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Solved Examples in fluid mechanics final
- 7. Valve A B Sections “A” and “B” in pipeline (shown schematically in the figure given below) are at the same elevation of 2.50 m above datum. A valve lies in- between “A” and “B”. The flow parameters at “A” are: Velocity head = 0.50 m Pressure head = 2.50 m, and Valve loss is 0.20 m The Piezometric head at “B” is: (a) 5.50 m (b) 5.30 m (c) 5.00 m (d) 4.80 m
- 12. Water is pumped from a reservoir through 150 mm diameter pipe and is delivered at a height of 15 m from center line of the pump through a 100 mm nozzle connected to a 150 mm discharge line (see the attached figure). If the pressure at pump inlet is 210 kN/ m2 absolute, inlet velocity of 6 m/s and the jet is discharged into atmosphere, determine: The energy supplied by the pump, (Assume atmospheric pressure as 101.3 kN/m2 and no friction)
- 14. The suction and delivery pipe diameters of a pump are 200 mm and 100 mm respectively. If the inlet and outlet pressures are 70 kN/m2 and 210 kN/m2 respectively and pump delivers 40 kw power to the fluid, find: The discharge of water flowing through the pump. (Assume exit to be 0.5 m above the inlet)
- 17. Fig ( ) shows a Venturi-meter with its axis vertical and arranged as suction device. The throat and outlet areas of the Venturi- are 0.00025 m2 and 0.001 m2 respectively. If the Venturi discharges into atmosphere, determine: The minimum discharge in the Venturi-meter at which flow will occur up the suction pipe.
- 18. Water 1.0 m 2.0 m 1.9 m 1 3 Water 2
- 19. Apply Bernoulli’s equation between sections “2” and “3” 𝑍2 + 𝑃2 𝜌 𝑔 + 𝑉2 2 2 𝑔 = 𝑍3 + 𝑃3 𝜌 𝑔 + 𝑉3 2 2 𝑔 …………………. (1) However according to continuity equation, 𝐴2 ∙ 𝑉2= 𝐴3 ∙ 𝑉3 or 0.00025 × 𝑉2= 0.001 × 𝑉3 i.e. 𝑉2= 4 𝑉3 and since pressure at “3” is atmospheric, take 𝑃3 = 𝑃𝑎𝑡𝑚. = o, and Eq. (1) gives 2.0 ? .90 0 𝑍2 + 𝑃2 𝜌 𝑔 + 𝑉2 2 2 𝑔 = 𝑍3 + 𝑃3 𝜌 𝑔 + 𝑉3 2 2 𝑔 16−1 𝑉2 2 2 𝑔 = 15 𝑉2 2 2 𝑔 = − 0.10 − 𝑃2 𝜌 𝑔 …………………. (2)
- 20. Further, since water will rise to section “2” 𝑍1 + 𝑃1 𝜌 𝑔 + 𝑉1 2 2 𝑔 = 𝑍2 + 𝑃2 𝜌 𝑔 + 𝑉2 2 2 𝑔 or, 𝑃2 𝜌 𝑔 = − 𝑍2 − 𝑍1 = −1.0 …………………. (2) From Eqs. “2” & “3” 15 𝑉2 2 2 𝑔 = − 0.10 − 𝑃2 𝜌 𝑔 = − 0.10 − +1.0 = 0.90 ∴ 𝑉2 = 2×9.81×0.90 15 = 1.85 𝑚 𝑠 The minimum discharge in the Venturi—meter 𝑄 = 𝐴 ∙ 𝑉 = 0.001 × 1.085 = 1.85 × 10−3 𝑚3 𝐿𝑠 = 1.085 𝐿 𝑠 = 0 = 0 = 0
- 29. Fig. 4.25
- 31. ُ قررت نأ تكون لكمىت انوبعن « » «Have a Dream ُ دلي ُملح » . حكيت للطالب ،الكنديني من صولأ كثرية من بيهنم ،يونرصم حاكيتني يفتنيرط . احلاكية وىلال عن « سطورةأ الكهف » كام تصورها ُ يلسوفلفا يقىرإغلا فأ الطون . حني تصور نأ بعض شخاصال حمبوسون داخل كهف منذ مودلمه . مقيدون ابلسالس ل وجوههم ُمةبمصو حنو حائط ل يرون ،اهوس تنعكس عليه ُ ظالل نَم ُ رييس ون ابخلارج من رشب اانتووحي . يتصورف لئكوأ ُ شخاصال ابلكهف نأ تكل الظالل ىه ُ لا ُ رشب ليسو ظاللها . فلو تصوران نأ ا ً خشص من لئكوأ شخاصال جنح ىف رسك القيود ُ و السالسل وخرج من الكهف ،املظل وشاهد ضوء الشمس ولل مرة ىف حياته . سوف تشفيك ب عد برهة نأ ما ظل اهري هو ورفاقه الوط عامرمهأ ليس إلا ،ظالل نأو احلقيقة ُمءيش ُ خرأ مت اًما . ذكل هو « التفكري خارج الصندوق » ، وعدم تسالمسالا للفرضيات املغلوطة همام ُ ع شتش داخلنا اتونس وعهود . ذكل هو ادلرس ولال اذلى عىل بنائناأ مهتعل ىف مقب ل ايهممأ .
- 33. هذان هام ادلرسان لذلانا ُ نيتمت نأ يدركهام التالميذ ،الكنديون وهام ذاهتام ُ رادل سان لذلانا حكهيامأ للتالميذ ينيرصامل اليوم عىل مشارف اجلامعة ا ًيضأ . ُيللوُىفُروايةولوُكومهُهوُماُقاهلُابالوُماُادلرسُالثالثأو « ميياىئخلا » . ُ إنسانلإنُحلُاا ُاًمُحل جلُحتقيقهُمعهُباكمهلُمنُأمرُالكونُتآ،ُعىلُحتقيقهرصأوُاادمنُبهُجأوُ،ما . ُمكرذكأو ُنيةغبآ I «Have a Dream» ُيقرلف ABBA ُالىت تقول : « ُ دلي ُملح / ُمةنيغأ غنهياأ / ىك تساعدىن عىل نأ اجهوأ ىأ ىشء . إذاا َُ شاهدت ال عجائب ىف حاكايت اتنيجلا / سوف تطيعتس نأ تصنع تقبلملسا / حىت لو خفقتأ م ًُةر / انأ ؤمنأ ملالئكةاب / ُ اليشء ُ الطيب ىف لك يشء اهرأ / انأ ؤمنأ ابملالئكة / حيامن ُ عرفأ نأ الوقت ىف صاحلى / سوف ُ عربأ الشالل ُ جتازأو الهنر / نل ُ دلي اًمحل » .
- 35. A rectangular tank is filled to the brim with water « مملوء حيت احلافة ابملاء » . When a hole at its bottom is unplugged, the tank is emptied in time T. How long will it take to empty the tank if it is half filled? Now let us consider that at certain moment the height of water level be ”h” and the velocity of water emerging through the orifice of cross sectional area ”a” at the bottom of the tank be “V”. As the surface of water and the orifice are in open atmosphere, then by Bernoulli's theorem we have: 𝑉 = 2 𝑔 ℎ . Let ”dh” represents decrease in water level during infinitesimally small time interval ”dt” when the water level is at height ”h”. So the rate of decrease in volume of water will be − A (dh/ dt). Again the rate of flow of water at this moment through the orifice is given by 𝑎 × 𝑉 = 𝑎 ∙ 2 𝑔 ℎ.
- 36. These two rate must be same by the principle of continuity. Hence 𝑎 ∙ 2 𝑔 ℎ = − 𝐴 𝑑ℎ 𝑑𝑡 → 𝑑𝑡 = 𝐴 𝑎 ∙ 2 𝑔 ℎ− 0.5 𝑑ℎ If the tank is filled to the brim then height of water level will be ”H” and time required to empty the tank “T” can be obtained by integrating the above relation. 𝑇 = 𝑜 𝑇 𝑑𝑡 = − 𝐴 𝑎 ∙ 2 𝑔 ∙ 0 𝑇 ℎ− 0.5 𝑑ℎ 𝑇 = 𝐴 𝑎 ∙ 2 𝑔 ∙ ℎ0.5 0.5 0 ℎ 𝑇 = 𝐴 𝑎 ∙ 2 𝑔 ∙ ℎ0.5
- 37. If the tank be half filled with water and the time to empty it be 𝑇′ then So 𝑇′ 𝑇 = 1 2 ∴ 𝑇′ = 𝑇/ 2 𝑇′ = 𝐴 𝑎 ∙ 2 𝑔 ∙ ℎ 2 0.5