FSI Introduction to Bridging Course Program FUNDAMENTALS OF AERODYNAMICS PDF

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FSI Introduction to Bridging Course Program FUNDAMENTALS OF AERODYNAMICS Lecturer: Engr. Roberto R. Renigen A. THE INTERNATIONAL STANDARD ATMOSPHERE The atmosphere is the mechanical mixture of gases surrounding the earth. Atmospheric Constituents Nitrogen ------------------ 78.03% Oxygen -----------...

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FSI Introduction to Bridging Course Program FUNDAMENTALS OF AERODYNAMICS Lecturer:

Engr. Roberto R. Renigen

A. THE INTERNATIONAL STANDARD ATMOSPHERE The atmosphere is the mechanical mixture of gases surrounding the earth. Atmospheric Constituents Nitrogen ------------------ 78.03% Oxygen ------------------- 20.99% Argon --------------------- 0.94% Carbon Dioxide ---------- 0.03% Hydrogen ------------------ 0.01% Helium --------------------- 0.004% Neon ----------------------- 0.0012% and a small amount of water vapor and other gases Four (4) Layers of Earth’s Atmosphere 1. 2. 3. 4.

Troposphere Stratosphere Ionosphere Exosphere

STANDARD VALUES FOR AIR AT SEA LEVEL Pressure PO= 14.7 lb/in2 = 2,116.8 lb/ft2 = 29.92” Hg = 76 cm Hg = 760 mm Hg = 101,325 Pa = 1 atm Density 3

ρo = 0.002377 slug/ft = 1.225 kg/m

3

Temperature TO= 519 °R = 288K Coefficient of Dynamic Viscosity

µo= 3.7372 x 10-7 slug/ft-sec = 1.7894 x 10-5 kg/m-sec

1

SEA LEVEL UP TO TROPOPAUSE 1. TEMPERATURE VARIATION WITH ALTITUDE

T = To + ah ; θ =

T To

Where:

θ=

T = temperature ratio To

T = temperature at any altitude above sea level up to tropopause in oR or K To = 519 °R or 288K a = temperature gradient or lapse rate (=-0.003566 °R/ft or -0.00651 K/m or -6.51 K/km) h = any altitude above sea level up to tropopause in ft or m or km

2. PRESSURE VARIATION WITH ALTITUDE P  ah   δ= = 1 + Po  To 

5.26

Where:

δ=

P = pressure ratio Po

P = pressure at any altitude above sea level up to tropopause in lb / ft2 or Pa Po = 2,116.8 lb/ft2 or 101,325 Pa a = -0.003566 °R/ft or -0.00651 K/m or -6.51 K/km h = any altitude above sea level up to tropopause in ft or m or km

3. DENSITY VARIATION WITH ALTITUDE ρ  ah   σ= = 1 + ρo  To 

4.26

Where:

σ=

ρ = density ratio ρo

ρ = density at any altitude above sea level up to tropopause in slug/ft3 or kg/m3 3

ρ o = 0.002377 slug/ft or 1.225kg/m

3

a = -0.003566 °R/ft or – 0.00651 K/m or -6.51 K/km h = any altitude above sea level up to tropopause in ft or m or km

2

ABOVE TROPOPAUSE UP TO STRATOPAUSE 1. TEMPERATURE VARIATION WITH ALTITUDE T = 390.15 °R or 216.5K (constant from tropopause up to stratopause)

2. PRESSURE VARIATION WITH ALTITUDE English System

δ=

P 1.26 = −5 Po e 4.805x10 h

P

= pressure at any altitude above tropopause up to stratopause in lb/ft

Where:

Po= 2,116.8lb/ft h

2

= any altitude above tropopause up to stratopause in feet

Metric System P 1.26 = −4 Po e1.578x10 h

δ= Where:

P = pressure at any altitude above tropopause up to stratopause in N/m2 (Pa)

Po = 101,325 Pa h

= any altitude above tropopause up to stratopause in meters

3. DENSITY VARIATION WITH ALTITUDE English System

σ=

ρ 1.68 = 4.805 x10 ρo e

−5

h

Where:

ρ= density at any altitude above tropopause up to stratopause in slug /ft ρ0= 0.002377 slug / ft h

3

= any altitude above tropopause up to stratopause in feet

Metric System

σ=

ρ 1.68 = 1.578 x10 ρ0 e

−4

h

3

3

2

Where:

ρ= density at any altitude above tropopause up to stratopause in kg/m

3

3

ρ o = 1.225 kg/m h = any altitude above tropopause up to stratopause in meters

ALTIMETERS An altimeter is a pressure gauge which indicates an altitude in the standard atmosphere corresponding to the measured pressure. Pressure altitude, hp – is the altitude given by an altimeter set to 29.92 “Hg. Density altitude, hd– is the altitude corresponding to a given density in the standard atmosphere. Temperature altitude, hT – is the altitude corresponding to a given temperature in the standard atmosphere.

Problems: 1. Calculate the pressure, density and temperature at 25,000 ft and 60,000 ft altitudes in the standard atmosphere. Ans. P25,000ft = 785.50 lb/ft2 , ρ25,000ft = 0.001065 slug/ft3 , T = 429.85 oR P60,000ft = 149.27 lb/ft2 , ρ60,000ft = 0.000223 slug/ft3 , T= 390.15 oR 2. On a hot day, the measured temperature and pressure are 38°C and 29.0 “Hg, respectively. Calculate the density and the density ratio. Ans. ρ = 1.100 kg/m3 , σ = 0.898 3. A standard altimeter reads 4,500 meters when the ambient temperature is 275K. What are the density altitude and the temperature altitude? Ans. hd = 5,061.98 m ,hT = 1,996 93 m 4. At a certain altitude, a standard altimeter reads 3,000 meters. If the density altitude is 2,500 meters, find the true temperature. Ans. T = 255.18 K

BASIC AERODYNAMIC PRINCIPLES AND APPLICATIONS I. CONTINUITY EQUATION – Conservation of mass along a streamtube, such as in a wind tunnel.

PRINCIPLE OF MASS CONSERVATION (LAW OF CONTINUITY) The mass flow of fluid that passing to one section of the tube in one second is equal to the mass flow of fluid that passing to the other section of the tube in one second.

4

ᵒ m

o

m = ρAV Where: o

m = mass flow of fluid in slug/sec or kg/sec 3

ρ = density of fluid in slug/ft or kg/m

3

2

A = cross-sectional area of tube in ft or m V = velocity of fluid in ft/s or m/s

2

a. For incompressible fluid, ρ= constant (M < 0.3 approximately). A 1V 1 = A 2V 2 or AV = constant Differential form:

dV V

=

−

dA A

b. For compressible fluid, ρ ≠ constant (M ≥ 0.3 approximately).

ρ1 A1V1

= ρ 2 A2V2

or

ρAV = constant Differential form:

dV V

+

dA A

+

dρ ρ

=

0

5

Problems: 1. A pipe is gradually tapering in size, diminishing by 0.1 sq.ft. per foot run. What is the change in velocity per foot run where the pipe is 4 sq.ft. in cross section, if the velocity there is 90 ft. per sec? Is the velocity increasing or decreasing? Ans. dV/dS = 2.25 fps per ft , increasing 2. A circular pipe, 100 ft. long tapers from 3 ft. in diameter at one end to 2 ft.in diameter at the other. Fluid is flowing from the bigger toward the smaller end. What is the rate of increase in velocity at the entrance if the velocity there is 80 ft. per sec? Ans. dV/dS = 0.444 fps per ft 3. Air having the standard sea level density has a velocity of 100 fps at a section of a wind tunnel. At another section having an area half as great as the first section the flow velocity is 400 mph. What is the density at the second section? Ans. ρ2 = 0.000810 slug per cu ft

II. INCOMPRESSIBLE BERNOULLI EQUATION – conservation of energy along a streamline.

BERNOULLI’S PRINCIPLE States that the total energy of a particle in motion is constant at all points on its path in a steady flow. In a continuous flow of fluid, as velocity increases, pressure decreases; and as velocity decreases, pressure increases. For incompressible fluid, ρ= constant (M < 0.3 approximately). V2 2

+

P ρ

V12 2

+

P1 ρ

=

constant

or =

V2 2 2

+

P2 ρ

Problems: 1. A horizontal pipe, 1 ft. in diameter, tapers gradually to 8 in. in diameter. If the flow is 500 cu ft. of water per minute, what is the difference between the pressures at the two sections? Ans. PA – PB = 443.63lb per sq ft 2. Water flows through a horizontal pipe at a velocity of 50 ft. per sec. Owing to the pipe gradually expanding to a larger size, the velocity decreases to 35 ft. per sec. What is the difference between the pressures at two points, one in each size of pipe? Ans. PB – PA = 1,236. 75 lb per sq ft 3. The diameter of a horizontal tube is 4 in., in which tetrabromoethane (spec. grav. = 0.30) is flowing at the rate of 0.50 cu ft. per sec. The pressure is 30 lb. per sq. in. (gage). If the tube gradually decreases to 3 in. in diameter, what is the pressure there? Ans. PB = 6,416.16 lb per sqft or 44.56 lb per sq in.

6

4. Alcohol (spec. grav. = 0.80) is flowing through horizontal pipe, which is 10 in. in diameter, with a velocity of 40 ft. per sec. at a smaller section of the pipe, there is 6 lb. per sq. in. less pressure. Assuming that the flow is smooth, what is the diameter there? Ans. dB = 0.73 ft or 8.76 in. 5. Air is flowing horizontally at a speed of 100 mph through a duct 4 sq.ft. in cross section. The duct gradually narrows down to a throat section. If a U-tube shows a difference in pressure between the throat and main sections of 7 in. of water, what is the cross-sectional area of the throat? (Assume that the air is non-compressible and has a density of 0.002377 slug per cu ft.) Ans. AB = 2.57 sq ft 6. A horizontal water pipe is reduced in size from 18in. in diameter at point A to 6 in. in diameter at point B. The flow in the pipe is 10 cu ft. per sec, and the pressure at A is 20 lb. per sq. in. (gage). If it is assumed that there is no loss in energy due to friction, what is the pressure at B? Ans. PB = 2,511.85 lb per sq ft or 17.44 lb per sq in. 7. Air flows through a horizontal pipe at the rate of 3,000 cu ft. per sec. If the pressure is 30 lb per sq. in. (gage) where the diameter is 3 ft., what is the pressure where the diameter is 2 ft.? Ans. PB = 5,567.10 lb per sq ft or 38.66 lb per sq in. 8. Water flows through a horizontal pipe at the rate of 800 gal per min. What is the difference in pressure between a point where the diameter is 2 in. and a point where the diameter is 1 in.? Ans. PA – PB = 96,856.38 lb per sq ft or 672.61 lb per sq in 9. A horizontal pipeline enlarges from a diameter of 6 in. at point A to a diameter of 12 in. at point B. The flow of water is 20 cu ft. per sec, and the pressure at A is 10 lb. per sq. in. What is the pressure at B? Ans. 3,995.32 lb per sqft or 27.75 lb per sq in. 10. The diameter at section (1) is 0.3m. The diameter at section (2) is 0.15m. What is the flow rate of a substance (sp. gr. = 0.85) if the pressure between sections (1) and (2) is 12.7 cm Hg? Ans. Q = 0.115 cu meter per sec. 11. Consider a low-speed subsonic wind tunnel with a 12/1 contraction area ratio for the nozzle. If the flow in the test section is at a standard sea level conditions with a velocity of 50 m /s, calculate the height difference in a U-tube mercury manometer with one side connected to the nozzle inlet and the other to the test section. ρHg = 1.36 x 103 kg/ m3. Ans. ∆h = 0.114 m 12. The wind tunnel shown in Fig. below has a smallest section (test section) measuring 1.22 m by 1.0m, and a largest section of 2 m square. At a certain tunnel speed the manometer reading is 0.72m. The manometer liquid has a specific gravity of 0.85. Calculate the airspeed in the test section. Assume incompressible flow and standard sea-level conditions. Ans. VB = 73.50 meters per sec.

7

13. A pitot-static tube is used to measure the airspeed at the test section of a wind tunnel. If the pressure difference across the pitot-static tube is 0.11 m of water, what is the airspeed at the test section? If the ratio of the cross-sectional area between the largest section and the test section is 100:1, what is the airspeed at the largest section? Assume incompressible flow at standard sea level conditions. Ans. V2 = 41.97 meters per sec. V1 = 0.4197 meter per sec. 14. Consider water flowing through a smooth pipe whose diameter is decreasing. At one location, the diameter is 12 cm. If the velocity there is 10 m/s. (a) Find the mass flow rate. (b) At a station farther down the pipe, the diameter is 4cm. Find the velocity at this station. o

= 113. 10 kg/sec. , VB = 90 meters per sec.

Ans. m

Venturi Tube The Venturi tube is a convergent–divergent tube with a short cylindrical throat or constricted section. This device determines the rate of flow of fluid through the tube by measuring the difference in pressure between the throat section and the entrance section.

Practical application is made of Bernoulli Equation. P1 ρ

+

V12 2

(

=

P2 ρ

1 P1 − P2 = ρ V22 − V12 2

+

V2 2 2

)

By the law of continuity, Q being the rate of flow, for non-compressible fluids, Q = A1 V1 =

V12 =

Q2 A12

A2 V2

;

V22 =

Then, 2 2  ρ  Q   Q       P1 − P2 =  − 2  A 2   A1    

8

Q2 A 22

2 2      ρ Q    A2   P1 − P2 =  1− 2  A 2    A1       

2

 Q    = 2( P1 − P2 )   2    A2  A 2     ρ 1−   A1    

Q = A2

Q=

πd 22 4

2( P1 − P2 )   A 2  ρ 1 −  2     A1    

2( P1 − P2 )   d 4  ρ 1 −  2     d1    

Where: Q

3

3

= rate of flow in ft /s or m /s 2

P1 = pressure at section 1 in lb /ft or Pa 2

P2 = pressure at section 2 in lb/ft or Pa A1

2

= cross-sectional area of section 1 in ft or m 2

A2 = cross-sectional area of section 2 in ft or m

2

2

d1 = diameter of section 1 in ft or m d2 = diameter of section 2 in ft or m

ρ

3

= density in slug/ft or kg/m

3

Problems: 1. A Venturi tube narrows down from 4 in. in diameter to 2 in. in diameter. What is the rate of flow of water if the pressure at the throat is 2 lb. per sq. in. less than at the larger section? Ans. Q = 0.388 cu ft per sec. 2 2. A Venturi tube in 6 in. in diameter at the entrance, where the pressure is 10 lb/in (gage). The 2

throat is 4 in. in diameter; there the pressure is 6 lb/in (gage). What is the flow of water? Ans. Q = 2.73 cu ft per sec. 3. Consider a Venturi with a throat-to-inlet area ratio of 0.8 mounted in a flow at standard sea level conditions. If the pressure between the inlet and the throat is 335 Pa, calculate the velocity of the flow at the inlet. Ans. V1 = 31.18 m/s

9

III. COMPRESSIBILITY EFFECTS When the air is regarded as compressible, the internal energy of the air mass must be considered.

A. ISENTROPIC EQUATION OF STATE

P

ργ P1 ρ1γ

= cons tan t =

P2

P2 P1

,

ρ2 γ

ρ  =  2   ρ1 

γ

THERMODYNAMIC PARAMETER RELATIONSHIP γ

γ  T2  γ −1

P2  ρ2  =   =   P1  ρ1   T1 

w  =  2   w1 

γ

2γ

 Va = 2  Va  1

 γ −1   

Problems: 3

1. Air at standard pressure and temperature, has density of 1.225 kg/m . If the air is compressed adiabatically to 3 atm, what are the specific weight, density and the temperature? Ans. w2 = 26.33 N/m3 , ρ2 = 2.685 kg/m3 , T2 = 394.20 K 2. Air at standard pressure and temperature is permitted to expand adiabatically to one-half atmospheric pressure. What are a) the density and b) the temperature? Ans. (a) ρ2 = 0.747 kg/m3 (b) T2 = 236.26 K 3. Air at standard pressure and temperature is adiabatically compressed to 50 lb. per sq. in. (gage) pressure. What is the temperature? Ans. T2 = 792.59 oR 4. Air at standard pressure and temperature is permitted to expand adiabatically until it is one-half standard density. a) What is the pressure? b) What is the temperature? Ans. (a) P2 = 38,395 Pa , (b) T2 = 218.26 K

B. THE SPEED OF SOUND ,Va

Va =

γP ρ

English System Va = γgRT

Where:

γ = 1.4 R = 53.342 ft/ °R

10

Va = 49.02 T

Where: Va = speed of sound in air in ft/s T = absolute temperature in °R

Metric System Va = 20.05 T

Where: Va = speed of sound in air in m/s T = absolute temperature in K

Problems: 1. Find the speed of sound in air at standard sea level conditions. Ans. Va = 340.26 m/s 2. Find the speed of sound in air at 7,000 ft. altitude in the standard atmosphere. Ans. Va = 1,089.57 ft/sec

C. COMPRESSIBLE BERNOULLI EQUATION V2 γ P + = constant 2 γ −1 ρ

or: V12 2

+

γ P1 V2 2 γ P2 = + γ − 1 ρ1 2 γ − 1 ρ2

Problems: 3

1. In an undisturbed airstream the pressure is 101,325 Pa, the density is 1.225 kg/m , and the velocity is 150 m/s. What is the pressure if the velocity is 190 m/s? Ans. P = 93,236.71 Pa 2. In an undisturbed airstream, where the pressure is 14.7 lb. per sq. in. and the temperature is 59oF, the velocity is 520 ft. per sec. Where the velocity is 600 ft. per sec., what is the local pressure? Ans. P = 2,012.23 lb per sq ft 3. In an undisturbed airstream, where the pressure is 14.7 lb. per sq. in. and the temperature is 59oF, the velocity is 550 ft. per sec. What is the velocity where the pressure is 13.9 lb. per sq. in.? Ans. V = 633.83 ft/sec

11

IV. AIRSPEED MEASUREMENT (a) LOW-SPEED AIRSPEED INDICATORS (INCOMPRESSIBLE FLOW)

PVρ Pt Vt ρ t

Pitot – Static Tube Where: PVρ = parameters of airflow initially Ptvtρt= parameters of the airflow at the stagnation point

Pitot-static tube- an instrument consisting of two tubes, having their points of origin in the open end unobstructed airstream. One of the tube ends is closed but slotted on the side in such a manner so that static air pressure is maintained in the tube and the other tube having an end open to the airstream so that it receives the full impact pressure of the airstream. Static pressure (P) – the force per unit area exerted by a fluid on a surface at rest relative to the fluid. Stagnation pressure or total pressure (Pt) – the pressure at stagnation point wherein the velocity of the medium is equal to zero. Applying the incompressible Bernoulli equation: V 2 P Vt2 Pt + = + 2 ρ 2 ρ

1 Pt − P = ρV 2 2 Pt = P +

1 ρV 2 2

Pt = P + q

Where:

1 q = ρV 2 = dynamic pressure 2 V=

2(Pt − P ) ρ 12

Application: Definition of equivalent airspeed Ve: Ve =

2(Pt − P ) ρ0

Where: Pt= total pressure or stagnation pressure P = static pressure ρo= density at sea level Definition of true airspeed V:

V =

Ve σ

Where:

σ=

ρ , density ratio ρ0

Problems: 1. An airplane is flying at standard sea level conditions at 45 meters per second. What is the difference between total and static pressure? Ans. Pt - P = 1,240.31 Pa 2. An airplane is flying at standard sea level conditions at airspeed of 75 meters per second. What is the total pressure? Ans. Pt = 104,770.31 Pa 3. An airplane is flying at standard sea level, the difference between total and static pressure is 1,750 Pa. What is the airspeed in meters per second? Ans. V = 53.45 m/s

(b) HIGH-SPEED AIRSPEED INDICATORS (COMPRESSIBLE FLOW) Mach number (M) – is the ratio of the local velocity of the fluid to the velocity of sound at that same point.

M=

V Va

Where: M = Mach number V = velocity of fluid Va= ambient speed of sound

13

Three different regimes of aerodynamic flows 1. If M 1, the flow is supersonic.

Two ...