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MEKANIKA FLUIDA SOAL PENYELESAIAN KANTINULAR ALIRAN NINDAYU NURUL AMANAH 20110110115

MEKANIKA FLUIDA SOAL PENYELESAIAN KANTINULAR ALIRAN NINDAYU NURUL AMANAH 20110110115. Air mengalir dari kamar mandi lantai dasar melalui pipa menuju kamar mani lantai 1 dengan kecepatan 8 m/s, dengan d= 70cm, hitunglah kecepatan aliran pada pipa dengan d= 120cm jawab :

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MEKANIKA FLUIDA SOAL PENYELESAIAN KANTINULAR ALIRAN NINDAYU NURUL AMANAH 20110110115

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  1. MEKANIKA FLUIDASOAL PENYELESAIAN KANTINULAR ALIRANNINDAYU NURUL AMANAH 20110110115

  2. Air mengalirdarikamarmandilantaidasarmelaluipipamenujukamarmanilantai 1 dengankecepatan 8 m/s, dengan d= 70cm, hitunglahkecepatanaliranpadapipadengan d= 120cm • jawab : • D1= 70cm = 0,070m • V1 = 8 m/det • D2 = 120cm = 1,2m • V2 = ???

  3. Q1 = Q2 A1.V1 = A2.V2 ¼ π .d².V1 = ¼ π.d².V2 ¼ π (0,070)².8 = ¼π(1,2)².V2 0.56= 1.44V2 V2 = 0.39 m/det

  4. Air mengalirdarikolamAmenujukolamBmelaluipipasepanjang150 m dan diameter  15 cm. Perbedaanelevasimuka air keduakolamadalah3 m. Koefisiengesekan Darcy – Weisbachf = 0,025.Hitung alirankehilangantenagasekunderdiperhitungkan :

  5. jawab:Panjangpipa                  =             L = 150 m Diameter pipa               =             D = 15 cm = 0,15 mKoefisiengerakan            =             f = 0,025Kehilangantenaga            =             H = 3,0 m

  6. Kehilangantenagaterjadipadasambunganantarapipadankolam( titik P dan Q ), dandisepanjangpipa. H=hep +  hf  + heQ 3=0,5 V2/2g+ 0,025 *150/15* V2/2g+ V2/2g 3=26,5 *V2/2g                                 V = 1,49 • Debit aliran Q = AV = Pi / 4 * ( 0,15 )2 * 1,49 = 0,0263 m3 /d =26,3  liter / detik

  7. 3. Air bersihmengalirdarirumahjonimelaluipipa 1,2,3,4, diameter ppa 1=10mm, pipa 2=45mm,pipa 3=65,mm pipa 4=80mm, kecepatandialiranpipa 2= 4m/s danpipa 3=1,5m/s, dan debit dipipa 3 adalah 2X debit alirandipipa 4,hitunglah Q1, V1, Q2,D3,dan V4

  8. Jawab: Q2 = A2.V2 =¼.π.0,045².4 =0.006358m³/det =6,358l/det Q1=Q2=6,358 l/det A1.V1=6,358 l/det V1=6,358.10ˉ³ ÷ ¼. Π. 0,01² = 0.81 m/s Q2=Q3+Q4 =Q3+ 0,5 Q3 =1,5 Q3

  9. Q3 = Q2 ÷ 1,5 = 0,006358 ÷ 1,5 = 0,004233m³/s = 4,233l/det A3 = Q3 ÷ V3 = 0,004233 ÷ 1,5 = 0,002822 m² D3 = √A3 ÷ ¼ π = √4.0,002822 ÷ π =0,059 m Q4 = A4 . V4½ . V3. A3 = A4 .V4 = ½ .1,5 .0,002822 ÷ ¼. π . 0,03² = 0,23 m/det

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