nasoslarning to'la bosimini aniqlash.

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8-amaliy mashgulot. nasosning to'la bosimini aniqlash(masalalar) h=n g+∑ hw+ v yu .s 2 −v p .s 2 2g ; (1.3) bu erda , ng= yu.s.s.-p.s.s.=53-22=31 m yuqorida keltirilgan (1.4) va (1.5) formulalar bilan mahalliy va uzunlik bo'yicha gidravlik qarshiliklarga bosim isroflarini topish uchun har xil kesim yuzalardagi oqimning tezliklarini aniqlaymiz: v p .s= q ωp .s =10 ,3 20 ,5 =0,5 m/s; v p .s 2 2 g = 0,52 19 ,62 =0 ,013 m v yu .s= q ωyu .s =10 ,3 5,7 =1,8 m/s; v yu .s 2 2 g = 1,82 19 ,62 =0 ,177 m v = 4q πd2 = 4⋅10 ,3 3 ,14⋅2,52 =2,1 m/s; v 2 2g = 2,12 19 ,62 =0 ,225 m v 1= 4q πd1 2 = 4⋅10 ,3 3 ,14⋅1,72 =4 ,54 m/s; v 1 2 2 g = 4 ,542 19 ,62 =1 ,05 m v 2= 4q πd2 …

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⋅0 ,36=0 ,031 m;  kon=0,088 qabul qilinadi (1-ilova) 5) uzunlik l3 uchun: hl3 =λ l3 d4 ¿ v 4 2 2g =0 ,089⋅7,1 2,3 ⋅0 ,36=0 ,01 m bu erda, -ishqalanish koeffitsienti, reynolds soni re ga bog'liq. reynolds soni re kinematik yopishqoqlik koeffitsienti =0,0115 cm2/c (t=15c), bo'lganda, re= v 4⋅d4 ν =264⋅230 0 ,0115 =5297042>2320=rekr demak, oqimning turbulent harakatli tartibi uchun 0089,0 )52,15297042lg8,1( 1 )52,1relg8,1( 1 22      6) konfuzor 2 uchun hkon2=ξkon2 v 3 2 2g =0 ,04⋅0 ,451=0 ,018 m. chunki tg θ ° 2 = d4−d3 2l5 =2,3−2,1 2⋅3,5 =0 ,03 =327 bo'lganda, 1-ilovadan  kon2=0,04 qabul qilinadi. so'rish truboprovodidagi bosim isroflari yig'indisi : hws=hpan+hkir+hkon1+hbur1+hl3+hkon2= =0,035+0,113+0,014+0,031+0,01+0,018=0,221m bosimli truboprovoddagi bosim isroflarini aniqlaymiz: 1) surma qulfak uchun: hkul=ξkul v 1 2 2g =0 ,07⋅1 ,05=0 ,074 m bu erda  kul=0,07 - qulfakni to'la ochiq holati uchun olinadi (1-ilova). 2) ikkita burilish uchun (=30). …

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2 2g (1.18) hvak=5,18 m. hman=hx+∑ hwx− d1 2 −0,3+ v yu .s . 2 −v 1 2 2g (1.20) hman=25 ,85m bu erda hx -geodezik haydash balandligi hx=yu.s.s - (p.s.s + hs)=53-(22+4)=27 m bosimlar o'lchash nuqtalari orasidagi balandlik 625,0525,03,085,0 4 3,0 2 31  dd z m. bundan tashqari vx=v1 va vs=v3 qabul qilinadi. aniqlangan qiymatlarni (1.8) formulaga qo'yib nasosning to'la bosimini tekshirib ko'ramiz: n=5 ,18+25 ,85+0 ,625+ 4 ,542−2 ,982 19 ,62 =32 ,20 m demak, (1,3) va (1.8) formulalar bilan topilgan nasosning to'la bosimi bir xil qiymatga ega (n=32,25 m). masala . bosim isroflari hw=19 m bo'lgan nasos qurilmasi pastki suv sathi p.s.s=10 m belgidan yuqori suv sathi yu.s.s.=24m belgiga suv haydayotgan bo'lsa so'rgich va uzatkich qismlariga o'rnatilgan vakuummetr bosim 18.2 m va manometrik bosim 10, 4m ko'rsatishlari asosida uning to'la bosimni aniqlang va kesimdagi bosim isroflari to'la bo'yicha bosimni tekshiring. hisoblarda z=6 m, pastki va yuqori …

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8-amaliy mashgulot. nasosning to'la bosimini aniqlash(masalalar) h=n g+∑ hw+ v yu .s 2 −v p .s 2 2g ; (1.3) bu erda , ng= yu.s.s.-p.s.s.=53-22=31 m yuqorida keltirilgan (1.4) va (1.5) formulalar bilan mahalliy va uzunlik bo'yicha gidravlik qarshiliklarga bosim isroflarini topish uchun har xil kesim yuzalardagi oqimning tezliklarini aniqlaymiz: v p .s= q ωp .s =10 ,3 20 ,5 =0,5 m/s; v p .s 2 2 g = 0,52 19 ,62 =0 ,013 m v yu .s= q ωyu .s =10 ,3 5,7 =1,8 m/s; v yu .s 2 2 g = 1,82 19 ,62 =0 ,177 m v = 4q πd2 = 4⋅10 ,3 3 ,14⋅2,52 =2,1 m/s; v 2 2g = 2,12 19 ,62 =0 ,225 m …

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