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6.25

Superheated vapor ammonia enters an insulated nozzle at 20°C, 800 kPa, shown

in Fig. P6.25, with a low velocity and at the steady rate of 0.01 kg/s. The

ammonia exits at 300 kPa with a velocity of 450 m/s. Determine the temperature

(or quality, if saturated) and the exit area of the nozzle.

Solution:

C.V. Nozzle, steady state, 1 inlet and 1 exit flow, insulated so no heat transfer.

Energy Eq.6.13: q + hi + V2

i/2 = he + V2

e/2,

Process: q = 0, Vi = 0

Table B.2.2: hi = 1464.9 = he + 4502/(2×1000) ⇒ he = 1363.6 kJ/kg

Table B.2.1: Pe = 300 kPa Sat. state at −9.2°C :

he = 1363.6 = 138.0 + xe × 1293.8,

=> xe = 0.947, ve = 0.001536 + xe × 0.4064 = 0.3864 m3/kg

A

e = m

.

eve/Ve = 0.01 × 0.3864 / 450 = 8.56 × 10-6 m2

Inlet

Low V

Exi

t

Hi V

Hi P, A Low P, A

cb

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Borgnakke and Sonntag

6.28

A diffuser, shown in Fig. P6.28, has air entering at 100 kPa, 300 K, with a

velocity of 200 m/s. The inlet cross-sectional area of the diffuser is 100 mm2. At

the exit, the area is 860 mm2, and the exit velocity is 20 m/s. Determine the exit

pressure and temperature of the air.

Solution:

Continuity Eq.6.3: m

.

i = AiVi/vi = m

.

e = AeVe/ve,

Energy Eq.(per unit mass flow)6.13: hi + 1

2Vi2 = he + 1

2Ve2

h

e - hi = 1

2 ×2002/1000 − 1

2 ×202/1000 = 19.8 kJ/kg

T

e = Ti + (he - hi)/Cp = 300 + 19.8/1.004 = 319.72 K

Now use the continuity equation and the ideal gas law

v

e = vi

AeVe

AiVi = (RTi/Pi)

AeVe

AiVi = RTe/Pe

P

e = Pi

Te

Ti

AiVi

AeVe = 100

319.72

300

100 × 200

860 × 20 = 123.92 kPa

Inlet

Low V

Exi

t

Hi V

Hi P, A

Low P, A

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Borgnakke and Sonntag

6.46

A small turbine, shown in Fig. P 6.46, is operated at part load by throttling a 0.25

kg/s steam supply at 1.4 MPa, 250°C down to 1.1 MPa before it enters the turbine

and the exhaust is at 10 kPa. If the turbine produces 110 kW, find the exhaust

temperature (and quality if saturated).

Solution:

C.V. Throttle, Steady, q = 0 and w = 0. No change in kinetic or potential

energy. The energy equation then reduces to

Energy Eq.6.13: h1 = h2 = 2927.2 kJ/kg from Table B.1.3

C.V. Turbine, Steady, no heat transfer, specific work: w = 110

0.25 = 440 kJ/kg

Energy Eq.: h1 = h2 = h3 + w = 2927.2 kJ/kg

⇒ h3 = 2927.2 - 440 = 2487.2 kJ/kg

State 3: (P, h) Table B.1.2 h < hg

2487.2 = 191.83 + x3 × 2392.8

T

v

1

2

3

⇒ T = 45.8°C , x3 = 0.959

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