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Modern Engineering Thermodynamics

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500 CHAPTER 13: Vapor and Gas Power Cycles<br />

When regeneration, interstage compressor cooling, and interstage turbine reheat are added to the Brayton cycle,<br />

it approximates the more efficient Ericsson cycle, as shown in Figure 13.39. The major focus of modern gas turbine<br />

development centers on increasing the turbine inlet temperature through the development of new highstrength,<br />

high-temperature materials. <strong>Modern</strong> gas turbine heat engines are used mainly in small- to medium-size<br />

stationary power-generating stations and as prime movers throughout the transportation industry.<br />

EXAMPLE 13.13<br />

Under static ground testing at sea-level conditions, the Pratt & Whitney JT3D-3B Turbofan engine has the actual internal<br />

temperatures and pressures as shown in Figure 13.47.<br />

Fuel<br />

Combustion<br />

chamber<br />

V inlet ≈ 0<br />

Compressor<br />

Turbine<br />

V exhaust = 1560 ft/s<br />

m exhaust = 270. lbm/s<br />

Station 3<br />

14.7 psia<br />

520. R<br />

Station 4<br />

200. psia<br />

1175 R<br />

Station 1<br />

190. psia<br />

2060 R<br />

Station 2<br />

28.0 psia<br />

1350 R<br />

FIGURE 13.47<br />

Example 13.13.<br />

Determine<br />

1. The engine’s static thrust.<br />

2. The compressor and turbine isentropic efficiencies for<br />

a. The Brayton cold air standard cycle.<br />

b. The Brayton hot air standard cycle using the gas tables for air, Table C.16a.<br />

3. The ASC and actual thermal efficiencies for<br />

a. The Brayton cold air standard cycle.<br />

b. The Brayton hot air standard cycle using the gas tables for air, Table C.16a.<br />

c. The maximum work Brayton cold ASC thermal efficiency.<br />

Solution<br />

1. The engine’s static thrust is given directly by Eq. (13.29) as<br />

2a. The compressor’s isentropic efficiency is given by<br />

T = _m ðV exhaust − V inlet Þ/g c<br />

= ð370: lbm/sÞð1560: ft/s − 0Þ½32:174 lbm⋅ft/ ðlbf ⋅s 2 ÞŠ<br />

= 17,900 lbf<br />

ðη s Þ c<br />

=<br />

<br />

_W c isentropic<br />

= T 4s − T 3<br />

_W c<br />

T 4 − T 3<br />

actua1<br />

and, using k = 1.40 = constant for the cold ASC, we have<br />

T 4s = T 3 ðp 4s /p 3 Þ ðk−1Þ/k = ð520:Þð200:/14:7Þ ð1:4−1Þ/1:4 = 1100 R = 640:°F<br />

so that the compressor’s isentropic efficiency using constant specific heats is<br />

1100 − 520:<br />

ðη s Þ compressor<br />

= = 0:886 = 88:6%<br />

1175 − 520:<br />

ðconstant<br />

specific heatsÞ

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