Stagnation Properties

Problem Statement

Air flow through a device such that stagnation i 0 6 MP th t ti t t ipressure is 0.6 MPa, the stagnation temperature is

400°C, and the velocity is 570 m/s. Determine the static pressure and temperature of the air at thisstatic pressure and temperature of the air at this state.

Given:

To = 400°C, Po = 0.6 MPa, V = 570 m/s

To Find:

T, P

To, PoSolution:Solution:

T, P(static)

At the stagnation point, the air velocity is zero, and the kinetic energy (V2/2) is converted to thermal energy. Thekinetic energy (V /2) is converted to thermal energy. The temperature and pressure (To, Po) also change at the stagnation point.

If th th l i th f t i h th th th l

2V

If the enthalpy in the free stream is h, then the enthalpy at the stagnation point is

2oVh h= +

Remember h = h(T) for an ideal gas.

In this problem, stagnation properties are given, therefore2V( )2

2oVh h T= −

⎡ ⎤( ) ( )( )

?

? ? (from the table A-17)

o o

o

h T h KkJhkg

⎡ ⎤= ⎣ ⎦

⇒ =g

2

2oVh h= −

2

2 2

2

?1000

?2

m kJ kgs m skJh

k

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⇒ = −2

?

kgkJhkg

=

?T K= ←from table A-17 ?T K= ←from table A 17,

An isentropic process relates the stagnation andAn isentropic process relates the stagnation and static properties.

( )1k kP TP T

−⎛ ⎞

⇒ = ⎜ ⎟⎝ ⎠o oP T⎝ ⎠

( )1.4 1.4 1?? KP MPa

−⎛ ⎞

⇒ ⎜ ⎟??

?

P MPaK

P MPa

⇒ = ⎜ ⎟⎝ ⎠

⇒ = ←