Introduction to fluid mechanics solution manual 7th edition

It starts with basic equations, then plainly states assumptions, and finally, relates results to expected physical behavior. Many of the steps involved in analysis are simplified by using Exceed. For those who have any questions, or want a receive a sample section before your purchase, please contact us via email: support getbooksolutions.

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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. Roberts, M. Hennessy, David A. Editionselected solutions student solution Solution manual to Fundamentals of Engineering Thermodynamics 6th Edition by.

Please contact the content providers to delete files if any and email us, we'll remove relevant links or contents immediately. Login Join User. Free Download Link1 Download Link 2. Fox Introduction to fluid mechanics 7th edition, by robert w. Fox , philip j.

Pritchard , alan t. Fox and McDonald 's Philip J.