fluid mechanics
TRANSCRIPT
Government Engineering College - Bhuj
COMPRESSIBLE FLUID FLOW• STUDENT NAME:- PARMAR UMANG R.• SUDENT ENROLLMENT NO.:-130150119076
FLUID MECHANICS
Example :- 1 An air vessel is flying at 215 m/s at a low altitude
where the velocity of sound is 325 m/s. At a certain point just outside the boundary layer of the wings, the velocity of air relative to the plane is 305 m/s Determine the pressure drop on the wing surface near this position. Assume frictionless adiabatic flow and take pressure of ambient air 102 KN/ and
2m .4.1
1
1
2
2
1
1
121
2
1
1
2
1
121
22
22
2
221
1
1
&,
112
2121
P
PPaSince
P
PPVV
VPVP
1
1
2
2
1
1
121
2
1
1
2
1
121
22
22
2
221
1
1
&,
112
2121
P
PPaSince
P
PPVV
VPVP
22
12
4.0
4.1
1
1
2
2
22
21
21
22
1
1
2
1
1
221
21
22
/7305410721.0721.0
721.0911.0911.0
911.0
325
215305
2
14.11
2
11
112
mKNPP
P
P
a
VV
P
P
P
PaVV
Where a = velocity of sound
Hence, pressure drop on the wing
......./46.28
54.731022 AnsmkN
Example :- 2 Find the velocity of a bullet fired in standard air if
mach angle is 30 degree. Take R=287J/Kg=1.4 and T=15degree C.
soln.:
KT
K
KKgJR
angleMachGive
288
4.1
/287
30,
smC
C
KRTC
/25.340
2882874.1
WE know that velocity of sound ‘C’ is
......../50.68030sin
25.340
25.34030sin
sin
AnssmV
V
V
V
CAlso
Example :- 3 A flying object is flying at a speed 1100 km/hour at sea level at a temperature of . Take K=1.4 and Find the Mach number at a point on this flying object. Soln.: Speed of object, V = 1100km/hour =1100*1100/60*60 = 305.55m/s
C20 ./287 KKgJR
.......89.011.343
55.305
/11.343
2932874.1
/287,4.1
29327320
AnsC
VMBut
smC
C
KRTC
KKgJRK
Ktetemperatur
We know that, sound velocity C is
Example :- 4 Find the speed of sound in air at an altitude of 5000 m.
At 5000 m, T = 255.7 K.
CkJ
kg KK
ms
kJkg
m
s
14 0 287 25571000
3205
2
2
. ( . )( . )
.
Notice that the temperature used for the speed of sound is the static (normal) temperature.
Example :- 5 An aircraft flies in air at 5000 m with a velocity of 250
m/s. At 5000 m, air has a temperature of 255.7 K and a pressure of 54.05 kPa. Find To and Po.2
2
2
2
250255.7
2 2(1.005 ) 1000
(255.7 31.1)
286.8
oP
kJmV kgs
T T KkJ mCkg K s
K
K
kPa
K
K
T
TPP
kk
77.80
7.255
8.28605.54
)14.1(4.1
)1(
Thank You…