Chapter 12compressible Flow 12 6air At 300 K Is

Chapter 12 Compressible Flow

Chapter 12

COMPRESSIBLE FLOW

Stagnation Properties

12-1C The temperature of the air will rise as it approaches the nozzle because of the stagnation process.

12-2C Stagnation enthalpy combines the ordinary enthalpy and the kinetic energy of a fluid, and offers

convenience when analyzing high-speed flows. It differs from the ordinary enthalpy by the kinetic energy

term.

12-3C Dynamic temperature is the temperature rise of a fluid during a stagnation process.

12-4C No. Because the velocities encountered in air-conditioning applications are very low, and thus the

static and the stagnation temperatures are practically identical.

12-5 The state of air and its velocity are specified. The stagnation temperature and stagnation pressure of

air are to be determined.

Assumptions 1 The stagnation process is isentropic. 2 Air is an ideal gas.

Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4.

Analysis The stagnation temperature of air is determined from

K 355.8=

⎟

⎠

⎞

⎜

⎝

⎛

⋅×

+=+= 22

2

0/sm 1000

kJ/kg 1

KkJ/kg 005.12

m/s) 470(

K 9.245

2p

c

V

TT

Other stagnation properties at the specified state are determined by considering an isentropic process

between the specified state and the stagnation state,

kPa 160.3=

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

=

−

−)14.1/(4.1

)1/(

0

0K 245.9

K 355.8

kPa) 44(

kk

T

PP

Discussion Note that the stagnation properties can be significantly different than thermodynamic

properties.

Chapter 12 Compressible Flow

12-6 Air at 300 K is flowing in a duct. The temperature that a stationary probe inserted into the duct will

read is to be determined for different air velocities.

Assumptions The stagnation process is isentropic.

Properties The specific heat of air at room temperature is cp = 1.005 kJ/kg⋅K.

Analysis The air which strikes the probe will be brought to a complete stop, and thus it will undergo a

stagnation process. The thermometer will sense the temperature of this stagnated air, which is the

stagnation temperature, T0. It is determined from

p

c

V

TT 2

2

0+=

AIR

300 K

V

(a) K 300.0=

⎟

⎠

⎞

⎜

⎝

⎛

⋅×

=2

s/

2

m 1000

kJ/kg 1

KkJ/kg 005.12

2

m/s) (1

+K 300

0

T

(b) K 300.1=

⎟

⎠

⎞

⎜

⎝

⎛

⋅×

=22

2

0s/m 1000

kJ/kg 1

KkJ/kg 005.12

m/s) (10

+K 300T

(c) K 305.0=

⎟

⎠

⎞

⎜

⎝

⎛

⋅×

=22

2

0s/m 1000

kJ/kg 1

KkJ/kg 005.12

m/s) (100

+K 300T

(d) K 797.5=

⎟

⎠

⎞

⎜

⎝

⎛

⋅×

=22

2

0s/m 1000

kJ/kg 1

KkJ/kg 005.12

m/s) (1000

+K 300T

Discussion Note that the stagnation temperature is nearly identical to the thermodynamic temp erature at

low velocities, but the difference between the two is very significant at high velocities,

Chapter 12 Compressible Flow

12-7 The states of different substances and their velocities are specified. The stagnation temperature and

stagnation pressures are to be determined.

Assumptions 1 The stagnation process is isentropic. 2 Helium and nitrogen are ideal gases.

Analysis (a) Helium can be treated as an ideal gas with cp = 5.1926 kJ/kg·K and k = 1.667. Then the

stagnation temperature and pressure of helium are determined from

C55.5°=

⎟

⎠

⎞

⎜

⎝

⎛

°⋅×

+°=+= 22

2

0s/m 1000

kJ/kg 1

CkJ/kg 1926.52

m/s) (240

C50

2p

c

V

TT

MPa 0.261=

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

=

−

−)1667.1(/667.1

)1(/

0

0K 323.2

K 328.7

MPa) 25.0(

kk

T

PP

(b) Nitrogen can be treated as an ideal gas with cp = 1.039 kJ/kg·K and k =1.400. Then the stagnation

temperature and pressure of nitrogen are determine d fr om

C93.3°=

⎟

⎠

⎞

⎜

⎝

⎛

°⋅×

+°=+= 22

2

0s/m 1000

kJ/kg 1

CkJ/kg 039.12

m/s) (300

C50

2p

c

V

TT

MPa 0.233=

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

=

−

−)14.1/(4.1

)1/(

0

0K 323.2

K 366.5

MPa) 15.0(

kk

T

PP

(c) Steam can be treated as an ideal gas with cp = 1.865 kJ/kg·K and k =1.329. Then the stagnation

temperature and pressure of steam are determined from

K 685C411.8 =°=

⎟

⎠

⎞

⎜

⎝

⎛

°⋅×

+°=+= 22

2

0s/m 1000

kJ/kg 1

CkJ/kg 865.12

m/s) (480

C350

2p

c

V

TT

MPa 0.147=

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

=

−

−)1329.1/(329.1

)1/(

0

0K 623.2

K 685

MPa) 1.0(

kk

T

PP

Discussion Note that the stagnation properties can be significantly different than thermodynamic

properties.

12-8 The inlet stagnation temperature and pressure and the exit stagnation pressure of air flowing through a

compressor are specified. The power input to the compressor is to be determined.

Assumptions 1 The compressor is isentropic. 2 Air is an ideal gas.

Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4.

100 kPa

27°C

AIR

0.02 kg/s

900 kPa

&

W

Analysis The exit stagnation temperature of air T02 is determined from

K 562.4

100

900

K) 2.300( 4.1/)14.1(

/)1(

01

02

0102 =

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

=

−

−kk

P

TT

From the energy balance on the compressor,

)( 0120in hhmW −= &

&

or,

kW 5.27=300.2)KK)(562.4kJ/kg 5kg/s)(1.00 02.0()( 0102in −⋅=−= TTcmW p

&

Discussion Note that the stagnation properties can be used conveniently in the energy equation.

Chapter 12 Compressible Flow

12-9E Steam flows through a device. The stagnation temperature and pressure of steam and its velocity are

specified. The static pressure and temperature of the steam are to be determined.

Assumptions 1 The stagnation process is isentropic. 2 Steam is an ideal gas.

Properties Steam can be treated as an ideal gas with cp = 0.4455 Btu/lbm·R and k =1.329.

Analysis The static temperature and pressure of steam are determined from

F663.7°=

⎟

⎠

⎞

⎜

⎝

⎛

°⋅×

−°=−= 22

2

0s/ft 25,037

Btu/lbm 1

FBtu/lbm 4455.02

ft/s) (900

F700

2p

c

V

TT

psia 105.5=

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

=

−

−)1329.1/(329.1

)1/(

0

0R 1160

R 1123.7

psia) 120(

kk

T

PP

Discussion Note that the stagnation properties can be significantly different than thermodynamic

properties.

12-10 The inlet stagnation temperature and pressure and the exit stagnation pressure of products of

combustion flowing through a gas turbine are specified. The power output of the turbine is to be

determined.

Assumptions 1 The expansion process is isentropic. 2 Products of combustion are ideal gases.

Properties The properties of products of combustion are cp = 1.157 kJ/kg⋅K, R = 0.287 kJ/kg⋅K, and k =

1.33.

Analysis The exit stagnation temperature T02 is determined to be

100 kPa

1 MPa

750

°

C

K 9.577

1

0.1

K) 2.1023( 33.1/)133.1(

/)1(

01

02

0102 =

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

=

−

−kk

P

TT

STEAM W

Also,

K kJ/kg157.1 133.1

K) kJ/kg287.0(33.1 1

)(

⋅=

−

⋅

=

−

=⎯→⎯−== k

kR

cRckkcc ppvp

From the energy balance on the turbine,

)( 0120out hhw −=−

or,

kJ/kg 515.2=K.9)5772K)(1023.kJ/kg 157.1()( 0201out −⋅=−= TTcw p

Discussion Note that the stagnation properties can be used conveniently in the energy equation.

Chapter 12 Compressible Flow

2-11 Air flows through a device. The stagnation temperature and pressure of air and its velocity are

specified. The static pressure and temperature of air are to be determined.

Assumptions 1 The stagnation process is isentropic. 2 Air is an ideal gas.

Properties The properties of air at an anticipated average temperatu re of 600 K are cp = 1.051 kJ/kg⋅K and

k = 1.376.

Analysis The static temperature and pressure of air are determined from

K 518.6=

⎟

⎠

⎞

⎜

⎝

⎛

⋅×

−=−= 22

2

0s/m 1000

kJ/kg 1

KkJ/kg 051.12

m/s) (570

2.673

2p

c

V

TT

and

MPa 0.23=

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

=

−

−)1376.1/(376.1

)1/(

02

2

022 K 673.2

K 518.6

MPa) 6.0(

kk

T

PP

Discussion Note that the stagnation properties can be significantly different than thermodynamic

properties.