工程热力学第三版答案【英文】第17章

17-5

Air at 320 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

V2

T0 T . The results for each case are calculated below:

2cp

1 kJ/kg (1 m/s)2 320.0 K (a) T0 320 K+

222 1.005 kJ/kg K m/s 1000(10 m/s)2 1 kJ/kg

K (b) T0 320 K+ 320.1

2 1.005 kJ/kg K 1000 m2/s2 (100 m/s)2 1 kJ/kg

(c) T0 320 K+ 325.0 K

2 1.005 kJ/kg K 1000 m2/s2 (1000 m/s)2 1 kJ/kg

(d) T0 320 K+ 817.5 K 222 1.005 kJ/kg K 1000 m/s

Discussion Note that the stagnation temperature is nearly identical to the

thermodynamic temperature at low velocities, but the difference between the two is significant at high velocities.

17-8

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 temperature 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

(570 m/s)2V2 1 kJ/kg

T

T0 673.2 518.6 K 222cp2 1.051 kJ/kg K 1000 m/s

and

T2

P2 P02 T

02

k/(k 1)

K 518.6

(0.6 MPa)

K 673.2

1.376/(1.376 1)

0.23 MPa

Discussion Note that the stagnation properties can be significantly different than

thermodynamic properties.

17-21

The Mach number of a passenger plane for specified limiting operating conditions is to be determined.

Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.287 kJ/kg·K. Its specific heat ratio at room temperature is k = 1.4. Analysis

From the speed of sound relation

293 m/s

1000 m2/s2

c (1.4)(0.287 kJ/kg K)(-60 273 K) 1 kJ/kg

Thus, the Mach number corresponding to the maximum cruising speed of the plane is

Ma

Vmax(945/3.6) m/s

0.897 c293 m/s

Discussion Note that this is a subsonic flight since Ma < 1. Also, using a k value at

-60 C would give practically the same result.

17-25

The inlet state and the exit pressure of air are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The properties of air are R = 0.287 kJ/kg·K and k = 1.4. The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded.

Analysis The final temperature of air is determined from the isentropic relation of ideal gases,

P2

T2 T1 P

1

(k 1)/k

0.4 MPa

(333.2 K)

1.5 MPa

(1.4 1)/1.4

228.4 K

Treating k as a constant, the ratio of the initial to the final speed of sound can be expressed as

Ratio

c2

c1

k1RT1k2RT2

12

.2.4

1.21

Discussion Note that the speed of sound is proportional to the square root of thermodynamic temperature.

17-40

The critical temperature, pressure, and density of air and helium are to be determined at specified conditions.

Assumptions Air and Helium are ideal gases with constant specific heats at room temperature.

Properties The properties of air at room temperature are R = 0.287 kJ/kg·K, k = 1.4, and cp = 1.005 kJ/kg·K. The properties of helium at room temperature are R = 2.0769 kJ/kg·K, k = 1.667, and cp = 5.1926 kJ/kg·K. Analysis (a) Before we calculate the critical temperature T*, pressure P*, and density *, we need to determine the stagnation temperature T0, pressure P0, and density 0.

(250 m/s)2V2 1 kJ/kg

T0 100 C 100+ 131.1 C

2cp2 1.005 kJ/kg C 1000 m2/s2

T

P0 P 0

T

k/(k 1)

404.3 K

(200 kPa)

K 373.2

1.4/(1.4 1)

264.7 kPa

0

P0264.7 kPa

2.281 kg/m3 3RT0(0.287 kPa m/kg K)(404.3 K)

Thus,

2 2 T* T0 (404.3 K) 337 K

k 11.4+1

2

P* P0

k 1

k/(k 1)

2

(264.7 kPa)

1.4+1

1.4/(1.4 1)

140 kPa

2

* 0

k 1

1/(k 1)

2

(2.281 kg/m)

1.4+1

3

1/(1.4 1)

1.45 kg/m3

(300 m/s)2V2 1 kJ/kg

40+(b) For helium, T0 T 2cp2 5.1926 kJ/kg C 1000 m2/s2

48.7 C

T

P0 P 0

T

k/(k 1)

321.9 K

(200 kPa)

K 313.2

1.667/(1.667 1)

214.2 kPa

0

P0214.2 kPa

0.320 kg/m3 3RT0(2.0769 kPa m/kg K)(321.9 K)

Thus,

2 2 T* T0 (321.9 K) 241 K

k 1 1.667+1

2

P* P0

k 1

k/(k 1)

2

(214.2 kPa)

1.667+1

1.667/(1.667 1)

104.3 kPa

2

* 0

k 1

1/(k 1)

2

(0.320 kg/m)

1.667+1

3

1/(1.667 1)

0.208 kg/m3

Discussion These are the temperature, pressure, and density values that will occur at the throat when the flow past the throat is supersonic.

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