# 53 Worked Examples: Bluff Body Flows

These worked examples have been fielded as homework problems or exam questions.

Worked Example #1

We are told the component is a bluff body with and an area ft. The drag will be

where slugs/ft based on MSL ISA conditions. Therefore,

remembering to convert from mph to ft/s, we must multiply by 1.467. So it is now a matter of plotting (in lb) versus wind speed (in mph) over the range asked for using the previous equation, as shown in the plot below.

Worked Example #2

A payload crate is dropped by parachute from a military transport airplane 2,000 m above the ground. The parachute quickly opens, and the crate descends vertically toward the ground. The crate has a mass of 76 kg, a projected cross-sectional area of 0.4 m, and a drag coefficient of 1.5 based on this area. The parachute has an area of 52 m and a drag coefficient of 1.87 based on its projected area. You may assume the average air density during the descent is 1.19 kg/m. (a) Determine the vertical airspeed at which the crate will parachute to a landing. (b) How long will it take (approximately) for the crate to reach the ground?

(a) In a steady descent, then the net drag will be the sum of the drag produced by the parachute plus the drag of the crate, and this net drag must equal the weight of the crate, i.e.,

where is the drag coefficient of crate and is the drag coefficient of the parachute. We are asked to solve for the vertical airspeed so

Rearranging and solving for the airspeed gives

Inserting the known values gives

(b) The time for the decent is found using

Worked Example #3

A general aviation aircraft is fitted with a ballistic parachute recovery system for improved safety in the case of a severe emergency. Consider a scenario where the aircraft’s pilot has to deploy this parachute system from an altitude of 3 km. Assume the aircraft has a net mass of 1,200 kg, a projected cross-sectional area of 25 m, and a corresponding drag coefficient of 1.8 based on this area. The parachute has an area of 150 m and a drag coefficient of 1.5 based on this area. Assume an average air density of 1.01 kg m during the descent. Disregard the weight of the parachute. Determine the vertical airspeed at which the aircraft will parachute to a landing. Approximately how long will the decent take?

The aircraft’s given drag coefficient is two orders of magnitude greater than expected during normal flight. In normal flight, the airplane is streamlined, with minimal pressure drag and boundary layer separation. However, during descent with a parachute, the airplane behaves as a bluff body with large amounts of flow separation and corresponding pressure drag. During a steady descent, the net vertical drag on the airplane will equal the weight of the airplane. The net drag will be the sum produced by the parachute plus that from the airplane itself.

The net drag on the airplane and parachute will be equal to the weight, so

where is the rate of descent, is the projected cross-sectional area of the airplane, is the drag coefficient of the airplane based on this reference, is the projected cross-sectional area of the parachute, is the drag coefficient of the parachute based in this reference area. Simplifying gives

and rearranging to solve for gives

Substituting in the known numerical values gives

To compute the descent time , we can assume that the aircraft travels at this vertical speed for the entire change in altitude so that.

Worked Example #4

A reentry vehicle has a weight of 7,000 lbs. At an altitude of 2,000 ft, a certain number of parachutes (to be determined) must open, each with a diameter of 115 ft as the vehicle descends into the ocean. Based on their projected area, these parachutes have a drag coefficient of 1.6. Assume that after the parachutes are deployed, the vehicle quickly reaches a descent equilibrium velocity of . During this time, the average air density is 0.0023 slug ft. Ignore drag forces on the vehicle itself.

Determine an equation for the vehicle’s rate of descent with the parachutes open. For example, if the maximum allowable impact speed of the vehicle with the ocean is 10 ft/s, how many parachutes are required?

During a steady descent, the net vertical drag on the reentry vehicle from the parachutes must equal the vehicle’s weight, i.e., , where is the drag of any one parachute. The aerodynamic drag on any one parachute (a bluff body) will be

where is the rate of descent, is the projected cross-sectional area of any parachute, and is the drag coefficient of the parachute based on this reference area. The net drag on the parachutes will be equal to the weight of the vehicle, so the force equilibrium equation is

Rearranging to solve for gives

which will determine the rate of descent for a given number of parachutes.

We are given the maximum allowable rate of descent, so further rearranging the force equilibrium equation to solve for gives

Substituting in the known numerical values gives

so we will need four (4) parachutes to reduce the rate of descent to 10 ft/s.

Worked Example #5

As part of a design of a utility plant in Florida, a solid hemisphere must be tested in the wind tunnel to measure its drag up to hurricane speeds. The hemisphere is 19 inches in diameter. It is to be tested in two conditions, the first with the rounded side into the wind and the second with the flat side into the wind. As part of the test plan, the drag force on the hemisphere must be estimated before it is mounted on the wind tunnel balance. Make a plot of the estimated drag force on this hemisphere, in both conditions, as a function of wind speed up to 145 mph. You may assume MSL ISA conditions.

The drag on the hemisphere will be given by the usual formula, i.e.,

where the reference area is

The drag is to be measured in two orientations, the first with the rounded side to the wind, i.e.,

where according to information given . In the second case with the flat side into the wind then

where .

We are told the airspeed range is up 145 mph, which is 212.67 ft/s. Therefore, we must plot two curves for and as a function of airspeed. Substituting the numbers, and using MSL ISA conditions that = 0.002378 slugs/ft gives

and

where would be in units of ft/s. The results show in the plot below that that drag will vary up to about 45 lb with with the rounded side to the wind and up to about 115 lb with the flat side into the wind. Notice that the wind speed has been converted back to mph for the purposes of plotting.

Worked Example #6

A helium-filled, spherical-shaped balloon with a smooth surface is tethered in a horizontal wind flow of 10.3 mph, as shown in the figure below. The diameter of the balloon is 55 cm and the mass of the balloon material (not including the gas inside) is 20 grams. The pressure inside the balloon is 120 kPa. Assume MSL ISA atmospheric ambient conditions and for helium = 2,077.0 J Kg K.

- Draw a free-body diagram to show the forces acting on the balloon at its equilibrium condition.
- Determine the density of the helium contained inside the balloon at its equilibrium condition.
- What is the Reynolds number for the flow around the balloon?
- Determine the drag coefficient of the balloon. Hint: Use the chart for for a sphere.
- Calculate the value of the equilibrium angle .

The buoyancy force, weight, aerodynamic drag, and tension from the string should be shown on a free-body diagram, such as shown below.

1. Using the equation of state, the density of the helium gas in the balloon at 15C or 288.15 K is

2. The wind speed in base units of m/s is

Therefore, the Reynolds number based on the diameter of the balloon and the wind speed is

3. At , the value of is approximately 0.49, per the standard drag chart for a sphere.

4. The projected frontal cross-sectional area of the balloon is

This means that the drag force on the balloon is

5. The volume of the spherical balloon is

In the vertical direction, the upward buoyancy force and weight must be considered. The buoyancy force is

The weight of the balloon, , including the helium gas inside, is

Inserting the known numerical values gives

Finally, the equilibrium angle is determined from

Therefore,