Livro resolvido, fox mcdonolds, Exercícios de Mecânica dos fluidos

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1.1 A number of common substances are Tar Sand ‘‘Silly Putty’’ Jello Modeling clay Toothpaste Wax Shaving cream Some of these materials exhibit characteristics of both solid and fluid behavior under different conditions. Explain and give examples.

Given: Common Substances

Tar Sand “Silly Putty” Jello Modeling clay Toothpaste Wax Shaving cream Some of these substances exhibit characteristics of solids and fluids under different conditions.

Find: Explain and give examples.

Solution: Tar, Wax, and Jello behave as solids at room temperature or below at ordinary pressures. At high

pressures or over long periods, they exhibit fluid characteristics. At higher temperatures, all three liquefy and become viscous fluids. Modeling clay and silly putty show fluid behavior when sheared slowly. However, they fracture under suddenly applied stress, which is a characteristic of solids. Toothpaste behaves as a solid when at rest in the tube. When the tube is squeezed hard, toothpaste “flows” out the spout, showing fluid behavior. Shaving cream behaves similarly. Sand acts solid when in repose (a sand “pile”). However, it “flows” from a spout or down a steep incline.

1.3 Discuss the physics of skipping a stone across the water surface of a lake. Compare these mechanisms with a stone as it bounces after being thrown along a roadway.

Open-Ended Problem Statement: Consider the physics of “skipping” a stone across the water surface of a lake. Compare these mechanisms with a stone as it bounces after being thrown along a roadway. Discussion: Observation and experience suggest two behaviors when a stone is thrown along a water surface:

1. If the angle between the path of the stone and the water surface is steep the stone may penetrate the water surface. Some momentum of the stone will be converted to momentum of the water in the resulting splash. After penetrating the water surface, the high drag *^ of the water will slow the stone quickly. Then, because the stone is heavier than water it will sink.
2. If the angle between the path of the stone and the water surface is shallow the stone may not penetrate the water surface. The splash will be smaller than if the stone penetrated the water surface. This will transfer less momentum to the water, causing less reduction in speed of the stone. The only drag force on the stone will be from friction on the water surface. The drag will be momentary, causing the stone to lose only a portion of its kinetic energy. Instead of sinking, the stone may skip off the surface and become airborne again. When the stone is thrown with speed and angle just right, it may skip several times across the water surface. With each skip the stone loses some forward speed. After several skips the stone loses enough forward speed to penetrate the surface and sink into the water. Observation suggests that the shape of the stone significantly affects skipping. Essentially spherical stones may be made to skip with considerable effort and skill from the thrower. Flatter, more disc-shaped stones are more likely to skip, provided they are thrown with the flat surface(s) essentially parallel to the water surface; spin may be used to stabilize the stone in flight. By contrast, no stone can ever penetrate the pavement of a roadway. Each collision between stone and roadway will be inelastic; friction between the road surface and stone will affect the motion of the stone only slightly. Regardless of the initial angle between the path of the stone and the surface of the roadway, the stone may bounce several times, then finally it will roll to a stop. The shape of the stone is unlikely to affect trajectory of bouncing from a roadway significantly.

1.4 The barrel of a bicycle tire pump becomes quite warm during use. Explain the mechanisms responsible for the temperature increase.

Open-Ended Problem Statement: The barrel of a bicycle tire pump becomes quite warm during use. Explain the mechanisms responsible for the temperature increase. Discussion: Two phenomena are responsible for the temperature increase: (1) friction between the pump piston and barrel and (2) temperature rise of the air as it is compressed in the pump barrel. Friction between the pump piston and barrel converts mechanical energy (force on the piston moving through a distance) into thermal energy as a result of friction. Lubricating the piston helps to provide a good seal with the pump barrel and reduces friction (and therefore force) between the piston and barrel. Temperature of the trapped air rises as it is compressed. The compression is not adiabatic because it occurs during a finite time interval. Heat is transferred from the warm compressed air in the pump barrel to the cooler surroundings. This raises the temperature of the barrel, making its outside surface warm (or even hot!) to the touch.

Given: Dimensions of a room

Find: Mass of air

Solution:

Basic equation: (^) ρ (^) Rp = air ⋅T

Given or available data (^) p = 14.7psi T = ( 59 + 460 )R Rair =53.33 ⋅ lbm Rft lbf⋅⋅

V = 10 ft⋅ × 10 ⋅ ft× 8 ⋅ft V =800 ft 3

Then (^) ρ (^) Rp = (^) air ⋅T ρ 0. lbm ft 3

= ρ 0.00238 slug ft 3

= ρ 1.23 kg m 3

M = ρ ⋅V M = 61.2 lbm M = 1.90 slug M =27.8 kg

Given: Mass of nitrogen, and design constraints on tank dimensions.

Find: External dimensions.

Solution: Use given geometric data and nitrogen mass, with data from Table A.6.

The given or available data is: (^) M = 10 lbm⋅ p = ( 200 + 1 ) atm⋅ p =2.95 × 10 3 ⋅psi

T = ( 70 + 460 ) K⋅ T = 954 R⋅ RN2 = 55.16 ⋅ lbm Rft lbf⋅⋅ (Table A.6)

The governing equation is the ideal gas equation (^) p = ρ ⋅R (^) N2⋅T and (^) ρ =MV

where V is the tank volume (^) V π^ D^

⋅^2

= 4 ⋅L where^ L =2 D⋅

Combining these equations:

Hence (^) M = V⋅ ρ Rp V⋅ = (^) N2 ⋅T p RN2 ⋅T

π ⋅D^2 = ⋅ 4 ⋅L p RN2 ⋅T

π ⋅D^2 = ⋅ 4 ⋅ 2 ⋅D p ⋅π ⋅D^3 = 2 R⋅ (^) N2⋅T

Solving for D D ⎛⎜⎝2 R⋅^ N2 p ⋅π⋅T^ ⋅M⎞⎟⎠

1 3 = D (^) π^2 × 55.16⋅ (^) lbm Rft lbf⋅⋅ × 954 ⋅ K× 10 ⋅ lbm × 29501 in

2 ⋅ (^) lbf ft 12 in⋅

2

⎡⎢⎣ × ⎤⎥⎦

1 3 =

D = 1.12 ft⋅ D = 13.5 in⋅ L = 2 D⋅ L =27 in⋅

These are internal dimensions; the external ones are 1/4 in. larger: L = 27.25 in⋅ D =13.75 in⋅

or (^1) W kV e V Wk 1

kgtW kgtW − = = L− N

MM O Q

− (^) ; e− PP

But V→V (^) t as t→∞, so Vt = Wk. Therefore (^) VVt 1 e

kgtW = − −

When (^) VVt= 0.95, then e 0. −kgtW = and kgtW = 3. Thus t = 3 W/gk

1.9 Consider again the small particle of Problem 1.8. Express the distance required to reach 95 percent of its terminal speed in terms of g , k , and W.

Given: Small particle accelerating from rest in a fluid. Net weight is W, resisting force is F D = kV, where

V is speed.

Find: Distance required to reach 95 percent of terminal speed, Vt.

Solution: Consider the particle to be a system. Apply Newton's second law.

Basic equation: ∑F y = ma y Assumptions:

1. W is net weight.
2. Resisting force acts opposite to V.

Then, (^) ∑ F y (^) = W − kV = ma (^) y = m dVdt =Wg VdVdy or 1 − (^) Wk V=V dVg dy

At terminal speed, ay = 0 and V = Vt = Wk. Then 1 − VVg =^1 g VdVdy

Separating variables V^1 t

V dV g dy

1 − V =

Integrating, noting that velocity is zero initially

[ ]

0.95 0. 2 (^00)

2 2 2 2 2 2 2 2

1 ln^1

0.95 ln (1 0.95) ln (1)

0.95 ln 0.05 2.

Vt V^ t t t t t t t t t t t

gy V dV^ VV V V

V V

V

gy V V V

gy V V

y V W

g gt

= = ⎡⎢^ − − ⎛⎜^ − ⎞⎟⎤⎥

− ⎢⎣^ ⎝^ ⎠⎥⎦

∫

Using the given data

0 0.01 0.

t (s)

V (m/s)

The time to reach 95% of maximum speed is obtained from (^3) ⋅M gπ ⋅⋅ μ⋅ d 1 e

− 3 ⋅ π⋅ μ⋅d

⋅ ⎛⎜⎝ −^ M^ ⋅t⎞⎟⎠=0.95 V⋅ max

so t = − 3 ⋅ πM⋅μ ⋅d⋅ln 1⎛⎜⎝ −0.95 V⋅^ max M g⋅⋅^3 ⋅^ π⋅^ μ⋅d⎞⎟⎠ Substituting values t =0.0133 s

The plot can also be done in Excel.

Given: Data on sphere and formula for drag.

Find: Diameter of gasoline droplets that take 1 second to fall 25 cm.

Solution: Use given data and data in Appendices; integrate equation of motion by separating variables.

The data provided, or available in the Appendices, are:

μ 1.8 × 10 −^5 N s⋅ m^2

= ⋅ ρw 999 kg m^3

= ⋅ SGgas = 0.72 ρgas = SGgas ⋅ρw ρgas 719 kg m^3

Newton's 2nd law for the sphere (mass M) is (ignoring buoyancy effects) (^) M ⋅ dVdt =M g⋅ − 3 ⋅ π⋅μ ⋅V ⋅d

so dV g −^3 ⋅π^ M⋅^ μ ⋅d⋅V

= dt

Integrating and using limits (^) V t( ) (^3) ⋅ M gπ⋅⋅μ ⋅d 1 e

− 3 ⋅ π⋅ μ⋅d

= ⋅⎛⎜⎝ −^ M^ ⋅t⎞⎟⎠

Integrating again (^) x t( ) (^3) ⋅M g π⋅⋅μ ⋅d t (^3) ⋅ πM⋅μ ⋅d e

− 3 ⋅ π⋅ μ⋅d

⎡⎢⎢+ ⋅⎛⎜⎝ M ⋅t− 1 ⎞⎟⎠

Replacing M with an expression involving diameter d (^) M ρgas^ π^ d

⋅^3

= ⋅ 6 x t( ) ρgas ⋅ d 2 ⋅g 18 ⋅μ t

ρgas ⋅d 2 18 ⋅μ e

− 18 ⋅μ ρgas ⋅d^2 ⋅^ t− 1

This equation must be solved for d so that x 1 s( ⋅) = 1 m⋅. The answer can be obtained from manual iteration, or by using Excel's Goal Seek. (See this in the corresponding Excel workbook.)

d =0.109 mm⋅

0 0.2 0.4 0.6 0.8 1

t (s)

x (m)

Note That the particle quickly reaches terminal speed, so that a simpler approximate solution would be to solve Mg = 3πμ Vd for d , with V = 0.25 m/s (allowing for the fact that M is a function of d )!

0 100 200 300 400 500

20

40

60

y(m)

V(m/s)

(c) For V ( t ) we need to integrate (1) with respect to t : M ⋅ dVdt =M g⋅ −k V ⋅ 2

Separating variables and integrating:

0

V M g⋅V V k V^

−^2

d 0

t =^ ⌠⎮⌡ 1 dt

so t 12 ⋅ (^) k gM⋅ ln

M g⋅ k +V M g⋅ k −V

= ⋅^12 ⋅ (^) k gM⋅ ln V Vt^ +V t −V

Rearranging V t( ) Vt e

2 ⋅ k g M⋅ ⋅t − 1

e

2 ⋅ k g M⋅ ⋅t

• 1

= ⋅ or V t( ) =Vt ⋅tanh V⎛⎜⎝ t ⋅ Mk⋅t⎞⎟⎠

0 5 10 15 20

20

40

60

t(s)

V(m/s)

V t( )

t

The two graphs can also be plotted in Excel.

Given: Data on sky diver: M = 70 kg⋅ kvert 0.25 N s

⋅^2

m 2

= ⋅ khoriz 0.05 N s

⋅^2

m 2

= ⋅ U 0 = 70 ⋅ms

Find: Plot of trajectory.

Solution: Use given data; integrate equation of motion by separating variables.

Treat the sky diver as a system; apply Newton's 2nd law in horizontal and vertical directions:

Vertical: Newton's 2nd law for the sky diver (mass M) is (ignoring buoyancy effects): (^) M ⋅ dVdt = M g⋅ −kvert ⋅V 2 (1)

For V ( t ) we need to integrate (1) with respect to t :

Separating variables and integrating:

0

V M g⋅V V kvert V

−^2

d 0

t =^ ⌠⎮⌡ 1 dt

so (^) t (^12) kM ⋅ (^) vert ⋅gln

M g⋅ kvert+V M g⋅ kvert−V

Rearranging orV t( ) (^) kM g⋅ vert

e

2 ⋅ kvert M⋅g⋅t − 1

e

2 ⋅ kvert M⋅g⋅t

• 1

= ⋅ so^ V t( ) = kM gvert⋅ ⋅tanh⎛⎜⎝ kvert M ⋅g⋅t⎞⎟⎠

For y ( t ) we need to integrate again: dydt = V or (^) y =^ ⌠⎮⎮⌡ V dt

y t( ) 0

t =^ ⌠⎮⌡ V t( )dt 0

t

kM gvert⋅ ⋅tanh⎛⎜⎝ kvert M ⋅g⋅t⎞⎟⎠ t

= d = kM gvert⋅ ⋅ln cosh⎛⎜⎝ ⎛⎜⎝ kvert M ⋅g⋅t⎞⎟⎠⎞⎟⎠

y t( ) = kM gvert⋅ ⋅ln cosh⎛⎜⎝ ⎛⎜⎝ kvert M ⋅g⋅t⎞⎟⎠⎞⎟⎠

0 20 40 60

500

1 10 ×^3

1.5 10 × 3

2 10 ×^3

t(s)

x(m)x t( )

t

Plotting the trajectory:

0 1 2 3

− 3

− 2

− 1

x(km)

y(km)

These plots can also be done in Excel.

Given: Data on sphere and terminal speed.

Find: Drag constant k , and time to reach 99% of terminal speed.

Solution: Use given data; integrate equation of motion by separating variables.

The data provided are: (^) M = 5 10⋅ −^11 ⋅kg Vt = 5 ⋅cms

Newton's 2nd law for the general motion is (ignoring buoyancy effects) (^) M ⋅ dVdt = M g⋅ −k V⋅ (1)

Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects) (^) M g⋅ = k V⋅ t so (^) k M g V⋅ = t

k = M g V⋅t 5 × 10 −^11 ⋅kg × 9.81m s^2

= ⋅ ×0.05 ms⋅ k =9.81 × 10 −^9 ⋅N s m⋅

dV g −M k⋅V To find the time to reach 99% of V (^) t , we need V ( t ). From 1, separating variables =^ dt

Integrating and using limits t = − Mk⋅ln 1⎛⎜⎝ −M gk⋅ ⋅V⎞⎟⎠

We must evaluate this when (^) V = 0.99 V⋅ t V =4.95 ⋅cms

t 5 × 10 −^11 ⋅ kg m 9.81 × 10 −^9 ⋅ N⋅s

× N s

⋅^2

× (^) kg m⋅ ln 1 9.81 10 ⋅ −^9 N s⋅ ⋅ (^) m

5 × 10 −^11 ⋅kg

× s

2 × (^) 9.81 m⋅ 0.0495 m⋅ × (^) s kg m⋅ N s ⋅^2

⎛⎜⎜ − ×

t =0.0235 s