AMEE 0 Introduction to Fluid Mechanics Instructor: Marios M. Fyrillas Email: m.fyrillas@frederick.ac.cy HOMEWORK ASSIGNMENT ON BERNOULLI S EQUATION. Conventional spray-guns operate by achieving a low pressure through high speed air passing through a nozzle (see figure) that pumps the liquid (paint) up into the stream of air. In an experiment, a compressor is used to supply air at kpa (gage). Determine the velocity of the air through the nozzle if the water in the container is elevated 0 cm. h= 0 cm 4
Point is the compressor tank where the pressure is 000 Pa. Take Bernoulli's equation between point and. p u gh p u gh 0 0 Because of the small density of air the gravity terms are negligible. Also the velocity in the compressor tank is 0. Hence p p u. Because the streamlines are parallel, only the hydrostatic pressure 0 need to be incorporated to relate the pressures between sections and. p gh p But p atm p atm (neglecting small pressure variation across the air stream) and combining above equations we obtain p p gh u u 000 000 9.80.. 8 m/s 000 p p gh atm. Water flows up ramp in a constant width rectangular channel at a rate Q 0.54 m /s. The upstream depth is y 0.7 m and the width of the channel is b m. a. Write the Bernoulli s equation for a streamline between sections and. Show the streamline b. Write the mass conservation for sections and. c. If the velocity downstream 0.79 m/s determine y and z. z
Consider Bernoulli Equation on the streamline joining the water interface at point and point : g g y z y z Eq. Mass conservation m m Q Q(incompressible) Q Q Q y b y bq Eq. Q 0.54 y b 0.7 Q Q yb y b 0.76 m/s Eq. 0.54 0.68 m 0.79 Assuming that the datum is at the level z, i.e. z 0 and substituting eq. and eq. into eq. we obtain 0.76 0.79 0.7 0+ 0.68 z 9.8 9.8 z 0.7+0.04-0.68-0.0=0.08. The flowrate of water under a sluice gate as shown in the figure is a function of the water depths upstream and downstream of the gate and the width of the sluice gate. Use the Bernoulli and continuity equations to determine the theoretical flowrate.
Consider Bernoulli Equation on the streamline joining the water interface at point and point : p p z z g g g g z z g g atm p p p Eq. Mass conservation m m (incompressible) Q z zbzb (same width b) z Substitude in eq. Q z z g z g z z g z z z z 0 g z z z / g z z olumetric flowrate Q A z. b z / z 4. A vertical nozzle projects a jet of water 50 m into the air. Use Bernoulli s equation to calculate the water velocity at the exit from the nozzle. Take Bernoulli's equation between point and. p u gh p u gh patm =0 patm 0 u gh u gh.4 m/s
5. In a rectangular channel the velocity is measured using a pitot tube as shown on Figure 4. Based on the measurements it was concluded that the height ( H ) is related to the depth ( y ) by the relation H 0. y. Determine: a. The velocity profile and the shear stress. b. If the cross-sectional area has dimensions 50 50 (cm) determine the volumetric flow rate. c. What second order correction should be included to the velocity equation in order to obtain a no shear stress condition on the surface y h. H () h y Flow in a channel p u gh p u gh h h p p u u 0 The hydrostatic pressure at points () and () can be obtained from statics: p g( h y) patm p g( H h y) patm Combine all three equations to obtain: g( H h y) patm g( h y) patm u Above equations can be simplified to obtain u gh u g 0. y u.98y but H 0. y
y 0.5 u.98y 0.99 u 0.99 0.5 Q area of triangular prism 0.5 0.4 m /s or Integrate 0.5 0.5 y 4 Q u d A u b d y.98y0.5 dy.980.5.980.5 0.4 m /s 0 0 u.98yay d u ( y 0.5) 0.98 ay.98 a 0.5 a.98 dy
6. Figure shows a pump that draws a liquid of density 90 kg/m from a sealed underground tank. In order to prime the pump with liquid prior to a pumping operation, the delivery valve is closed and a compressed air supply at a stagnation pressure of 5 bars is allowed to flow through the reducer shown in the figure. The tank and the pump body are connected as indicated on the figure. For the conditions shown, assume incompressible flow through the reducer and calculate the maximum values of h for which priming can be achieved. p =4.98 A Air in A 0.A Air h Figure 5: Reducer-nozzle Take Bernoulli's equation between points and. p u gh p u gh 500000 498000 0 0 Mass conservation between points and ua ua assume incompressible uaua ua0.ua u 0. u Hydrostatic pressure between points and p gh p
From Bernoulli's at point and mass conservation 500000 498000 p 0.u 500000 u 50000 0. 498000 From above and Bernoulli's equation at point p 500000 u 500000 50000 450000 From above and hydrostatic pressure equation h p p g 498000 450000 90 9.8 5. m 7. Air is drawn in the wind tunnel used for testing automobiles. Determine the manometer reading, h, when the velocity in the test section is 60 km/hr. Determine the difference between the stagnation pressure on the front of the automobile and the pressure in the test section. p atm Km/hr 900 kg/m Figure 5: Wind tunnel (Question ) 60 patm psection air.6 p gh p g 0.054 section water atm oil
8. A hovercraft (air cushion vehicle) is supported by forcing air into the chamber created by a skirt around the periphery of the vehicle as shown in Figure 7. The air escapes through the 0 cm clearance between the lower end of the skirt and the ground (or water). Assume the vehicle weighs 5000 kg and is essentially rectangular in shape, 0 b y 0 m. The volume of the chamber is large enough so that the kinetic energy of the air within the chamber is negligible. Determine the flowrate, Q, needed to support the vehicle. If the ground clearance were reduced to 4 cm, what flowrate would be needed? If the vehicle weight were reduced to 500 kg and the ground clearance maintained at 0 cm, what flowrate would be needed? Figure : Hovercraft The pressure inside the chamber must be greater than atmospheric, and provides a lift force mg 5000 9.8 F pa mg p( gage) A 0 0 Take a streamline from inside the chamber to the lower end of the skirt: p( gage) patm u pu Q uau (0000) 0
9. Water flows through the pipe contraction shown in the figure. For the given 0.-m difference in the manometer level, determine the flowrate as a function of the diameter of the small pipe, D. 0. Water flows through the pipe contraction shown in the figure. For the given 0.-m difference in the manometer level, determine the flowrate as a function of the diameter of the small pipe, D.
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. Water in a rectangular channel flows into a gradual contraction section as is indicated in the figure. If the flowrate is Q 0.708 m / s and the upstream depth is y 0.6 m, determine the downstream depth, y. Consider Bernoulli Equation on the streamline joining the water interface at point and point : y z y z g g Eq. Mass conservation m m (incompressible) Q Q =0.708 m /s (note that the width is not the same) 0.708 yb 0.6. Q y b y b Q 0.95 m/s Q 0.708 = Eq. y b y 0.94 Eq. Assuming that the datum is at the level z, i.e. z 0 0.95 0.708 y y y and substituting eq. and eq. into eq. we obtain 0.6+ 0 0 9.8 0.94 9.8 0.0 0.656= y Multiplying by y y 0.656y 0.0=0 and simplifying we obtain The solutions of the above equation are y -0.9, y 0.9, y 0.556
To determine which of the positive roots to choose we need to compare the values with the critical value. Q The specific energy head E y y g y b g de Q The minimum occurs when dy b g y Q y c c c 0 b g Note that because b is not the same the critical value is different at each section. 0.708 At section, the critical value is y c 0.5m.. g Since y y, it means that the flow upstream is subcritical. The flow will remain subcritical since the reduction in width is gradual, i.e. we will choose the biggest value, i.e. y 0.556m.