C1

PROBLEMS:

PROBLEMS:

1. A circular disc 3m in diameter is held normal to a 26.4 m/s wind of density 1.2 kg/m3 .what force is required to hold it at rest?Assume coefficient of drag of disc as 1.1.

Forc required is drag = CD A ( U2 2 = 1.1 x ( 32 x1.2 x 26.42 4 2 = 3251.7 N

2.calculate the total drag shear drag and the pressure exerted on 1m length of an infinite cylinder which has a diameter equal to 30mm,air of density 1.236 kg/m3 flowing past the cylinder with velocity 3.6 m per minute. Take total drag coefficient equal to 1.4 and shear drag coefficient equal to 0.185.Total drag = Cd A (U2 2 = 1.4 x (1x 0.03) x 1.236 x 1 (3.6/60) 2 2

= 9.344 x105 N

Shear drag = 0.185(.1x0.03) x 1.236 x 1 x(3.6/60) 2 2

= 1.235x105 NPressure drag = 9.344x105 - 1.235x105

= 8.109x10 N

3. A semi-tubular cylinder of 75mm radius with concave side upstream (drag coeff. = 2.3) is submerged in flowing water of velocity 0.6m/s.if the cylinder is 7.2m long, calculate the drag. Assume density of water as 1000kg/m3Drag D = Cd A ( U2 2

= 2.3x (0.075x7.2) x 1000 x 1x 0.62 = 223.56 N 4. The coefficient Cd of a 5 kN and 0.5 m diameter bomb as a function of the mach number is tabulated below. Linear interpolation may be used between tabulated values.

Mach number0.500.751.0010251.59

Drag.coeff:0.250.280.350.550.70

If the bomb is dropped from a flying plane find the velocity with which it hits the ground .Atmosphere pressure may be assumed 101KN/m2, temperature 290.15K ; gas constant for air 287Nm/Kg K and adiabatic index 1.4 Near ground velocity of sound a = ((rRT)

R = 287; r = 1.4; T = 341.44Therefore a = 341.44 m/sFor mach number M local velocity of the bomb U may be taken as = M x 341.44 m/s.

Air density near ground is calculated from perfect gas law

( = P/ RT P = 101x 103 ; R = 287; T = 290.15 Therefore = 1.213 Kg/m3 As the drag on the bomb exactly balances its weight in its fall

W = D

( U= Cd A 2 2

W = 5x10 ; ( = 1.213 And also substituting U in terms of mach number M

5x10 = Cd x ( x 0.52) 1.213 x 0.5 (M x 341.44) 2Simplyfing Cd = 0.3601/m2Solving by the help of given table we get

M = 1.0068 for U = 343.76

5. After boiling out from aero plane and before parachute is released max accelerates to terminal velocity .Assuming weight of mass as 800N, ( = 1.2 Kg/m3 area of fall 0.11m2 Cd = 1.8 .Determine the fall velocity For the freely falling parachute the weight is balanced by the drag forceTherefore

800 = Cd A ( U2 2

1.8 x 0.11 x 1.2 xU2 2

U = 82.06 m/s

6.The vertical component of the landing speed of a parachute is 6.0m/s. Treat the parachute as as open atmosphere and determine its diameter if the total weight to be carried is 1500N Assume ( to be 1.208 kg/m3Assume Cd = 1.33 of paracute

For the freely faslling falling parachute the weight is balanced by drag force

1500 = Cd A ( U2 2

= 1.33 x A x 1.208 x(62)/2 Therefore dia = 8.126m

7. A Kite has an effective area of 0.40.m2 and weights 2.0 N .in a wind of 40 Km/hr, the drag on the kite is 11.9 N .Determine the tension in the cord if the cord makes an angle of 45 degrees with the horizontal. Also find the lift coefficient.

Speed U = 40km/hr = 40x1000 60x60 = 11.11m/s drag = tension x Cos 45therefore tension = 11.9x (2 = 16.83 N

Lift on the kite L = W + TCos 45

= 2+ 16.83 x (1/(2)

= 13.9N

L = CL A ( U2 2

L = 13.9 ; ( = 1.208 ; U = 11.11 ; A = 0.40;

From which we get CL = 0.466

8. An aero plane weighting 22,500 N has a wing area of 22.5 M2 and span of 12.0 m what is the life coefficient if it travels at 320 Km/hr in the horizontal direction? Also compute the theoretical value of circulation and angle of attack measured from Zero axis

Assuming level flight lift exactly balances weight

W = L

= Cl A ( U2 2

W = 22500; A = 22.5; =1.208 U = 320 {convert to m/s }

We get Cl = 0.2095

Assuming lift curve slope as 2 i.e. CL = 2

9.An experimental plane is fitted with RAF 34 wing with total area of 96 m2 and chord to length ratio of 1: 6 If the plane weights 5 x 105 N , determine the minimum take off speed .How is the runway length related to this speed ?also determine the power required.(Assuming for RAF-43 CL is maximum at 19 degree angle of attack for which CD= 0.16 and CL =1.3)assuming level flight ,lift balances weight

5x105 = 1.3 x 96 x 1.208 x u2 2

u = 81.44 m/s

As the take off speed corresponding to maximum value of CL

Take off speed = 81.44m/s

The power produced by the aero engine is such that it accelerates the aeroplane to achieve this velocity well within the runway of length S .for M is the mass of aeroplane whose velocity changes from 0 to U in distance S

Acceleration a can be written in 2 ways

a = u2 and a = drag /M

2s

s = u2 u2 x M 2s 2 x D

power required = drag x velocity

= CL A ( U2 xU 2 = 0.16 x96 x 1.208 x 81.43x 0.5

= 5011.2 x 103watts

10. for an aircraft of following characteristics determine the angle of attack that will ensure take off at 30 m/s and power required for take off.

Weight = 13,500 N Wing area = 30 m

Take off speed = 30m/s

A model test has shown that CL and Cdof the wing vary with angle of attack as

CL = 0.36(1 + 0.2 z)

Cd = 0.008(1 + z)

Assuning level flight W = L

= CL A ( U2 2

13500 = CL x 30 x 1.208 x 302x 0.5 CL = 0.828 Given CL = 0.36(1 + 0.2 z)

From which we get z = 6.5 degrees

Also given Cd = 0.008(1 + z)

From which we get Cd = 0.06

Power required = drag x velocity

= CL A ( U2 xU 2

= 0.6 x 30 x 1.205 x 303 x .5 = 29.35 x 103 watts11.A water craft is fitted with a hydrofoil of area to provide necessary lift to support the weight of the craft .If the drag and lift coefficients of the hydrofoil are 0.50 and 1.60 respectively, and weight of the craft is 18000 n, determine thee minimum speed when ware craft will be fully supported. What will be the power required to overcome hydrofoil resistance in water?

If the craft is fitted with 90kW engine estimate the top speed.

The hydrofoil boats or water crafts filted with hydrofoils which are submerged in water.As the boat speeds up, the lift on the hydrofoil increases just like an aerofoil .It seema at certain speed , the fukk weight of boat with crew is supported by the lift of thww hydrofoil keeping the boat above water level in that condition W = L

18000 = CL A ( U2 2

= 1.6 x .70 x 1000 x U2 2

U = 5.675 m/s

Drag = CL x lift = (0.5/1.6) x 18000

Cd = 5625 N Power minimum = D x velocity

= 5625 x 5.675

= 31.92 x 103 watts

at 90 Kw power

90000 = drag x velocity

= CL A ( U2 2

= 0.5 X 0.70 X 1000 X U2 2

from this we get top speed u = 8.017 m/s 12. Aircraft of mass M is slowed after landing by two parachutes employed from the rear. Each parachute is of diameter D and drag coefficient Cd if the landing speed of the air craft is U0 estimate the time required for the aircraft to decelerate to a velocity of U0 /4 . Assume air resistance of aircraft be negligible The resistance of a parachute may be taken as the drag of the parachute

= CL A ( U2 2

resistance force on aircraft = mass x acceleration

- M du = CL A ( U2 dt 2

du /dt =

13. a light plane weighting 12.0 kN has a wing span of 9.0 m and chord length of 1.2 m assuming the lift characteristics of thw wing determine the angle of attack for a take off speed of 144 km/hr .Also determine the power required at take off speed if the parasite drag coefficient is 0.02 .further determine the stall speed .For angles of attack between 10 & 14 degrees a liner variation of CL from 0.8 to 1.1 may be assumed for angle of attack between 10 and 14 degrees the linear variation of CL from o.8 to 1.1 may be assumed .also for the airplane Cd = 0.07 and Cf = 0.02

for level flight conditions at sea level ,

W = lift

12000 = X 9 x 1.5 x 1.208 x [(144 x 1000)/60x60]

From which CL = 0.92Corresponding angle of attack is 12 degrees

Total drag coefficient = 0.07+0.02 = 0.09

Power required = drag x velocity

= CL A ( U2 xU 2

substituting the values we get = 46.97 x 103 watts if U corresponds to stalling condition 12000 = 1.3 x (9 x 1.5 ) x 1.208 x U2 2

u = 33.65 m/s

14.a man weighting 80kg decents from an aeroplane using 5.5m diameter parachute .Determine maximum terminal velocity that can be attained .Assume drag coefficient As 1.33

The drag of parachute will be exactly balance thw weight of the airman

80 x x9.81 = CL A ( U2 2

= 1.33 x A x 5.5 x 1.208 x U2 2

U = 6.434 m/s 15. determine the rate of deceleration that will be experienced by a blunt nosed projectile of drag coefficient 1.22 when it is moving horizontally at 1600 km/hr .The projectile has a diameter of 0.5 m and weights 3000N

horizontal velocity = 1600 km/hr

= 444.44m/s

drag force on the projectile is equal to mass multiplied by acceleration

CL A ( U2 = - M x acceleration

2

1.22 x (3.14/4) x 0.52 x x1.208 x 444.442 x .5 = (-300/9.81)x a

declaration a = 93.369 m/s2]16. A flat plate 1m x 1m moves at 6.5 m/s to its plane .Compute the resistance of the plate of when the surrounding fluid is

a) air with mass density 1.2 kg/m3

b) water with a density 1000kg/m3

assume drag coeffienient = 1.15 for both the cases

the resistance or drag = Cd A ( U2 2

a) for air = 1.15 x (1 x 1 ) x 1.2 x 6.52 x .5

= 29.13N

b) for water = 1.15 x (1 x 1 ) x 1000x 6.52 x .5

= 24293.75N

17.Atruck having a projected area of 6.5 m2 traveling at 70km/hr has a total resistance of 2000N of this 20% is due to friction and 10% due to surface friction .The rest is due to form drag .Make calculations for coefficient of drag .Take density = 1.22 kg/m3 for air total resistance to motion = 2000 N

resistance of rolling friction = 2000 x (20/100) = 400 N

resistance of surface friction = 2000 x (10/100) = 200 N

resistance due to form drag = 2000 (400+200)

= 1400N

= CD A ( U2 2

= CD x 6.5 x 1.22 x 05 x [(70 x 1000)/(60 x 60)] we get CD = 0.934

the coefficient of form drag = 0.984

18. Air blows over a cylinder smoke stack of 5 cm diameter with a velocity of 0.2 m/s work out the total aerodynamic drag. shear drag ,pressure drag, in a 1m length of smoke stack .For the flow situation , the total and shear drag coefficients are 1.25 and 0.18 respectively .take air density as 1.20total drag = CD A ( U2 2

= 1.25 x (1 x 0.05) x 1.20 x 0.22 x .5 = 0.1498 Nshear drag = 0.18 x (1 x 0.05) x 1.2 x .22 x .5

= 0.0216N

pressure drag = 0.1498 0.0216

= 0.1282 N

19. A man weighting 100 kg descends to the ground froma an aeroplane with the help of a parachute agsinst the resistance of air .the ahpe of the parachute is hemispherical of 2m diameter .Find the velocity of the parachute with which it comes down .Assume cd = 0.5, density 0.00125 g/cm and v = 0.15 stokes .For parachute decending down the weight balances with the air drag

Projected area of hemisphere = 3.14 x 22/4

Air density = 0.00125 g/cm3 = 1.25 kg/m3weight of parachute decending down = 100 kgf = 100 x 9.81 N

W = CD A ( U2 2

100 x 9.81 = 0.5 x A x 1.25 x U2/2

U = 31.62m/s

20. Experiments were conducted in a wind tunnel with wing speed of 50 km/hr on a flat plate 2 m long and 1 m wide .The density of air is 1.15kg/m3And the plate is kept ay such an angle that the coefficients of lift and drag are 0.75 and 0.15 respectively .determine :

1) lift force

2) drag force

3) resultant force

4) power exterted by the air stream on the plate

lift = CL A ( U2 2

= 0.75 x (2 x 1) x 1.15 x 0.5 [(50 x 1000)/(100 x 60)]drag D = Cd A ( U2 2 = 0.15 x (2 x 1) x 1.15 x .5 x [(50 x 1000)/(100 x 60)] = 166.11 NResultant force (L2 + D2 ) 0.5 = 169.44 N

Inclination of resultant with free stream flow

= tan-1 (L/D) (166.11/33.22)

= tan-1(166.11/33.22)

= 78.42 degrees

Power exerted = drag X velocity

= 33.22 x [(50 x 1000 )/(60 x 60 )] = 462.5 watts

21.A kite weighting 1 kg having an area 1 square meters makes an angle of 70 degree to horizontal when flying in a wind of 36 km/hr .if pull on the sting attached to the kite is 5 kgf and it is inclined to the horizontal at 45degrees, calculate the lift and drag coefficients .

under the action of lift ,drag,weightand tension the kite is said to be in equilibrium

D = T cos 45

= 5 x cos 45

= 3.53 x 9.81 N

= Cd A ( U2 2

= Cd x 1 x .5 x [(36 x 1000)/(60 x 60 )] 2 = 0.557

L = Tsin45 + W

= (5sin 45 +1 )9.81 N

= 4.53 x 9.81 N

= CL A ( U2

2

= CL x 1 x 1.2 x .5 x [(936 x 1000)/(60 x 60)]

we get CL = 0.740

22.A kite of dimensions 0.9 x 0.9m and weighting 7.5N is maintained in air at an angle of 10degrees to the horizontal .The string attached to the kite makes an angle of 45 degrees to the horizontal and at this position, the drag and lift coefficient are estimated to be 0.6 and 0.8 respectively. Make calculations for the wind speed and tension in the string. Take density as 1.2 kg/m2Under the action of lift ,drag,weight and tension ,the kite is said to be in equilibrium

D = T cos 45

Cd A ( U2 = T cos 45

20.6 x (0.9 x 0.9 ) x 1.2 x 0.5 x U2 = T cos 45

0.3276 U2 = T /(21/2 )

L = Tsin45 + W

0.8 x (0.9 x 0.9 ) x 1.2 x 0.5 x U2 = t sin 45 + 7.5

0.4368 U2 = T / (21/2 )+ 7.5

From the two equations we get the values of U and T

As

u = 8.287 m/s T = 31.82 N

23. The wind velocity is commonly measured with a cup anemometer which consists of two hollow hemispheres mounted in opposite directions at the ends of a horizontal rod which turns freely about a vertical axis. The accompanying sketch shows one such anemometer .If the wind velocity is 60km/hr, what torque is required to hold the rotating member stationary. Cd=1.32 for hemisphere hollow on the up stream

=0.35 for hemisphere hollow on the down stream

And for air (=1.22kg/m3

Drag force ( hollow stream ) D1 = Cd A( U2 2

= 1.3 x [(3.14 x 0.12)] x 1.22 x .5 [(60 x 1000)/(60 x60)] 2 = 1.765 N

Drag force (hollow down stream ) = 0.35 x [(3.14 x 0.12)] x 1.22 x .5 [(60 x 1000)/(60 x60)] 2 = 0.47 NTorque required = (D1-D2) x radius

= (1.765-0.47) x 0.5/2

= 0.3225N

24.Estimate the pressure on a maruthi van , 1.4m wide ,1.5 m high and 2.5 m long , traveling at 60km/h through air of ( =1.2 kg /m3 v= 15 x106 m/s2 .Assume that a turbulent boundary layer starts immediately at the leading edge . Cd =0.75.

Profile drag or total drag on the body is made of a pressure drag and skin friction drag

Pressure drag = Cd A ( U2 = 0.75 2

A = 1.5 x 1.4 ; Cd = 0.75

U = 60km/hr = 16.67m/sec

Pressure drag = 0.75 x (1.5 x 1.4 ) x 1.2 x 16.672 x 0.5

= 262.5 N

Reynolds number for the flow taking longest dimension

= UL/v

assuming for air Re = (16.67 x 2.5 )/0.000015

= 2.78 x 106

the turbulent boundry layer is assumd to develop on two sides and top .The skin friction drag for turbulent boudry layer Cf = 0.455/(log Re ) 2.58 Cf = 0.455/(log 2.78 x106) 2.58 = 7.128 N

Assuming skin friction drag =

= 2.78 x106 x 2.5 x [1.6 x 2 + 1.4 ] x 1.2 x 16.672x 0.5 = 7.128 N

Profile drag = 262.5 +7.128

= 269.6128 N

25.A kite weighing 0.25 kg can be assumed to be a flat plate of face area 0.5 m2 and is flying in air ((=1.2 kg/m3).The wind velocity is 10 m/s in the horizontal direction .The tension in the string holding the kite is 20KN and it makes an angle of 40 to the horizontal .Calculate the lift and drag coefficients.

Since the kite is in equilibrium ,the drag and lift forces are balanced by the tension in the string and the weight og the kite

Horizontal forces D = T cos 40

Vertical forces L = T sin 40 + W

D = 20 cos 40 = 15.32 N

= Cd A( U2 2

= Cd x 0.5 x 1.2 x 100 x 0.5

Cd = 0.511

L = 20 sin 40 + (0.25 x 9.81)

= 15.308

= CL x 0.5 x 1.2 x 100 x 0.5

CL = 0.51

26.A wing of a small airplane is rectangular in plan (10m x 1.2).In the straight and level flight at 240 km/h, the total aerodynamic force acting on the wing is 20KN .If the lift/drag ratio is 10 , calculate the coefficient of lift and the total weight the plane can carry.

The aerodynamic force acting on the wing has two components ,vertical component lift L, and horizontal component drag D) Such that the aerodynamic force is equal to (L2 + D2) 0.5 Also given L/D = 10 ; for level flight the eight of air craft exactly balances lift L

20 x 103 = (L2 + D2) 0.5 substituting D = L/10

we get 20 x 103 = L (1.010.5)

L = CL A ( U2

2

L = CL x (10 x 1.2 ) x 1.208 x 0.5 x[(240x 1000)/3600] 2From this we get CL = 0.622The weight of aero plane = 19.9x103 Nthe steel ball will occupy a position a position such that the string makes an angle to the vertical through the point of suspension in the equilibrium position of the ball .

Reynolds number of the fluid flow taking dis of ball as characteristic length Re = ( U dia = (1.2 x 12 x0.01)/(1.85 x 10-5 )

v = (24/Re)+(3/Re0.5) + 0.34 = 0.337

D = Cd A( U2 2

= 0.377 x 3.14 x (0.0052) x1.2 x 122 x 0.5

= 25.59 N

For equilibrium neglecting the string tension

D cos 0 = m g sin 0 Where m is the mass of steel ball

m = (4/3) x 3.14 x (0.005) 3 x 7600 = 0.398 x 10-2 kg

(Assuming density of steel as 7600 kg/m3) angle = tan-1(d/mg)

= 3.75 degrees

28. a spherical balloon of 4 m in diameter is filled with hydrogen ((=0.1 kg/m3).And the total mass of the balloon skin and the load it carries is 30 kg.Estimate the upward terminal velocity of the balloon when for air (= 1.0 kg/m3 and U =1.8 x 10-5 pa-s. When the balloon is moving upwards the buoyant force will be acting upwards and it is balanced by drag force and weight on the balloon, both acting downwards.

Upward buoyant force = weight of the displaced air = (4/3) x 3.14 = (4/3) x 3.14 x 23 x 1 x 9.81

= 328.78 N

weight of the balloon = 30 + weight of the hydrogen in the balloon

= [30 +(4/3) x 3.14 x R3 x (] x g

= [30 +(4/3) x 3.14 x 23 x 0.1] 9.81 = 327.178 NDrag force acting down wards = 328.78 327.178

= 1.502 N

= Cd A( U2 2

= 0.4 x 3.14 x R2 x 1x 0.5 x U2 from which U = 0.733 m/s

the upward velocity is 0.773 m/s for the balloon

29. a paratrooper weighing 800N descends with a terminal velocity of 5m/s in atmosphere of (=1.2 kg/m3 . Estimate the diameter of the parachute if the drag coefficient is 1.3.For a desenting parachute the drag force acting upwars is balanced by the weight

W = D

= Cd A( U2 2

800 = 1.30 x (3.14 x Dia2 ) x 1.2 x 5 x 5 / 4 2 Dia = 7.22 m30. The wing of an airplane has a span of 12 m and chord length of 2.5 m. It produces a lift of 100 KN at a speed of 500 km/h. A 1/25 the model of the wing is tested in a wind tunnel at 400 m/s and( =5.3 kg/m3 .The total drag measured is 380 N assuming that the test is based on wings if the infinite span, calculate the total drag for the prototype when the lift distribution is elliptical and ( For air =1.2 kg/ m3.It is assumed that the wind tunnel test is based on wings of infinite span and so the effect of induced drag is absent .Hence in calculation of total drag of prototype induced drag is to be added.

For a 1/25th size model wing area can be taken as [1/25] 2. The Times the wing area of the prototype. Wing are of model = [1/25] 2 x wing area of the prototype

= [1/25] 2 x 2.5 x1.2 = 30/625 m2 Cd from wind tunnel test of model = D/[(A x ( xU2/ 2]

= 380/[1.2 x (30/ 625) x 4002 ]

= 0.0187

For level flight assuming lift balances weight of prototype

W = L

L = CL A ( U2

2

= CL x 30 x 1.2 x [(500 x 1000)/60 x 60 ] 2

CL = 0.288

For elliptical lift distribution across span the coefficient of induced drag

Cd = CL 2 / ( 3.14 x AR)

= 0.288 /[3.14 x (12/2.5)]

= 0.55 x 10-2

total drag coefficient Cd = 0.0187 + 0.55 x 10-2

= 0.0242

total drag of wing = Cd A( U2 2

= 0.0242 x (12 x 2.5) x 1.2 [(500 x 1000)/(60 x 60)]

= 8401.7 N

31.A 25 cm diameter round nosed projectile travels at a speed of 650 m/s through standard atmosphere at an altitude of 6 km.Find the drag.

(For a round nosed projectile be assumed 0/6 at or near mach no: 2)

Temperature and air density at 6 Km altitude from standard table -23 degrees and 0.66 Kg/m3

Local velocity of projectile M = V/A = (650/314.914)

= 2.05

Mach no: being near 2 , Cd may be taken as 0.6 for the case

Drag = Cd A( U2 2

substituting we get = 4.107 N