Introduction to flight 7th edition pdf

Hence, from Eq. However, the flow velocities in the present problem require a compressible analysis. Make certain to examine the solutions of both Problems 4. At the point in question, the velocity is very near zero, and hence the point is nearly a stagnation point. Thus, combining Eqs. Since it costs money to produce a pressure difference say by running compressors or vacuum pumps , then a diffuser, the purpose of which is to improve the aerodynamic efficiency, allows the wind tunnel to be operated more economically.

Hence, p 7. Hence, the Rayleigh Pitot tube formula must be used. Then use Eq. This is a definition; it applies to all flows — subsonic or supersonic. Hence, Eq. The stagnation pressure at the nose of a body in a Mach 2 stream is the total pressure behind a normal shock wave, which is lower than the total pressure of the freestream, as calculated above.

This stagnation pressure at the nose of a blunt body is given by Eq. The required expansion ratio of The required reservoir pressure of The required reservoir temperature of K is tremendously large, especially when you remember that the surface temperature of the sun is about K.

For a continuous flow hypersonic tunnel, such high reservoir temperatures can not be handled. To obtain Mach 20, i. High hypersonic Mach numbers demand wind tunnels with very large exit-to-throat ratios.

In practice, this is usually obtained by designing the nozzle with a small throat area. The higher Mach number of 20 is achieved not by a large increase in exit velocity but rather by a large decrease in the speed of sound at the exit. This is characteristic of most conventional hypersonic wind tunnels — the higher Mach numbers are not associated with corresponding increases in the test section flow velocities.

This is considerably different than the velocity squared law. The comparison is only qualitative, however, because the Reynolds numbers in Problem 4. In any event, the bottom line of Problems 4. This speed is relative to the gas. Since the gas velocity is zero, then this is also the wave velocity relative to the pipe. One wave travels to the right along the positive x-axis with the velocity Since the gas itself is moving relative to the pipe, then the wave speed relative to the pipe is the sum of the wave speed relative to the gas and the speed of the gas relative to the pipe.

After 0. The weak pressure distributions, however, continue to propagate to the right and left at the speed of sound relative to the gas. The density of the element of air, assuming it has experienced an isentropic process, is, from Eq. This situation is a stable atmosphere because the isentropically perturbed element will tend to sink back down to its originally lower altitude.

Chapter 5 5. Make certain to read the code properly, and use only the unflapped data, as given above. Use cd for this value. From Problem 5. However, the conditions of Problem 5. Using the Prandtl-Glauert Rule, the high speed pressure coefficient is C p0 Hence, it is not valid for compressible flow. In the present problem, check the Mach number. Hence, the desired pressure coefficient is a low speed value, C p0. Thus, 0. When a critical Mach number is stated for a given airfoil in the literature, it is usually for a small cruising angle of attack.

Moreover, assume the wings can be approximated by a thin flat plate. In flight, both the top and bottom of the plate will experience skin friction, hence that total skin friction coefficient is 2 0.

However, in case b where the angle of attack is higher, the wave drag coefficient 0. In this case, wave drag dominates the overall drag for the plate. Compare the results of Problems 5. Hence L Hence, the total drag is simply the profile drag.

This is the profile drag coefficient, the sum of skin friction drag and pressure drag due to flow separation. To estimate the skin friction drag coefficient, from Eq. For this case, the pressure drag due to flow separation is negligible. The large increase in cd at higher values of cl, hence a, seen in Appendix D is due almost entirely to an increase in pressure drag due to flow separation.

The transition Reynolds number is , This implies that the turbulent skin friction coefficient based on the running length of surface from the leading edge to the transition point, xcr, is 0. Note: This problem is similar to worked Example 4. In Example 4. In our solution here for Problem 5. We do not need to calculate forces. Indeed, in the statement of the problem there is no velocity, density, or surface area given, so we can not calculate forces; none is needed.

It is instructive to work out Example 4. You get the same answer as was obtained in Example 4. If the boundary layer were laminar, then the fraction of drag due to friction is 0. If the boundary layer were turbulent, then the fraction of drag due to friction is 0.

These results highlight the large effect of a turbulent boundary on the drag of an airfoil as compared to a laminar boundary layer. Chapter 6 6. Our calculations have not taken the large drag rise at drag-divergence into account. Hence, the maximum velocities calculated above are somewhat higher than reality. The power required and power available are plotted below. For example, if the PR curve is constructed from 30 points instead of the six or eight points as above, the subsequent results for rate-of-climb and absolute ceiling will be more accurate than obtained above.

Some leeway on the students' answers is therefore advised. In my own experience, I am glad when the students fall within the same ballpark. This is the velocity at which the average force is evaluated. The engine produces horsepower supercharged to 17, ft. Above that altitude, we assume that the power decreases directly as the air density. However, the drag is the retarding force acting vertically upward, and it is 31, lb. At Mach one, the drag far exceeds the maximum downward force.

Hence, it is not possible for the airplane to achieve Mach one. However, the drag is the retarding force acting vertically upward, and it is 47, lb. Hence it is not possible for the airplane to achieve Mach 1. This is the required combined thrust from both engines. Assuming the engine thrust is proportional to density, as discussed in Section 6.

Hence, the total drag coefficient is that for zero lift. The airplane can not accelerate through Mach one going straight up at sea level. Hence, at all altitudes, the airplane can not accelerate through Mach one going straight up. Hence, the mass of the F is kg. Thus, F But it must be solved by trial and error. The difference is 0.

Chapter 7 7. To examine whether or not the model is balanced, first calculate CM 0. Marsden, A. Allen, Douglas R. McCormac and Russell H. Kulakowski , F. Kothari, I. Woodson, James R. Boyce and Richard C. Cover, Joy A. Solutions Manual by William H. Hayt Jr. Crowe, Donald F. Elger, John A. Roberson and Barbara C.

Meriam, L. Reitz, Frederick J. Solutions Manual by C. Alexander M. Moran, Howard N. Munson, Donald F. Young, Theodore H. Okiishi, Wade W. Incropera D. Petrucci; William S. Lang and G. Sepe, W. Milic and Z. Bertsekas and John N. Friedberg , Arnold J.

Insel , Lawrence E. Callister, Jr. Simon , Lawrence E. Schulz, Ajit D. Ahuja , Thomas L. Magnanti , James B. Steven C. I 6th Ed.