Mekanika Fluida Minggu 03 Priyambodo, ST, MT Mochamad Yusuf Santoso, ST, MT, MSc

Fluid Statics – Statika Fluida Hydrostatics no relative motion between fluid particles, no shear stresses present (a shear results from a velocity gradient). Does not mean that the fluid particles are not moving, but they are not moving relative to one another; if they are moving, as in a can of water rotating about its axis, they move as a solid body. tidak ada gerak relatif antara partikel fluida, tidak ada tegangan geser yang terjadi (akibat geseran dari gradien kecepatan). Tidak berarti bahwa partikel fluida tidak bergerak, tetapi mereka tidak bergerak relatif terhadap satu sama lain, jika mereka bergerak, seperti dalam sekaleng air berputar pada porosnya, mereka bergerak sebagai benda padat.

Fluid Statics – Statika Fluida Hydrostatics The only stress involved in fluid statics is the normal stress, the pressure. It is the pressure acting over an area that gives rise to the forces in problems involving fluid statics. Three types of problems (1) fluids at rest, as in the design of a dam; (2) fluids undergoing linear acceleration, as in a rocket; and (3) fluids that are rotating about an axis. tegangan yang terlibat dalam statika fluida adalah tegangan normal, tekanan. Ini adalah tekanan yang bekerja di area yang menimbulkan kekuatan dalam masalah yang melibatkan statika fluida. Tiga jenis hal yang dibahas (1) fluida saat diam, seperti dalam desain bendungan; (2) fluida mengalami percepatan linear, seperti dalam roket, dan (3) fluida yang berputar terhadap suatu sumbu.

Fluid Statics – Statika Fluida Hydrostatics Hydrostatics is fundamental to hydraulics, the engineering of equipment for storing, transporting and using fluids. geophysics and astrophysics (understanding plate tectonics and the anomalies of the Earth's gravitational field), meteorology, medicine (in the context of blood pressure), Hydrostatics offers physical explanations why atmospheric pressure changes with altitude, why wood and oil float on water, why the surface of water is always flat and horizontal whatever the shape of its container. Hidrostatik adalah dasar hidrolika, untuk menyimpan, mengangkut dan menggunakan fluida geofisika dan astrofisika (memahami lempeng tektonik dan anomali medan gravitasi bumi), meteorologi, kesehatan (dalam konteks tekanan darah), Hydrostatics menjelaskan perubahan tekanan atmosfer dengan ketinggian, mengapa kayu dan minyak mengapung di atas air, mengapa permukaan air selalu datar dan horisontal apapun bentuk wadahnya.

Fluid Statics – Statika Fluida Pressure Variation ΔP = -Δh = -ρgΔh 1. Convert 230 kPa to millimeters of mercury, inches of mercury, and feet of water P = ρgh 230000 N/m2 = 13.6 x 1000 kg/m3 x 9.81 m/s x h h = 1.724 m Hg h = 1726 mm Hg h = 67.87 inch Hg 230000 N/m2= 1000 kg/m3 x 9.81 m/s x h h = 23.445 m Water h = 76.923 ft Water 2. If the pressure at the bottom of the tank is 231.3 kPa, what is the specific gravity of olive oil? S SAE 30 oil = 0.89 S water = 1 S Hg = 13.6 Patm+PSAE 30 Oil+Pwater+Polive oil +PHg = 231.3 kPa 101.03 kPa+ρgh SAE 30 Oil+ρgh water+ρgh olive oil + ρgh Hg = 231.3 kPa 101.03 kPa+(0.89)(9.81)(1.5)+ (1)(9.81)(2.5)+(S) (9.81) (2.9) +(13.6) (9.81) (0.4) = 231.3 kPa ρ olive oil = 1.38

Fluid Statics – Statika Fluida 3. Find FD FA.lA = FB.lB ; FB = 843.3N PC = PD = FC/AC = FD/AD FD = 15835.3N specific weight  = ρg (Gamma) 4.Find h if gasoline = 6670 N/m3, air = 11.8 N/m3 Full of Gasoline, Pgage = gasoline h = 6670 x 0.32 = 2134.4 Pa Contaminated by water, Pgage = water h water + gasoline h gasoline + air hair 2134.4 Pa = 9810 x 0.03 + 6670 x (0.32-0.03-h) + 11.8 x h h = 0.014 m

Fluid Statics – Statika Fluida 5. Find PA, PB, PC, PD PA = ρwater ghA = 1000 x 9.81 x - (0.4+0.4) = - 7.848 kPa (vacuum) PB = ρwater ghB = 1000 x 9.81 x (0.5) = 4.905 kPa PB = PC => abaikan udara PD = PC + ρoil ghD = 4.905 kPa + 0.9 x 1000 x 9.81 x (0.4+0.5+1) = 21.68 kPa Manometer P2 = Patm + ρgh

Fluid Statics – Statika Fluida 6. Find PA – PB PA + PSAE+ PHg = PB + PCCl4 PA+ρSAE ghSAE+ρHg ghHg= PB + ρCCl4ghCCl4 PA – PB = ρCCl4ghCCl4 - ρSAE ghSAE-ρHg ghHg PA – PB =1.59 x 1000 x 9.81 x 0.8 – 0.89 x 1000 x 9.81 x 1.1 – 13.6 x 1000 x 9.81 x 0.3 PA – PB = - 37150.47 N/m2 PB > PA

Fluid Statics – Statika Fluida 7. A manometer connects an oil pipeline and a water pipeline. Determine the difference in pressure between the two pipelines using the readings on the manometer. S oil =0.86 ; S Hg =13.6 p2=p3 pwater +  water x 0.04 = p4 +  Hg x 0.08 p4 = p5 p4 = p oil -  oil x 0.06 p water - p oil = -  water x 0:04 +  Hg x 0:08 -  oil x 0.06 p water – p oil = -9800 x 0.04 +(13.6 x 9800) 0.08 – (0.86 x 9800) 0.06 = 10 780 Pa 8. Find P P = Patm + ρgh P = 96 kPa + 0.85 x 1000 kg/m3 x 9.81 m/s2 x 0.55 m P = 96 kPa + 4586.175 N/m2 P = 96 kPa + 4.586 kPa P = 100.586 kPa