Examples – Bluff Body Flows

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

Worked Example #1

As part of the design of a building structure in Florida, a component is to be tested in a wind tunnel to measure its drag up to Category 5 hurricane speeds. The component is in the form of a bluff body with an estimated drag coefficient of 0.98. The projected cross-sectional area of the component is 3.26 ft^2. As part of the test plan, the drag force on the component must be estimated before it is mounted on the wind tunnel balance system. Make a plot of the estimated drag force as a function of wind speed up to 156 mph. Assume MSL ISA conditions.

We are told the component is a bluff body with C_D = 0.98 and an area S = 3.26 ft^2. The drag D will be

    \[ D = \frac{1}{2} \rho_{\infty} V_{\infty}^2 S C_D \]

where \rho_{\infty} = 0.002378 slugs/ft^3 based on MSL ISA conditions. Therefore,

    \[ D = \frac{1}{2} \times 0.002378 \times 1.467^2 V_{\infty}^2 \times 3.26 \times 0.98 = 0.008175 V_{\infty}^2 \]

remembering to convert from mph to ft/s then we must multiply by 1.467. It is now a matter of plotting D (in lb) versus wind speed (in mph) over the range asked for by 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 at 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, and has a projected cross-sectional area of 0.4 m^2 and a drag coefficient of 1.5 based on this area. The parachute has an area of 52 m^2 and a drag coefficient of 1.87 based on its projected area. You may assume the average density of the air during the descent is 1.19 kg/m^3. (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.,

    \[ D = W = M g = \frac{1}{2} \rho V^2 S_{\rm body} \, C_{D_{1}} + \frac{1}{2} \rho V^2 S_{\rm parachute} \, C_{D_{2}} \]

where C_{D_{1}} is the drag coefficient of crate and C_{D_{2}} is the drag coefficient of the parachute. We are asked to solve for the vertical airspeed so

    \[ W = M g = \frac{1}{2} \rho V^2 \left( S_{\rm body} \, C_{D_{1}} + S_{\rm parachute} \, C_{D_{2}} \right) \]

Rearranging and solving for the airspeed gives

    \[ V = \sqrt{ \frac{2 \, M \, g}{\rho \left( S_{\rm body} \, C_{D_{1}} + S_{\rm parachute} \, C_{D_{2}} \right) } } \]

Inserting the known values gives

    \[ V = \sqrt{ \frac{2 \times 76.0 \times 9.81}{1.19 \times \left( 0.4 \times 1.5 + 52.0 \times 1.87 \right) } } = 3.58~\mbox{m/s} \]

(c) The time t for the decent is found using

    \[ t \approx \frac{2000.0}{3.58} = 559~\mbox{seconds} = 9.3~\mbox{minutes} \]

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 pilot of the aircraft 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^2, and a corresponding drag coefficient of 1.8 based on this area. The parachute has an area of 150 m^2 and a drag coefficient of 1.5 based on this area. Assume an average air density of 1.01 kg m^{-3} 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 what would be expected during normal flight. In normal flight, the airplane is streamlined, with minimal pressure drag and boundary layer separation. However, during decent 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 of that produced by the parachute plus that from the airplane itself.

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

    \[ W = M \, g = \frac{1}{2} \rho V_d^2 S_a C_{D_{a}} + \frac{1}{2} \rho V_d^2 S_p C_{D_{p}} \]

where V_d is the rate of descent, S_a is the projected cross-sectional area of the airplane, C_{D{a}} is the drag coefficient of the airplane based on this reference, S_p is the projected cross-sectional area of the parachute, C_{D{p}} is the drag coefficient of the parachute based in this reference area. Simplifying gives

    \[ M \, g = \frac{1}{2} \rho V_d^2 \left( S_a C_{D_{a}} + S_p C_{D_{p}} \right) \]

and rearranging to solve for V_d gives

    \[ V_d = \sqrt{ \frac{2 M \, g}{\rho \left(S_a C_{D_{a}} + S_p C_{D_{p}} \right)}} \]

Substituting in the known numerical values gives

    \[ V_d = \sqrt{ \frac{2 \times 1,200.0 \times 9.81 }{1.01 \times \left(25.0 \times 1.8 + 150.0 \times 1.5 \right)} } = 9.29~\mbox{m/s} \]

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

    \[ t_d = \frac{h}{V_d} = \frac{3,000}{9.29} = 323~\mbox{secs} = 5.4~\mbox{mins} \]

Worked Example #4

A reentry vehicle has weight of 7,000 lbs. At an altitude of 2,000 ft, a certain number of parachutes N_p (to be determined) must open, each with a diameter d of 115 ft as the vehicle descends into the ocean. These particular parachutes have a drag coefficient C_D of 1.6 based on their projected area. Assume that after the parachutes are deployed, the vehicle quickly reaches a descent equilibrium velocity of V_d. During this time the average air density is \rho = 0.0023 slug ft^{-3}. Ignore drag forces on the vehicle itself.

Determine an equation for the rate of descent of the vehicle with the parachutes open. 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 N_p parachutes must equal the weight of the vehicle, i.e., N_p D = W, where D is the drag of any one parachute. The aerodynamic drag on any one parachute (a bluff body) will be

    \[ D = \frac{1}{2} \rho V_d^2 S C_{D} \]

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

    \[ W = M g = N_p \left( \frac{1}{2} \rho V_d^2 S C_{D} \right) = \frac{1}{2} \rho V_d^2 S C_{D} N_p \]

Rearranging to solve for V_d gives

    \[ V_d = \sqrt{ \frac{2 M g}{\rho S C_D N_p} } \]

which will allow the rate of descent to be determined 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 N_p gives

    \[ N_p = \frac{ 2 M g}{\rho V_d^2 S C_D} = \frac{ 2 M g}{\rho V_d^2 (\pi d^2/4) C_D} \]

Substituting in the known numerical values gives

    \[ N_p = \frac{ 2 \times 7,000}{0.0023 \times 10.0^2 \, \pi \times (115.0^2/4.0) \times 1.6} = 3.66 \]

so we will need 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 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.,

    \[ D = \frac{1}{2} \rho V^2 A_{\rm ref} \, C_D \]

where the reference area A_{\rm ref} is

    \[ A_{\rm ref} = \frac{ \pi D^2}{4} = \frac{ \pi (19/12)^2}{4} = 1.97~\mbox{ft$^2$} \]

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

    \[ D_1 = \frac{1}{2} \rho V^2 A_{\rm ref} \, C_{D_{1}} \]

where according to information given C_{D_{1}} = 0.42. In the second case with the flat side into the wind then

    \[ D_2 = \frac{1}{2} \rho V^2 A_{\rm ref} \, C_{D_{2}} \]

where C_{D_{2}} = 1.17.

We are told the airspeed range is up 145 mph, which is 212.67 ft/s. Therefore, we must plot two curves for D_1 and D_2 as a function of airspeed. Substituting the numbers, and using MSL ISA conditions that \rho = 0.002378 slugs/ft^3 gives

    \[ D_1 = \frac{1}{2} \rho V^2 A_{\rm ref} C_{D_{1}} = \frac{1}{2} \times 0.002378 \times 1.97 \times 0.42 V^2 = 0.000984 V^2 \]

and

    \[ D_2 = \frac{1}{2} \rho V^2 A_{\rm ref} C_{D_{2}} = \frac{1}{2} \times 0.002378 \times 1.97 \times 1.17 V^2 = 0.00274 V^2 \]

where V 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 V_\infty =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 R_{\rm He} = 2,077.0 J Kg^{-1} K^{-1}.

  1. Draw a free-body diagram to show the forces acting on the balloon at its equilibrium condition.
  2. Determine the density of the helium contained inside the balloon at its equilibrium condition.
  3. What is the Reynolds number Re for the flow around the balloon?
  4. Determine the drag coefficient C_D of the balloon. Hint: Use the chart for C_D = f (Re) for a sphere.
  5. Calculate the value of the equilibrium angle \theta.

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 15^\circC or 288.15 K is
    \[ \rho_{He} = \frac{p}{R_{He} \  T} = \frac{120,000}{2,077 \times 288.15} = 0.201~\mbox{kg m$^{-3}$} \]

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

    \[ V_\infty = 10.3\times 0.447 = 4.60~\mbox{m/s} \]

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

    \[ Re_D = \frac{\rho_{\rm air} V_\infty D}{\mu_{\rm air}} = \frac{1.225 \times 4.60 \times (55.0/100.0)}{1.789 \times 10^{-5}} = 173,394 \approx 173,400 \]

3. At Re = 173,400, the value of C_D is approximately 0.49, per the standard drag chart for a sphere.

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

    \[ A = \frac{\pi D^2}{4} = \frac{3.14 \times 0.55^2}{4} = 0.2376~\mbox{m$^2$} \]

This means that the drag force on the balloon is

    \[ D = \frac{1}{2}\rho_{\rm air} \, V_\infty^2 \, A \, C_D = 0.5 \times 1.225 \times 4.60^2 \times 0.2376 \times 0.49 = 1.06~\mbox{N} \]

5. The volume of the spherical balloon is

    \[ {\cal{V}} = \frac{4}{3}\pi \left(\frac{D}{2} \right)^3 = 0.0871~\mbox{m$^3$} \]

In the vertical direction, the upward buoyancy force F_{\rm up} and weight W must be considered. The buoyancy force is

    \[ F_{\rm up} = \rho_{\rm air}\, g \, {\cal{V}} = 1.225 \times 9.81 \times 0.0871 = 1.05~\mbox{N} \]

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

    \[ W = W_{b} + W_{\rm He} = M_{b} \, g + \rho_{\rm He} \, g \, {\cal{V}} \]

Inserting the known numerical values gives

    \[ W  = (20.0/1000.0) \times 9.81 + 0.201 \times 9.81 \times 0.0871 = 0.20 +  0.17 =  0.37~\mbox{N} \]

Finally, the equilibrium angle \theta is determined from

    \[ \tan \theta = \frac{F_{\rm up} - W}{D} = \frac{1.05 - 0.37}{1.06} = 0.6435 \]

Therefore, \theta = 32.8^\circ

 

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