MAE 101B Homework 5: Solutions

5/18/2018

1

Munson 9.47 – How fast do small water droplets of 0.06 μm (6 × 10−8 m) diameter fall through the air under standard sea-level conditions? Assume the drops do not evaporate. Repeat the problem for standard conditions at 5000 m altitude. Given Model Diagram 𝐷 = 6 × 10−8 m Droplet is a sphere Stokes flow Standard atmosphere Basic Equations No evaporation

∑ �⃗⃗� = 𝑚�⃗⃗� Terminal velocity reached 𝑅𝑒𝑥 = 𝑈𝑥/𝜈

𝐶𝔇 =𝔇

1

2𝜌𝑈2𝐴

𝐴 =𝜋

4𝐷2

∀=𝜋

6𝐷3

Analysis At terminal velocity, �⃗⃗� = 0, a force balance exists. Consider the 𝑧 (vertical) force balance.

∑𝐹𝑧 = 𝔇 + 𝐹𝐵 − 𝑊 = 0

𝔇 = 𝑊 − 𝐹𝐵

Assume 𝑅𝑒 ≪ 1 , which is likely due to the very small size, allowing for the use of Stoke’s flow (low 𝑅𝑒

analysis). From Table 9.4 or Figure 9.23a:

𝐶𝔇 =24

𝑅𝑒=

𝔇

12𝜌𝑈2𝐴

𝔇 =24𝜇𝑎𝑖𝑟

𝜌𝑎𝑖𝑟𝑈𝐷

1

2𝜌𝑎𝑖𝑟𝑈

2𝜋

4𝐷2

𝔇 = 3𝜋𝐷𝑈𝜇𝑎𝑖𝑟

Substitute the results of Stokes flow into the force balance,

3𝜋𝐷𝑈𝜇𝑎𝑖𝑟 = 𝑊 − 𝐹𝐵

3𝜋𝐷𝑈𝜇𝑎𝑖𝑟 = ∀𝑔(𝜌H2O − 𝜌𝑎𝑖𝑟)

𝑈 =∀𝑔

3𝜋𝐷𝜇𝑎𝑖𝑟(𝜌H2O − 𝜌𝑎𝑖𝑟)

The volume of a sphere is ∀=𝜋

6𝐷3

𝑈 =𝑔𝐷2

18𝜇𝑎𝑖𝑟(𝜌H2O − 𝜌𝑎𝑖𝑟)

At sea level At ℎ = 5000 m

𝑈 =(9.807 m/s2)(6×10−8 m)

2

18(1.789×10−5 N∙s/m2)(999

kg

m3 − 1.225kg

m3 ) 𝑈 =(9.781 m/s2)(6×10−8 m)

2

18(1.628×10−5 N∙s/m2)(103 kg

m3 − 0.7364kg

m3 )

𝑈 = 1.0939 × 10−7 m/s 𝑈 = 1.2014 × 10−7 m/s

2

Now checking our assumption of 𝑅𝑒 ≪ 1:

𝑅𝑒 =𝜌𝑎𝑖𝑟𝑈𝐷

𝜇𝑎𝑖𝑟 𝑅𝑒 =

𝜌𝑎𝑖𝑟𝑈𝐷

𝜇𝑎𝑖𝑟

𝑅𝑒 =(1.225 kg/m3)(1.0939×10−7 m/s)(6×10−8 m)

(1.789×10−5 N∙s/m2) 𝑅𝑒 =

(0.7364 kg/m3)(1.2014×10−7 N∙s/m2)(6×10−8 m)

(1.628×10−5 N∙s/m2)

𝑅𝑒 = 4.5 × 10−10 𝑅𝑒 = 3.26 × 10−10

Discussion Both flows have small Reynolds numbers. As a result, employing Stokes flow (low 𝑅𝑒 flow over a sphere) is justified. This problem is similar to the iterative pipe flow problems. Since the velocity is unknown we could not evaluate the drag force directly, as a result we had to assume the flow behavior, calculate 𝑈 and 𝑅𝑒, and verify our assumption was correct. In this case 𝜌𝑎𝑖𝑟 ≪ 𝜌𝑤𝑎𝑡𝑒𝑟 making the buoyancy force negligible. However, if the fluids were reversed, air bubble in water, the buoyancy force would dominate, and the weight would be insignificant.

3

Munson 9.54 – A coal barge 1000 ft long and 100 ft wide is submerged a depth of 12 ft in 60 ℉ water. It is being towed at a speed of 12 mph. Estimate the friction drag on the barge. Given Model 𝑈 = 12 mph = 17.6 ft/s Flat plate B.L. flow 𝐿 = 1000 ft Uniform properties 𝑊 = 100 + 2(12) ft = 124 ft 𝑇 = 60 ℉ 𝜌 = 1.938 slugs/ft3 Diagram

𝜈 = 1.21 × 10−5 ft2/s Basic Equations 𝑅𝑒𝑥 = 𝑈𝑥/𝜈

𝐶𝐷𝑓 = 𝐷𝑓/1

2𝜌𝑈2𝐴

𝐶𝐷𝑓 =0.455

[log10(𝑅𝑒𝐿)]2.58 −1700

𝑅𝑒𝐿

Analysis The friction drag can be estimated by considering the barge sides and bottom as flat plates and assuming B.L. flow. Calculating Reynolds number confirms high 𝑅𝑒, boundary layer flow.

𝑅𝑒𝐿 =𝑈𝐿

𝜈=

(17.6 ft/s)(1000 ft)

1.21×10−5 ft2/s= 1.4545 × 109

Since 𝑅𝑒𝐿 ≫ 𝑅𝑒𝑐𝑟 the flow is turbulent at trailing edge of the barge. At the leading edge (front of barge) the

flow is laminar. At some location 𝑥𝑐𝑟 the Reynolds number exceeds 𝑅𝑒𝑐𝑟 = 5 × 105 and the B.L. flow transitions to a turbulent flow. From Table 9.3 the friction drag coefficient can be found:

𝐶𝐷𝑓 =0.455

[log10(𝑅𝑒𝐿)]2.58

−1700

𝑅𝑒𝐿

𝐶𝐷𝑓 = 0.0015

The drag coefficient is the drag force nondimensionalized by the dynamic pressure and area (𝐴 = 𝑊𝐿),

𝐶𝐷𝑓 =𝐷𝑓

1

2𝜌𝑈2𝐴

or 𝐷𝑓 = 0.5𝜌𝑈2𝑊𝐿𝐶𝐷𝑓

𝐷𝑓 = 0.5(1.938 slugs/ft3)( (17.6 ft/s)2(124 ft)(1000 ft)(0.0015)

𝐷𝑓 = 54546 lbf

Discussion: The B.L. flow transitions to turbulence at 𝑥𝑐𝑟 = 0.34 ft. Since 𝑥𝑐𝑟 ≪ 𝐿 the flow is essentially turbulent over the entire surface. For this reason, you could use either the second or third equation for 𝐶𝐷𝑓

in Table 9.3. Note the term 1700/𝑅𝑒𝐿 corrects for the laminar portion of the boundary layer. As 𝑥𝑐𝑟/𝐿 goes to 0, 𝑅𝑒𝐿 increases, and 1700/𝑅𝑒𝐿 ≈ 0, making the two equations converge at high 𝑅𝑒. In addition to the two turbulent B.L. correlations in Table 9.3, one could also use Fig. 9.15 (which plots these correlations), or the momentum integral results (HW4) for a turbulent B.L., e.g, Example 9.6. The friction drag coefficient, 𝐶𝐷𝑓, has the integrated effects of the wall shear stress built into it. This saves

the effort of performing integrations when 𝐶𝐷𝑓 data is available. If data is unavailable the more fundamental

equation 𝐷𝑓 = ∫𝜏𝑤𝑑𝐴 must be used.

Since the barge is rectangular it can be assumed that the flow over the submerged sides and bottom behaves as B.L. over a flat plate. Effects at the corners and leading/trailing edges are not accounted for and could result in deviations from the analysis. Given the size of the barge it is likely these effects are small, and we were able to obtain a first order estimate of the friction drag with a simplified model. The area of the front and back of the barge contribute to pressure drag only. There would be a high-pressure region acting on the front and a low-pressure wake behind the barge.

4

Munson 9.77 – As shown in Video V9.11 and Fig. P9.77, a vertical wind tunnel can be used for skydiving practice. Estimate the vertical wind speed needed if a 160 lb person is to be able to “float” motionless when the person (a) curls up as in a crouching position or (b) lies flat. See Fig. 9.32 for appropriate drag coefficient data. Given Model Diagram 𝑊 = 160 lbf Incompressible flow No acceleration Basic Equations

∑ �⃗⃗� = 𝑚�⃗⃗�

𝐶𝐷 =𝐷

1

2𝜌𝑈2𝐴

Analysis At terminal velocity, �⃗⃗� = 0, a force balance exists. Consider the 𝑧 (vertical) force balance. The buoyancy force is neglected since the weight of displaced air is small compared to that of the skydiver.

∑𝐹𝑧 = 𝐷 − 𝑊 = 0

𝑊 = 𝐷

Using the definition of the drag coefficient

𝐶𝐷 =𝐷

1

2𝜌𝑈2𝐴

or 𝐷 =1

2𝜌𝑈2𝐶𝐷𝐴

Substitute into the force balance

𝑊 =1

2𝜌𝑈2𝐶𝐷𝐴

Solving for the velocity 𝑈

𝑈 = √2𝑊

𝜌𝐶𝐷𝐴

From Table 9.32 it can be seen that 𝐶𝐷𝐴 = 9 ft2 when lies flat and 𝐶𝐷𝐴 = 2.5 ft2 when curled up.

Flat Crouching

𝑈 = √2(160 lbf)

(0.002377slugs

𝑓𝑡3)(9 ft2)

[slug∙ft

lbf∙s2 ]

𝑈 = 122.3 ft

s= 83.4

mi

hr

𝑈 = √2(160 lbf)

(0.002377slugs

𝑓𝑡3)(2.5 ft2)

[slug∙ft

lbf∙s2 ]

𝑈 = 232.1 ft

s= 158.2

mi

hr

Discussion Increasing the area by laying flat generates a larger low-pressure wake behind the skydiver resulting in an increased drag force compared to the crouching case. In actual skydiving, parachutes can be used to further increase drag and reduce velocity!

𝐷

𝑊

5

H5.1 – Answer the following questions (answers should be clear, explicit, and concise):

a. What is a favorable pressure gradient? What is an adverse pressure gradient? A favorable pressure gradient is one in which the pressure decreases in the flow direction. It is called “favorable” because it corresponds to a net pressure force in the flow direction. This helps keep the low momentum fluid in the B.L. moving and the flow attached. An adverse pressure gradient occurs when the pressure increases in the flow direction. It is called “adverse” because the net pressure force opposes fluid motion. This further reduces the momentum in the boundary layer and if sufficiently strong, may cause boundary layer separation. Favorable pressure gradient: ∂𝑃/𝜕𝑥 < 0 Adverse pressure gradient: ∂𝑃/𝜕𝑥 > 0

b. What is boundary layer separation? Explain underlying physics and implication on drag. Boundary layer separation is when fluid flowing near the surface of a body separates from the surface into the free-stream. Separation of the boundary layer occurs because the fluid in the boundary layer has lower momentum which may not be sufficient to overcome an adverse pressure gradient. In that case, the boundary layer flow will decelerate and at some point detach from the surface. At the point of separation the velocity gradient is zero (in laminar flow) and further increase in pressure (i.e. progressing downstream) backflow occurs. Boundary layer separation results in pressure drag which increases the total drag force on the body. The pressure in the wake at the rear of the body (i.e. the separation region) is lower than the pressure on the forward region (i.e. the stagnation point). This results in an asymmetric pressure distribution over the surface, and thus a net pressure force (pressure drag) is produced.

c. How does streamlining an object reduce drag? Will it always work? Explain. Total drag is a result of friction drag and pressure drag. Streamlining reduces the adverse pressure gradient, 𝜕𝑃/𝜕𝑥 which can delay/minimize boundary layer separation and thus reducing pressure drag. This does not always reduce total drag. Streamlining objects can increase the area exposed to the flow thereby increasing friction drag. For low Reynolds number flows there are no boundary layers and friction drag dominates pressure drag. In these flows separation is not important and the increased area due to streamlining increases total drag force.

6

H5.2 – For steady motion of a car on a level road, power must be delivered to the wheels to overcome the aerodynamic drag and the rolling resistance (product of vehicle weight and the rolling resistance coefficient). Consider a car with a total weight of 3500 lbf and frontal area of 26 ft2. The drag coefficient is 0.34, and the rolling resistance coefficient is 0.013. Do the following:

a. Plot the power (hp) required to overcome aerodynamic drag versus driving speed (consider a range of speeds 0 - 100 mph. Note: you may want to include curves in parts (b) and (c) on same plot.)

b. Plot the power (hp) required to overcome rolling resistance versus driving speed. c. Plot the total power (hp) required versus driving speed. d. Comment on above (a)-(c) results. What is the speed at which the rolling resistance is equal to the

aerodynamic drag force? e. Determine fuel economy (miles per gallon, mpg) for driving speeds of 55 mph and 80 mph. Assume the

density of gasoline to be 50 lbm/ft3 and the heating value of gasoline to be 20,000 Btu/lbm. Consider an overall engine efficiency of 23% and a drive train efficiency of 90%.

f. Perform a web search to determine actual values of 𝐶𝐷, frontal area, and weight for specific car models. Consider different vehicle types (sedan, race car, SUV, minivan, etc) with gas engines. For 2 of these vehicles, repeat e. to estimate corresponding variation in mpg values (can assume same values for other specs, e.g., engine efficiencies). Given Model Diagram 𝑚𝑔 = 3500 lbf Incompressible flow 𝐴 = 26 ft2 External flow over car 𝐶𝐷 = 0.34 Constant velocity 𝐶𝑅 = 0.013 Air at 𝑇𝑎𝑚𝑏 = 80 ℉ 𝜌 = 50 lbm/ft3 [𝐻𝑉] = 20,000 Btu/lbm 𝑈 𝜂𝑒𝑛𝑔 = 0.23

𝜂𝑑𝑡 = 0.9 Basic Equations

𝐶𝐷 =𝐹𝐷

1

2𝜌𝑈2𝐴

𝐶𝑅 =𝐹𝑅

𝑚𝑔

�̇�𝑎𝑐𝑡 =�̇�

𝜂𝑒𝑛𝑔𝜂𝑑𝑡

�̇� = 𝐹 ∙ 𝑉 Analysis

a. Plot the power (hp) required to overcome aerodynamic drag versus driving speed (consider a range of speeds 0 - 100 mph. Note: you may want to include curves in parts (b) and (c) on same plot.)

�̇�𝐷 = 𝐹𝐷𝑈 = 𝐶𝐷 (1

2𝜌𝑎𝑖𝑟𝑈

2𝐴)𝑈 =1

2𝐶𝐷𝜌𝑎𝑖𝑟𝑈

3𝐴

�̇�𝐷 =1

2(0.34) (2.286 × 10−3 slug

ft3) (𝑈

ft

s)3(26 ft2) [1

lbf

slug∙ft

s2

] [1 hp

550 ft∙lbf/s] = 1.837 × 10−5 𝑈3 [hp]

Note: The density of air was evaluated at 𝑇 = 80 ℉ since a temperature was not specified. 𝑈 is in ft/s to get

�̇�𝐷 in hp.

7

b. Plot the power (hp) required to overcome rolling resistance versus driving speed.

�̇�𝑅 = 𝐹𝑅𝑈 = 𝐶𝑅𝑚𝑔𝑈

�̇�𝑅 = 0.013(3,500 lbf) [1 hp

550 ft∙lbf/s]𝑈 = 0.0827 𝑈 [hp]

c. Plot the total power (hp) required versus driving speed.

∑𝐹𝑥 = 𝑚𝑎𝑥

𝐹𝐷 + 𝐹𝑅 − 𝑇 = 0 or 𝑇 = 𝐹𝐷 + 𝐹𝑅

�̇�𝑇 = 𝑇𝑈 = 𝐹𝐷𝑈 + 𝐹𝑅𝑈 = �̇�𝐷 + �̇�𝑅

�̇�𝑡𝑜𝑡 = �̇�𝐷 + �̇�𝑅

d. Comment on above (a)-(c) results. What is the speed at which the rolling resistance is equal to the aerodynamic drag force? At low velocities rolling resistance is dominant. At high velocities power to overcome drag dominates as it is proportional to 𝑈3. The two are equal at 𝑈 = 45 mph.

e. Determine fuel economy (miles per gallon, mpg) for driving speeds of 55 mph and 80 mph. Assume the density of gasoline to be 50 lbm/ft3 and the heating value of gasoline to be 20,000 Btu/lbm. Consider an overall engine efficiency of 23% and a drive train efficiency of 90%.

Power at 55 mph

Power at 80 mph

�̇�𝑡𝑜𝑡 = 16.32 hp

�̇�𝑓𝑢𝑒𝑙 =�̇�𝑡𝑜𝑡

𝜂𝑒𝑛𝑔𝜂𝑑𝑡= 78.8 hp

�̇�𝑡𝑜𝑡 = 39.38 hp

�̇�𝑓𝑢𝑒𝑙 =�̇�𝑡𝑜𝑡

𝜂𝑒𝑛𝑔𝜂𝑑𝑡= 190.2 hp

Heating Value [𝐻𝑉] – Chemical energy of fuel per unit mass.

𝐻𝑉 =𝐸𝑓𝑢𝑒𝑙

𝑚𝑓𝑢𝑒𝑙=

𝐸𝑓𝑢𝑒𝑙/Δ𝑡

𝑚𝑓𝑢𝑒𝑙/Δ𝑡=

�̇�𝑓𝑢𝑒𝑙

�̇�𝑓𝑢𝑒𝑙

�̇�𝑓𝑢𝑒𝑙 =�̇�𝑓𝑢𝑒𝑙

[𝐻𝑉]

Fuel Economy (𝐹𝐸) – Distance traveled per volume of fuel consumed

𝐹𝐸 =Δ𝑥

∀𝑓𝑢𝑒𝑙=

Δ𝑥/Δ𝑡

∀𝑓𝑢𝑒𝑙/Δ𝑡=

𝑈

𝑄𝑓𝑢𝑒𝑙=

𝜌𝑓𝑢𝑒𝑙𝑈

�̇�𝑓𝑢𝑒𝑙=

[𝐻𝑉]𝜌𝑓𝑢𝑒𝑙𝑈

�̇�𝑓𝑢𝑒𝑙

Fuel Economy at 55 mph Fuel Economy at 80 mph

𝐹𝐸 = (20000 Btu

lbm) (50

lbm

ft3) (55

mi

hr) (

1

78.8 hp) [

1 hp∙hr

2544 Btu] [

1 ft3

7.48 gal]

𝐹𝐸 = 36.7 mpg

𝐹𝐸 = (20000 Btu

lbm)(50

lbm

ft3)(80

mi

hr) (

1

190.2 hp) [

1 hp ∙ hr

2544 Btu] [

1 ft3

7.48 gal]

𝐹𝐸 = 22 mpg

8

f. Perform a web search to determine actual values of 𝐶𝐷, frontal area, and weight for specific car models. Consider different vehicle types (sedan, race car, SUV, minivan, etc) with gas engines. For 2 of these vehicles, repeat e. to estimate corresponding variation in mpg values (can assume same values for other specs, e.g., engine efficiencies).

Toyota Camry Honda Odyssey 𝐶𝐷 0.31 0.3 𝐴 24.4 ft2 37.2 ft2

𝑚𝑔 3241 lbf 4354 lbf 𝐹𝐸(55 mph) 41.4 mpg 29.1 mpg 𝐹𝐸(80 mph) 25.3 mpg 17.6 mpg

Discussion

Looking at the equation for �̇�𝑡𝑜𝑡 the power needed to overcome �̇�𝐷 and �̇�𝑅 is proportional to 𝑈 at low

speeds. As you increase in speed �̇�𝑡𝑜𝑡 become proportional to 𝑈3 and the fuel economy decreases as 1/𝑈2. The analysis above is simplified and incomplete, there are factors that are not included in the model. It’s unlikely that the drag coefficient will remain constant over such a large range of velocities. The engine efficiency is a function of engine speed (rpm) and zero at idle. Also, the engine must continue to supply power to the electronics on the car and keep the engine running. Performing a brief Google search reveals the true fuel economy measured for four vehicles.

https://blog.automatic.com/the-cost-of-speeding-save-a-little-time-spend-a-lot-of-money-5e8129899fec