White fluid mechanics solution manual pdf

Gage A reads kPa absolute. Determine a the height h in cm; and b the reading of gage B in kPa absolute. This is a little off, thus our mass estimate must have been a little off. Solution: a The analytic formula is found by integrating Eq. To To 2! Let go of the card. Will the card stay attached when the glass is upside down? Yes: This is essentially a water barometer and, in principle, could hold a column of water up to 10 ft high! Modify Eq.

Solution: a Convert 24, ft to m. Find the standard temperature from Eq. The manometer fluid is mercury. We will study loss coefficients in Chap. Example 2. What is the air pressure in the closed chamber B? Here we must account for reservoir volume changes. If d is not small, this is a considerable difference, with surprisingly large error. What is the pressure difference between points 1 and 2 in the pipe? Assume the reservoir is very large.

Of course, you could also go straight down to the bottom of the tube and then across and up. Find the gage pressure, in Pa, in the air gap in the tank. Neglect surface tension.

Assume no change in the liquid densities. The mercury in the left vertical leg will drop Determine the pressure at point A in pounds per square foot.

Normal levels, however, are 2. The manometer uses mercury and air as fluids. Determine the total pressure drop and also the part due to friction only. Which part does the manometer read? Is it higher or lower than Patmosphere?

Estimate the specific gravity of fluid X. The pump stops when it can no longer raise the water pressure.

Combining these two gives a quadratic equation for H: 0. Solution: a The vertical elevation of the water surface in the slanted tube is 1.

Stay with BG units. Some teachers 6 ft say it is more instructive to calculate these by direct integration of the pressure forces. Find the water force on the panel and its line of action. From Eq. The center of pressure is thus 3. The three bottom shapes and the fluids are the same. This is called the hydrostatic paradox. Explain why it is true and sketch a freebody of each of the liquid columns. In b side pressures are horizontal. In c upward side pressure helps reduce a heavy W.

For what water depth h is this condition reached? Neglecting atmospheric pressure, find the resultant hydrostatic force on panel BC, a from a single formula; b by computing horizontal and vertical forces separately, in the spirit of curved surfaces. What water level h will dislodge the gate? Neglect Fig.

The CG would be For what water depth h will the force at point B be zero? The centroid of Ans. This weight acts downward at the CG of the full gate as shown not the CG of the submerged portion.

Thus, W is 7. For Ans. If the water level is high enough, the gate will open. Compute the depth h for which this happens. The forces on AB and BC are shown in the freebody at right. This solution is independent of both the water density and the gate width b into the paper. Determine the required force P for equilibrium. Thus it is 3. Find the hydrostatic force on surface AB and its moment about C.

Could this force tip the dam over? Would fluid seepage under the dam change your argument? As shown in the figure, the line of action of F is 2. This moment is counterclockwise, hence it cannot tip over the dam. If there were seepage under the dam, the main support force at the bottom of the dam would shift to the left of point C and might indeed cause the dam to tip over.

The Fig. What horizontal force P is required at point B for equilibrium? The gate weight of N is assumed at the centroid of the plate, with moment arm 0. If not, Fig. Neglect the atmosphere. Because the actual plate force is not vertical. The hinge 15 cm hinge h is 15 cm from the centerline, as shown. Then the moment about the hinge is 0. Just iterate once or twice.

What sphere diameter is just right to close the gate? What is the water depth h which will first cause the gate to open? The hinge at A is 2 ft above the freshwater level.

Find h when the gate opens. Solution: Find the force on each panel and set them equal: Fig. Compute a the hydrostatic force of the water on the panel; b its center of pressure; and c the moment of this force about point B. Neglect atmospheric pressure.

CD is longer than AB, but its centroid is not as deep. If you have a great insight, let me know. Neglecting atmospheric pressure, determine the lowest level h for which the gate will open.

Is your result independent of the liquid density? The width into the paper is b. Solution: The critical angle is when the hydrostatic force F causes a clockwise moment equal to the counterclockwise moment of the dam weight W.

The negative sign occurs because the sign convention for dF was a downward force. Determine the horizontal and vertical components of hydrostatic force against the dam and the point CP where the resultant strikes the dam. Find the force F just sufficient to keep the gate from opening.

The gate is uniform and weighs lbf. It is cm wide into the paper. Find a the vertical and b horizontal water forces on the panel. This is a rectangle, 75 cm by cm, and its centroid is This is in two parts 1 the weight of the rectangular portion above the line AC; and 2 the little curvy piece above the parabola and below line AC.

Recall from Ex. Find the horizontal force P required to hold the gate stationary. The width b into the paper is 3 m. Neglect the weight of the gate. For the position shown, determine a the hydrostatic force on the gate per meter of width into the paper ; and b its line of action. Does the force pass through point O? This force passes, as expected, right through point O. What is the force in each bolt required to hold the dome down? If the end caps are neglected, compute the force in each bolt.

Derive an analytic expression for the hydrodynamic force F on the shell and its line of action. Which force is larger? Their horizontal forces equal the force on the projected plane AB. If the panel is 4 m wide into the paper, estimate the resultant hydrostatic force of the water on the panel. Compute the a horizontal and 1 m B b vertical water forces on the curved panel AB. Compute the horizontal and vertical hydrostatic forces on the gate and the line of action of the resultant force.

Water Determine a the horizontal and b the vertical hydrostatic forces on the hemisphere, in lbf. Compute the hydrostatic force on the conical surface ABC. Find a the hydrostatic force on the bottom; and b the force on a side panel. What is the specific gravity of fluid X? What is its weight in newtons?

Was it gold? When released into the U. Standard Atmosphere, at what altitude will it settle? If so, how many cubic centimeters will be exposed? Therefore the sphere is floating exactly half in and half out of the water.

Therefore it floats in gasoline. What diameter spherical balloon will just support the weight? Determine a the string tension; and b the specific gravity of the wood. What is the specific gravity of the rod material?

The rod is neutrally stable for any tilt angle! We see that the buoyancy exceeds the weight by 19 kN, or more than 2 tons. The sphere would float nicely. Estimate a the tension in the mooring line, and b the height in the standard atmosphere to which the balloon will rise if the mooring line is cut. They estimate the weight of the box and treasure in air at lbf.

Their plan is to attach the box to a sturdy balloon, inflated with air to 3 atm pressure. The empty balloon weighs lbf. Can you guess its name? Y This dimensionless parameter is commonly called the Capillary Number. If the outward net hydrostatic force on the cm by cm panel at the bottom is N, estimate a the pressure in the air space; and b the reading h on the manometer.

Solution: The force on the panel yields water gage pressure at the centroid of the Fig. It is hinged at B and rests against a smooth wall at A. Find the water level h which will just cause the gate to open. Pumps are in big trouble if the liquid vaporizes cavitates before it enters the pump.

Solution: a From Table A. Solution: Since power is a function of hp, Bernoulli is required. What happens? In order for international standards to be valid Fig. Tough calculation, no appendix tables for methane, should probably use EES. Download full file from buklibry.

Explain your vexation. Related Papers. By Luciano Teixeira de Morais. Fluid Mechanic White 5e Ch01 Solution. By caner celik. By Serhan Kasap. By Morteza Nemati. Download pdf.