Table B.2 Properties of the International Standard Atmosphere (ISA) between sea level and 20 kmgeopotential altitude9,0009,500Standard values at altitudeP (Pa)8Alt. (m)T (K)0288.15 1 340.29500 284.90 0.9887 338.371,000 281.65 0.9774 336.431,500 278.40 0.9662 334.492,000 275.15 0.9549 332.532,500 271.90 0.9436 330.563,000 268.65 0.9306 328.583,500 265.40 0.9210326.584,000 262.15 0.9098 324.584,500 258.90 0.8985 322.56255.65 0.8872 320.53252.40 0.8759 318.48249.15 0.8647 316.436,500245.90 0.8534 314.367,000 242.65 0.8421 312.27310.17308.065,0005,5006,0007,500 239.40 0.83088,000 236.15 0.81958,500305.9333,09941,060 0.4052 0.5895 0.481238,251 0.3775 0.5566 0.454435,599 0.3513 0.5252232.90 0.80830.32670.4951229.650.7970 303.7930,742 0.3040 0.4663226.40 0.7857 301.63 28,523 0.2815 0.43890.41270.387710,0000.36390.26550.22680.1851223.150.7744299.46 26,436 0.260910,500 219.90 0.7631 297.27 24,474 0.241511,000 216.65 0.7519 295.07 22,632 0.223412,000 216.65 0.7519 295.07 19,330 0.1908 0.310813,000 216.65 0.7519 295.07 16,510 0.162914,000 216.65 0.,7519 295.07 14,101 0.139215,000 216.65 0.7519 295.07 12,044 0.1189 0.1937 0.158116,000 216.65 0.7519 295.07 10,287 0.1015 0.1654 0.135017,000 216.65 0.7519 295.07 8,786 0.0867 0.1413 0.115318,000 216.65 0.7519 295.07 7,505 0.0741 0.1207 0.098519,000 216.65 0.7519 295.07 6,410 0.0633 0.1031 0.084220,000216.65 0.7519 295.07 5,475 0.0540 0.0880 0.07180a (m/s)p (kg/m³) 0μ/1₂111.22501.1673 0.9529 0.99121.1117 0.9075 0.98231.0581 0.8638 0.97351.0065 0.8216 0.96450.9569 0.7811 0.95560.9091 0.7421 0.94650.8632 0.7055101,325 195,461 0.942189,874 0.887084,556 0.834579,4950.784674,682 0.737170,108 0.691965,764 0.649061,640 0.608357,728 0.569754,020 0.533150,506 0.4985 0.6971 0.5691 0.900647,181 0.4656 0.6597 0.538544,034 0.4346 0.6238 0.50920.90990.8191 0.66860.7768 0.63410.7361 0.60090.93750.92830.91910.25370.2167P₁A₁V₁ = P₂A₂V₂VP₂+PZ=P₁+PZP/= P₁V²C₂T₁+2=C₂T₂+0.8911 V₂ =0.88180.87240.86280.4287 0.85320.4042 0.84360.3807 0.83390.3583 0.82410.81430.33690.31650.80440.29710.7944p=pRTa = √√YRTV₁ =Advanced Aircraft Design: Conceptual Design. Analysis and Optimization of Subsonic Civil Airplanes. First Edition. Egbert Torenbeck.2013 by Tighert Torenbeek. Published 2013 by John Wiley & Sons, Lad.C₂ =h = c₂Te = c₂TP₂P₁PozP₁=0.79440.7944 M0.79442(P₁-P₂)√P[1-(A₂/A₂)²)2(Po-P)P2(Po-P)PsYRY-1-) =)"Y/(-1)17/(-1).(y + 1)²M[4YM-2(y-1)]2 (P₂)(x-1)/y0.79440.7944Y-10.7944 dp = -pvdv0.79440.7944 T₁=To [1 + x2 ¹ Mi ¹0.7944²16) *-₁]Y-2a Po-Pi1-y+2yMY+1PD P² + 1) - / - 1-7/(Y-1)P₁ = Po [(1 + ² = 1¹ M³ ] "2P₁ = Po [(1 + ² = ¹ M²³)]*-1/(-1)(4)Re, =8 =CC₁:8 ="=Ca1 2M² y+15.2xPuVxHoo√Rex0.664a=q=Re1.328REL0.37xRe2ci-ci-c|_ci0.0592Re0.074a=dc₁/daC₂Cp=9004a√M²-1C₂CoatREAR(¹1μ= arcsin(+1)/(-1)1+57.3a0/(ne,AR)1M Consider the flow of air over a small flat plate that is 5 cmlong in the flow direction and 1 m wide. The free-streamconditions correspond to standard sea level, and the flowvelocity is 120 m/s. Assuming completely turbulent flow,calculate1. The boundary layer thickness at the downstream edge(the trailing edge)2. The drag force on the plate

Given:Length of the plate, L=5 cm=0.05 mWidth of the plate, w=1 mFree steam velocity of air, V∞=120…

Consider the flow of air over a small flat plate that is 5 cm long in the flow direction and 1 m wide. The free-stream conditions correspond to standard sea level, and the flow velocity is 120 m/s. Assuming completely turbulent flow, calculate 1. The boundary layer thickness at the downstream edge (the trailing edge) 2. The drag force on the plate

Consider the flow of air over a small flat plate that is 5 cm long in the flow direction and 1 m wide. The free-stream conditions correspond to standard sea level, and the flow velocity is 120 m/s. Assuming completely turbulent flow, calculate 1. The boundary layer thickness at the downstream edge (the trailing edge) 2. The drag force on the plate

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Related questions

Q: Atmospheric boundary layers can be estimated using turbulent flow approximations over a flat-plate…

A: Write down the given values along with suitable variables. u = 10 m/s k = 0.41 (Von Karman constant)…

Q: 9.40 A 180-1b man parachutes from a plane using a hemispherical parachute in air at 20 °F and 14.60…

A: The weight of a man is W=180 The air temperature is T=20°F=479.67°R The pressure is P=14.6 psi The…

Q: the boundary layer

A: given:dynamic viscosity μ =1.79 x 10-5 kg/msvelocity V=37m/slength L=1.6m

Q: The boundary layer formed as turbulent around the sphere is when ReD = 2.5*10^5 (a) Golf…

A: Re= 2.5×105Golf ball diameter, Dg= 43 mm= 0.043 mSoccer ball diameter, Ds= 222 mm= 0.222 m Note:…

Q: Two definitions of displacement thickness are given in this chapter. Write both definitions in your…

Q: The two spheres stand with a stop element as in the figure. The spheres are connected with…

A: Given: The specific gravity of steel ball= 5.98 The density of air= 1.225 kg/m3 The density of…

Q: b. A long, thin flat plate is placed parallel to a 6 m/s stream of water at 20°C. At what distance X…

Q: A gas at 20°C and 1 atm flow at 6 ft/s past a thin flat plate.At x = 3 ft, the boundary layer…

A: Given data: The velocity of fluid flow is V=6 ft/s. The distance from the leading edge is x=3 ft.…

Q: Assume that a radioactive dust particle approximates asphere of density 2400 kg/m3. How long, in…

A: (a) It is assumed that Stokes's law is applied for small particles. The dynamics viscosity is taken…

Q: Consider two free-falling parachutists with linear drag. The first jumper is 70 kg and has a drag…

A: Given data:

Q: Consider again a semi truck traveling at 55 mph on the freeway. The ambient air is at 1 atm and 80F.…

A: Write the formula for Reynolds number. Re=VLν Here, Re is the Reynolds number, V is the velocity of…

Q: 2- Air (v = 1.8×10° m/s) flows along a flat plate with a velocity of 150 km/hr. How long does the…

Q: Air equivalent to one of a standard height of 4000 m flows at 450 mi/h over a wing with a thickness…

A: Given, Velocity = V = 450 mi/h = 201.17 m/s Altitude = h = 4000 m Thickness = t = 18 cm = 0.18 m…

Q: boundary layer for which the velocity profile can be approximated by a power law equation, u/U =…

Q: For the shown continuously circular (i.e. not parabolic) arc airfoil with sharp corners of each 8 =…

A: The airfoil angle, The angle of attack, The Mach number,

Q: Air at 20°C flows at V = 80.0 m/s over a smooth flat plate of length L = 17.5 m. Plot the turbulent…

A: Write the expression of Reynolds Number, ReL=VLν The kinematic viscosity of air at 20 oC is taken…

Question

Table B.2 Properties of the International Standard Atmosphere (ISA) between sea level and 20 km
geopotential altitude
9,000
9,500
Standard values at altitude
P (Pa)
8
Alt. (m)
T (K)
0
288.15 1 340.29
500 284.90 0.9887 338.37
1,000 281.65 0.9774 336.43
1,500 278.40 0.9662 334.49
2,000 275.15 0.9549 332.53
2,500 271.90 0.9436 330.56
3,000 268.65 0.9306 328.58
3,500 265.40 0.9210
326.58
4,000 262.15 0.9098 324.58
4,500 258.90 0.8985 322.56
255.65 0.8872 320.53
252.40 0.8759 318.48
249.15 0.8647 316.43
6,500
245.90 0.8534 314.36
7,000 242.65 0.8421 312.27
310.17
308.06
5,000
5,500
6,000
7,500 239.40 0.8308
8,000 236.15 0.8195
8,500
305.93
33,099
41,060 0.4052 0.5895 0.4812
38,251 0.3775 0.5566 0.4544
35,599 0.3513 0.5252
232.90 0.8083
0.3267
0.4951
229.65
0.7970 303.79
30,742 0.3040 0.4663
226.40 0.7857 301.63 28,523 0.2815 0.4389
0.4127
0.3877
10,000
0.3639
0.2655
0.2268
0.1851
223.15
0.7744
299.46 26,436 0.2609
10,500 219.90 0.7631 297.27 24,474 0.2415
11,000 216.65 0.7519 295.07 22,632 0.2234
12,000 216.65 0.7519 295.07 19,330 0.1908 0.3108
13,000 216.65 0.7519 295.07 16,510 0.1629
14,000 216.65 0.,7519 295.07 14,101 0.1392
15,000 216.65 0.7519 295.07 12,044 0.1189 0.1937 0.1581
16,000 216.65 0.7519 295.07 10,287 0.1015 0.1654 0.1350
17,000 216.65 0.7519 295.07 8,786 0.0867 0.1413 0.1153
18,000 216.65 0.7519 295.07 7,505 0.0741 0.1207 0.0985
19,000 216.65 0.7519 295.07 6,410 0.0633 0.1031 0.0842
20,000
216.65 0.7519 295.07 5,475 0.0540 0.0880 0.0718
0
a (m/s)
p (kg/m³) 0
μ/1₂
1
1
1.2250
1.1673 0.9529 0.9912
1.1117 0.9075 0.9823
1.0581 0.8638 0.9735
1.0065 0.8216 0.9645
0.9569 0.7811 0.9556
0.9091 0.7421 0.9465
0.8632 0.7055
101,325 1
95,461 0.9421
89,874 0.8870
84,556 0.8345
79,495
0.7846
74,682 0.7371
70,108 0.6919
65,764 0.6490
61,640 0.6083
57,728 0.5697
54,020 0.5331
50,506 0.4985 0.6971 0.5691 0.9006
47,181 0.4656 0.6597 0.5385
44,034 0.4346 0.6238 0.5092
0.9099
0.8191 0.6686
0.7768 0.6341
0.7361 0.6009
0.9375
0.9283
0.9191
0.2537
0.2167
P₁A₁V₁ = P₂A₂V₂
V
P₂+PZ=P₁+PZ
P/= P₁
V²
C₂T₁+2=C₂T₂+
0.8911 V₂ =
0.8818
0.8724
0.8628
0.4287 0.8532
0.4042 0.8436
0.3807 0.8339
0.3583 0.8241
0.8143
0.3369
0.3165
0.8044
0.2971
0.7944
p=pRT
a = √√YRT
V₁ =
Advanced Aircraft Design: Conceptual Design. Analysis and Optimization of Subsonic Civil Airplanes. First Edition. Egbert Torenbeck.
2013 by Tighert Torenbeek. Published 2013 by John Wiley & Sons, Lad.
C₂ =
h = c₂T
e = c₂T
P₂
P₁
Poz
P₁
=
0.7944
0.7944 M
0.7944
2(P₁-P₂)
√P[1-(A₂/A₂)²)
2(Po-P)
P
2(Po-P)
Ps
YR
Y-1
-) =)"
Y/(-1)
17/(-1).
(y + 1)²M
[4YM-2(y-1)]
2 (P₂)(x-1)/y
0.7944
0.7944
Y-1
0.7944 dp = -pvdv
0.7944
0.7944 T₁=To [1 + x2 ¹ Mi ¹
0.7944
²16) *-₁]
Y-
2a Po-Pi
1-y+2yM
Y+1
PD P² + 1) - / - 1
-7/(Y-1)
P₁ = Po [(1 + ² = 1¹ M³ ] "
2
P₁ = Po [(1 + ² = ¹ M²³)]*
-1/(-1)
(4)
Re, =
8 =
C
C₁:
8 =
"
=
Ca
1 2
M² y+1
5.2x
PuVx
Hoo
√Rex
0.664
a=
q=
Re
1.328
REL
0.37x
Re2
ci-ci-c|_ci
0.0592
Re
0.074
a=dc₁/da
C₂
Cp=
900
4a
√M²-1
C₂
Coat
REAR
(¹
1
μ= arcsin
(+1)/(-1)
1+57.3a0/(ne,AR)
1
M

Consider the flow of air over a small flat plate that is 5 cm
long in the flow direction and 1 m wide. The free-stream
conditions correspond to standard sea level, and the flow
velocity is 120 m/s. Assuming completely turbulent flow,
calculate
1. The boundary layer thickness at the downstream edge
(the trailing edge)
2. The drag force on the plate

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 3 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Similar questions

Hand written plz asap i'll give you multiple upvote
3. Pressure Flow, Flat Plates. Adverse pressure gradients in flow over bodies lead to flow reversal & separation, wakes, and increased drag. To examine such behavior, consider fully developed, laminar flow between two flat pates a distance 8 apart. The bottom plate is fixed, and the top plate moves with speed U (representing the outer flow over a boundary layer of thickness 8). (a) Derive the critical adverse pressure gradient, (dp/dx)*, above which there will be flow reversal, by solving the N-S equation in the coordinates shown, to get a relation for u(y), du/dy, and the critical (dp/dx)* in terms of 8, U, and u above which there μ will be flow reversal at the lower plate. (b) If 8 = 5 mm and U = 10 m/s, evaluate (dp/dx)* if the fluid is air. Moving plate Fixed plate 2/ 1.0 0.8 0.6 0.4 0.2 Back- flow P=-3 0 -0.4 -0.2 0 0.2 0.4 u U (b) U 0.6 0.8 1.0 1.2 b 1.4 (a) FIGURE 6.30 Viscous flow between parallel plates with the bottom plate fixed and the upper plate moving (Couette flow): (a)…
The sphere is dropped in gasoline at 20 ° C.Ignoring its acceleration phase, what will it's terminal (constant)fall velocity be, from Fig. ?
Air at 15°C flows at 10 m/s over a flat plate of length 3 m. Using one-seventh power law of the turbulent flow, what is the ratio of local skin friction coefficient for the turbulent and laminar flow cases? (The kinematic viscosity of air is 1.470 × 10−5 m2/s.) (a) 4.25 (b) 5.72 (c) 6.31 (d ) 7.29 (e) 8.54
Question A3 a) With the help of sketches briefly describe the main differences between laminar and turbulent boundary layers in relation to friction drag and flow separation. A tug boat is pulling a log raft (Figure QA3) on the surface of a lake at a constant speed of Uboat = 5m/s. The log raft can be modelled as a flat plate (L=20m and W = 5.0m). u ug U boat 8 U Tug Boat rope boat Air Water L -X Log Raft Figure QA3 Assume the boundary layer forming on both sides (air side and water side) of the log raft is two-dimensional and laminar with a second order velocity distribution: W b) By using the Momentum Integral Equation find an expression for the wall shear stress in terms of Reynolds number and then estimate the tensile stress acting on the rope assuming the rope has a circular cross section with a diameter of 15 mm) and the power required to pull it. c) What would be the tensile stress on the rope if we assume the boundary layers on the log raft were turbulent? You may assume:…
For flow over a smooth plate, what approximately is the maximum length of the boundary layer if V_o = 9.0 m/s in the irrational uniform flow and the fluid is air? Water?