Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

Chapter 5 | 235

STEAM

V 2 = 900 ft/s

P 2 = 200 psia

T 1 = 700°F

P 1 = 250 psia

A 1 = 0.2 ft^2

m = 10 lbm/s

qout = 1.2 Btu/lbm

FIGURE 5–26A

Schematic for Example 5–5.

Substituting, we get

From Table A–17, the temperature corresponding to this enthalpy value is

Discussion This result shows that the temperature of the air increases by
about 20°C as it is slowed down in the diffuser. The temperature rise of the
air is mainly due to the conversion of kinetic energy to internal energy.

EXAMPLE 5–5 Acceleration of Steam in a Nozzle

Steam at 250 psia and 700°F steadily enters a nozzle whose inlet area is
0.2 ft^2. The mass flow rate of steam through the nozzle is 10 lbm/s. Steam
leavesthe nozzle at 200 psia with a velocity of 900 ft/s. Heat losses from the
nozzle per unit mass of the steam are estimated to be 1.2 Btu/lbm. Determine
(a) the inlet velocity and (b) the exit temperature of the steam.

Solution Steam enters a nozzle steadily at a specified flow rate and velocity.
The inlet velocity of steam and the exit temperature are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with
time at any point and thus mCV0 and ECV0. 2 There are no work
interactions. 3 The potential energy change is zero, pe 0.
Analysis We take the nozzleas the system (Fig. 5–26A). This is a control
volumesince mass crosses the system boundary during the process. We
observe that there is only one inlet and one exit and thus m

.

1 m

.

2 m

.

.

(a) The specific volume and enthalpy of steam at the nozzle inlet are

Then,

(b) Under stated assumptions and observations, the energy balance for this
steady-flow system can be expressed in the rate form as

Rate of net energy transfer Rate of change in internal, kinetic,

by heat, work, and mass potential, etc., energies

m

#

ah 1 

V 12

2

bQ

#

outm

#

ah 2 

V 22

2

b¬¬ 1 since W

#

0, and ¢pe 02

E

#

inE

#

out

E

#

in
E

#

out^ ^ dEsystem>dt^ ^0

V 1 134.4 ft/s

10 lbm>s

1

2.6883 ft^3 >lbm

¬ 1 V 121 0.2 ft^22

m#

1

v 1

V 1 A 1

P 1 250 psia

T 1 700°F

f¬

v 1 2.6883 ft^3 /lbm

h 1 1371.4 Btu/lbm

¬¬ 1 Table A–6E 2

T 2 303 K

303.14 kJ/kg

h 2 283.14 kJ/kg

0 
 1 200 m/s 22

2

a

1 kJ/kg

1000 m^2 /s^2

b

0 (steady)

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