Calculation of Velocity in Isentropic Flow of an Airfoil, Assignments of Aerodynamics

A step-by-step solution to calculate the velocity at a point on an airfoil surface in an isentropic flow, given the freestream conditions and the pressure at the point. The solution involves the use of the isentropic relation and the mach number.

Typology: Assignments

2023/2024

Uploaded on 04/18/2024

juliane-nicole-blen
juliane-nicole-blen ๐Ÿ‡ต๐Ÿ‡ญ

1 document

1 / 6

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
1. An airfoil is in a freestream where ๐‘ƒโˆž = 0.61 atm, ๐œŒโˆž = 0.819 kg/m3, and ๐‘‰โˆž = 300
m/s. At a point on the airfoil surface, the pressure is 0.5 atm. Assuming isentropic
flow, calculate the velocity at that point.
GIVEN:
๐‘ƒโˆž=0.61 atm
๐œŒโˆž=0.819 kg/m3
๐‘‰โˆž=300 m/s
P = 0.5 atm
REQUIRED:
V
SOLUTION:
๐‘ƒโˆž=๐œŒโˆž๐‘…๐‘‡โˆž
๐‘‡โˆž=๐‘ƒโˆž
๐œŒโˆž๐‘…
๐‘‡โˆž=(0.61 ๐‘Ž๐‘ก๐‘š)(101,325 ๐‘ƒ๐‘Ž
1 ๐‘Ž๐‘ก๐‘š )
(0.819๐‘˜๐‘”
๐‘š3)(287.08 ๐ฝ
๐‘˜๐‘”โˆ’๐พ
๐‘‡โˆž=262.88 ๐พ
๐‘‰๐‘Žโˆž=20.05 โˆš๐‘‡
๐‘‰๐‘Žโˆž=20.05 โˆš262.88 ๐พ
๐‘‰๐‘Žโˆž=325.08๐‘š
๐‘ 
๐‘€โˆž=๐‘‰โˆž
๐‘‰๐‘Ž
Activity
No.: 2
Name: ALDECOA, Mico Onil Ram
BLEN, Juliane Nicole A.
VILLARUEL, Pee-Jaye
Instructor: Engr. Roberto Renigen
Submission Date: 18 March 2024
Rating:
Page 1of 6 pages
Problem Solving
pf3
pf4
pf5

Partial preview of the text

Download Calculation of Velocity in Isentropic Flow of an Airfoil and more Assignments Aerodynamics in PDF only on Docsity!

  1. An airfoil is in a freestream where ๐‘ƒ

โˆž

= 0.61 atm, ๐œŒ

โˆž

= 0.819 kg/m

3

, and ๐‘‰

โˆž

m/s. At a point on the airfoil surface, the pressure is 0.5 atm. Assuming isentropic

flow, calculate the velocity at that point.

GIVEN:

โˆž

= 0. 61 atm

โˆž

= 0. 819 kg/m

3

โˆž

= 300 m/s

P = 0.5 atm

REQUIRED:

V

SOLUTION:

โˆž

โˆž

โˆž

โˆž

๐‘ƒ

โˆž

๐œŒ

โˆž

๐‘…

โˆž

( 0. 61 ๐‘Ž๐‘ก๐‘š)(

101 , 325 ๐‘ƒ๐‘Ž

1 ๐‘Ž๐‘ก๐‘š

)

( 0. 819

๐‘˜๐‘”

๐‘š

3

)( 287. 08

๐ฝ

๐‘˜๐‘”โˆ’๐พ

โˆž

๐‘Ž

โˆž

๐‘Ž โˆž

๐‘Ž

โˆž

๐‘š

๐‘ 

โˆž

๐‘‰

โˆž

๐‘‰

๐‘Ž

No.: 2

Name: ALDECOA, Mico Onil Ram

BLEN, Juliane Nicole A.

VILLARUEL, Pee-Jaye

Instructor: Engr. Roberto Renigen

Submission Date: 18 March 2024

Rating:

Problem Solving

โˆž

โˆž

๐‘ƒ

๐‘ƒ โˆž

๐‘€

โˆž

2

  • 5

๐‘€

2

2

  • 5
  1. 5

[

โˆž

โˆž

2

2

2

  1. 5

]

1

  1. 5

๐‘ƒ

๐‘ƒ โˆž

1

  1. 5

๐‘€ โˆž

2

  • 5

๐‘€ 2

2

  • 5

๐‘ƒ

๐‘ƒ

โˆž

1

  1. 5

2

2

โˆž

2

2

2

๐‘€

โˆž

2

  • 5

(

๐‘ƒ

๐‘ƒ

โˆž

)

1

  1. 5

2

2

๐‘€

โˆž

2

  • 5

(

๐‘ƒ

๐‘ƒ

โˆž

)

1

  1. 5

2

๐‘€

โˆž

2

  • 5

(

๐‘ƒ

๐‘ƒ

โˆž

)

1

  1. 5

2

( 0. 923 )

2

  • 5

(

  1. 5

  2. 61

)

1

  1. 5

2

No.: 2

Problem Solving

  1. An air tank with a nozzle has a pressure of 196.32 Kpa and a density of 1.9 kg/m

3

Outside the converging-diverging nozzle, the pressure is atmospheric and designed

to have a Mach No. of 1.0 and 1. 5 at the throat and exit, respectively. The area at

the throat is 0.11m

2

. Calculate the following:

(a) Temperature and speed of sound at the tank

(b) Pressure, density, temperature, and speed of sound at the throat

(c) Mass flow at the exit

GIVEN:

P

o

= 196.32 KPa = 196,320 Pa

ฯ o

= 1.9 kg/m

3

Area at the throat (๐ด

๐‘ก

) = 0.11m

2

REQUIRED:

(a) ๐‘‡

๐‘œ

๐‘Ž 0

(b) ๐‘ƒ

๐‘‡

๐‘‡

๐‘‡

๐‘Ž

๐‘‡

(c) แน

๐ธ

SOLUTION:

(a) ๐‘ƒ

๐‘œ

= ฯ o

R๐‘‡

๐‘œ

๐‘œ

๐‘ƒ ๐‘œ

๐œŒ๐‘…

196 , 320 Pa

( 1. 9

๐‘˜๐‘”

๐‘š

3

)( 287. 08

๐ฝ

๐‘˜๐‘”โˆ’๐พ

)

๐‘œ

= 359.92K

No.: 2

Problem Solving

๐‘Ž

๐‘Ž

๐‘‡

= 380.38 m/s

(b)

๐‘ƒ

๐‘ก

๐‘ƒ

0

= [

(๐‘€

1

)

2

  • 5

(๐‘€

2

)

2

  • 5

]

  1. 5

๐‘‡

= [

( 1 )

2

  • 5

( 1. 5 )

2

  • 5

]

  1. 5

[ 196. 32 ๐พ๐‘ƒ๐‘Ž]

๐‘‡

= 101.23 Kpa = 101,230 Pa

๐‘ ๐‘‡

๐‘ ๐‘œ

๐‘ƒ ๐‘‡

๐‘ƒ ๐‘œ

1

  1. 4

๐‘‡

๐‘ƒ

๐‘‡

๐‘ƒ

๐‘œ

1

  1. 4

(๐‘

๐‘œ

๐‘‡

  1. 23 ๐พ๐‘ƒ๐‘Ž

  2. 32 ๐พ๐‘ƒ๐‘Ž

1

  1. 4 ( 1. 9

๐‘˜๐‘”

๐‘š

3

๐‘‡

๐‘˜๐‘”

๐‘š

3

๐‘‡

๐‘‡

๐‘‡

๐‘‡

๐‘ƒ

๐‘‡

๐‘

๐‘‡

๐‘…

  1. 23 ๐พ๐‘ƒ๐‘Ž

( 1. 184

๐‘˜๐‘”

๐‘š

3

) ( 287. 08

๐ฝ

๐‘˜๐‘”โˆ’๐พ

)

๐‘‡

Speed of Sound at that section

Va T

๐‘‡

Va T

Va T

= 346.01 m/s

No.: 2

Problem Solving