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Q1 = Q2 A1.V1=A2.V2 ¼ π . d².V1=¼ π .d².V2 ¼ π ( 0,065)².5 =( ¼π . 1,5)².V2 0,021 = 2,25 V2

MEKANIKA FLUIDA MATERI : SOAL PENYELESAIAN KANTINULAR ALIRAN NAMA : YUSTI MARSELI NIM : 20110110110. Air mengalir dari kamar mandi lantai dasar melalui pipa menuju kamar mani lantai 1 dengan kecepatan 5 m/s, dengan d= 65cm, hitunglah kecepatan aliran pada pipa dengan d= 150cm jawab :

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Q1 = Q2 A1.V1=A2.V2 ¼ π . d².V1=¼ π .d².V2 ¼ π ( 0,065)².5 =( ¼π . 1,5)².V2 0,021 = 2,25 V2

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  1. MEKANIKA FLUIDAMATERI : SOAL PENYELESAIAN KANTINULAR ALIRANNAMA : YUSTI MARSELINIM : 20110110110 Air mengalirdarikamarmandilantaidasarmelaluipipamenujukamarmanilantai 1 dengankecepatan 5 m/s, dengan d= 65cm, hitunglahkecepatanaliranpadapipadengan d= 150cm jawab : D1= 65cm = 0,065m V1 = 5 m/det D2 = 150cm = 1,5m V2 = ???

  2. Q1 = Q2 A1.V1=A2.V2 ¼π.d².V1=¼ π.d².V2 ¼π(0,065)².5 =(¼π. 1,5)².V2 0,021 = 2,25 V2 V2 = 107,14 • Air mengalirdarikolamAmenujukolamBmelaluipipasepanjang150 m dan diameter  15 cm. Perbedaanelevasimuka air keduakolamadalah3 m. Koefisiengesekan Darcy – Weisbachf = 0,025.Hitung alirankehilangantenagasekunderdiperhitungkan : jawab:

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

  4. 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

  5. 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 • 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

  6. Q3 = Q2 ÷ 1,5 = 0,006358 ÷ 1,5 = 0,004233m³/s = 4,233l/detA3 = Q3 ÷ V3 = 0,004233 ÷ 1,5 = 0,002822 m²D3 = √A3 ÷ ¼ π = √4.0,002822 ÷ π =0,059 mQ4 = A4 . V4½ . V3. A3 = A4 .V4 = ½ .1,5 .0,002822 ÷ ¼. π . 0,03² = 0,23 m/det

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