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Tugas mekanika fluida Amalia septiani

Tugas mekanika fluida Amalia septiani. 20110110058 Teknik sipil Universitas muhammadiyah yogyakarta. Soal dan penyelesaian.

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Tugas mekanika fluida Amalia septiani

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  1. TugasmekanikafluidaAmaliaseptiani 20110110058 Tekniksipil Universitasmuhammadiyahyogyakarta

  2. Soaldanpenyelesaian • 1. air mengalirdidalampipaberdiameter 25 cm dengankecepatan 1m/d. padaujungpipatersebutterdapatpipadegan diameter 50 cm. berapakahkecepataaliranpadaujungtersebut ?

  3. penyelesaian Diameter pipa= 25 cm =0,25m Luastampangpipa =A1 = ¼ π DxD = ¼ π (0,25)x(0,25) = 0,0490 m² Diameter pipa= 50cm=0,5m Luastampangpipa =A2= 1/4 3,14 (0,5) (0,5) = 0,1963 Kecepatanaliran = v1 =1m/d Debit aliran = Q = A1xV1= 0,0490 m² Ditanyakan v2 ?

  4. A1xV1 = A2xV2 ¼ π D1² V1 = ¼ π D2² V2 ¼ 3,14 (0,5)² (1) = ¼ 3,14 V2 V2 = 0,0490 : 0,1963 = 0,249 m/d

  5. 2. Suatufluidamengalirmelaluisebuahpipaberjari-jari 6 cm dengankecepatan 4 m/s. Berapakah debit fluidatersebutdinyatakandalam m3/s dan m3/jam?

  6. r = 6 cm = 6 x 10-2 mv = 4 m/sDitanya: Q dalam m3/s dan m3/jam?Jawab:Q = v AA = πr2 = π(6 x 10-2 m)2= 0,0113 m2

  7. Sehingga,Q = v A= 4 m/s . 0,0113 m2= 0,045 m3/sSedangkan dalam satuan m3/jam:Q=(0,045m^3)/1s=(0,045m^3)/(1×1/3600 jam)=162 m^3/jam

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