π r12)(v1) = (π r22)(v2)
(r12)(v1) = (r22)(v2)
(122)(1 m/s) = (62)(v2)
144 = 36 v2
v2 = 144/36
v2 = 4m/s
Ingcindezi yamanzi epayipini 2 ibalwa kusetshenziswa i-equation kaBernoulli:
120.000 + ½ (1000)(12) + (1000)(10)(0) = p2 + ½ (1000)(42) + (1000)(10)(2)
120.000 + ½ (1000)(1) + (1000)(10)(0) = p2 + ½ (1000)(16) + (1000)(10)(2)
120.000 + 500 + 0 = p2 + (500)(16) + 20.000
120.000 + 500 = p2 + 8000 + 20.000
120.500 = p2 28.000 +
p2 = 120.500 - 28.000
p2 = Ama-Pascal angu-92.500
p2 = 92,5 kPa
Impendulo efanele ngu-D.
3. Bheka isithombe esilandelayo!
Ipayipi elikhulu lingama-5 m ngaphezu komhlabathi kanti ipayipi elincane lingama-1 m ngaphezu komhlabathi. Izinga lokugeleza kwamanzi epayipini elikhulu lingama-36 km/h.-1 ngokucindezela okungu-9,1 x 105 Pa, kuyilapho ingcindezi epayipini elincane ingu-2.105 Pa, khona-ke isivinini samanzi epayipini elincane…. (ubuningi bamanzi = 103 kg.m-3)
A. 10 ms-1
B. 20 ms-1
C. 30 ms-1
D. 40 ms-1
E. 50 ms-1
Ingxoxo
Kuyaziwa ukuthi:
Umfutho wamanzi emapayipini amakhulu (p)1) = 9,1 x 105 Ama-Pascal = ama-Pascal angu-910.000
Umfutho wamanzi emapayipini amancane (p)2) = 2 x 105 Ama-Pascal = ama-Pascal angu-200.000
Ijubane lamanzi emapayipini amakhulu (v)1) = 36 km/h = 36(1000)/(3600) = 36000/3600 =10 m/s
Ukuphakama kwepayipi elikhulu (h)1) = -4 amamitha
Ukuphakama kwepayipi elincane (h)2) = 0 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 ms-2
Ubuningi bamanzi (ρ) = 1000 kg.m-3
Kubuziwe: Ijubane lamanzi epayipini elincane (v)2)
Impendulo:
Ijubane lamanzi epayipini elincane (v)2) kubalwa kusetshenziswa i-equation kaBernoulli:
910.000 + ½ (1000)(102) + (1000)(10)(0) = 200.000 + ½ (1000)(v22) + (1000)(10)(-4)
910.000 + 50.000 + 0 = 200.000 + 500 v22 - 40.000
960.000 = 160.000 + 500 v22
800.000 = 500 v22
800.000 / 500 = v22
1600 = v22
v2 = √1600
v2 = 40m/s
Impendulo efanele ngu-D.
4. Kulesi sibalo, amanzi adonswa yi-compressor ngokucindezela okungu-120 kPa angene epayipini elingezansi (1) bese egeleza phezulu ngesivinini esingu-1 m/s.-1 (g = 10 ms-2 kanti ubuningi bamanzi buyi-1000 kg.m-3). Umfutho wamanzi epayipini eliphezulu (II) u….
A. 52,5 kPa
B. 67,5 kPa
C. 80,0 kPa
D. 92,5 kPa
E. 107,5 kPa
Ingxoxo
Kuyaziwa ukuthi:
Irediyasi enkulu yepayipi (r)1= 12cm
Irediyasi encane yepayipi (r)2= 6cm
Umfutho wamanzi emapayipini amakhulu (p)1) = 120 kPa = 120.000 amaPascal
Ijubane lamanzi emapayipini amakhulu (v)1) = 1 ms-1
Ukuphakama kwepayipi elikhulu (h)1= 0m
Ukuphakama kwepayipi elincane (h)2= 2m
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 ms-2
Ubuningi bamanzi (ρ) = 1000 kg.m-3
Kubuziwe: Umfutho wamanzi epayipini 2 (p)2)
Impendulo:
Ijubane lamanzi epayipini 2 libalwa kusetshenziswa i-Continuity Equation:
A1 v1 =A2 v2
(π r12)(v1) = (π r22)(v2)
(r12)(v1) = (r22)(v2)
(122)(1 m/s) = (62)(v2)
144 = 36 v2
v2 = 144/36
v2 = 4m/s
Ingcindezi yamanzi epayipini 2 ibalwa kusetshenziswa i-equation kaBernoulli:
120.000 + ½ (1000)(12) + (1000)(10)(0) = p2 + ½ (1000)(42) + (1000)(10)(2)
120.000 + ½ (1000)(1) + (1000)(10)(0) = p2 + ½ (1000)(16) + (1000)(10)(2)
120.000 + 500 + 0 = p2 + (500)(16) + 20.000
120.000 + 500 = p2 + 8000 + 20.000
120.500 = p2 28.000 +
p2 = 120.500 - 28.000
p2 = Ama-Pascal angu-92.500
p2 = 92,5 kPa
Impendulo efanele ngu-D.
5. Bheka isithombe esilandelayo! Indawo yepayipi elikhulu ingamamitha ama-5 ngaphezu komhlabathi kanti ipayipi elincane lingamamitha ama-1 ngaphezu komhlabathi. Ijubane lokugeleza kwamanzi epayipini elikhulu lingamamitha angama-36 ngehora.-1 ngokucindezela okungu-9,1 x 105 Pa, kuyilapho ingcindezi epayipini elincane ingu-2.105 Pa, khona-ke isivinini samanzi epayipini elincane…. (ubuningi bamanzi = 103 kg.m-3)
A. 10 ms-1
B. 20 ms-1
C. 30 ms-1
D. 40 ms-1
E. 50 ms-1
Ingxoxo
Kuyaziwa ukuthi:
Umfutho wamanzi emapayipini amakhulu (p)1) = 9,1 x 105 Ama-Pascal = ama-Pascal angu-910.000
Umfutho wamanzi emapayipini amancane (p)2) = 2 x 105 Ama-Pascal = ama-Pascal angu-200.000
Ijubane lamanzi emapayipini amakhulu (v)1) = 36 km/h = 36(1000)/(3600) = 36000/3600 =10 m/s
Ukuphakama kwepayipi elikhulu (h)1) = -4 amamitha
Ukuphakama kwepayipi elincane (h)2) = 0 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 ms-2
Ubuningi bamanzi (ρ) = 1000 kg.m-3
Kubuziwe: Ijubane lamanzi epayipini elincane (v)2)
Impendulo:
Ijubane lamanzi epayipini elincane (v)2) kubalwa kusetshenziswa i-equation kaBernoulli:
910.000 + ½ (1000)(102) + (1000)(10)(0) = 200.000 + ½ (1000)(v22) + (1000)(10)(-4)
910.000 + 50.000 + 0 = 200.000 + 500 v22 - 40.000
960.000 = 160.000 + 500 v22
800.000 = 500 v22
800.000 / 500 = v22
1600 = v22
v2 = √1600
v2 = 40m/s
Impendulo efanele ngu-D.
Umthombo wombuzo:
Imibuzo Yefiziksi Yezivivinyo Zikazwelonke Zesikole Samabanga Aphezulu/Isikole Samabanga Aphezulu Sokufundela Umsebenzi
