Question 13.13: Under static ground testing at sea-level conditions, the Pra...

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

T = \dot{m} (V_\text{exhaust} − V_\text{inlet}) /g_c

= (370.  \text{lbm/s}) ( 1560.  \text{ft/s} − 0) [ 32.174  \text{lbm.ft/( lbf.s²})]

= 17,900  \text{lbf}

2a. The compressor’s isentropic efficiency is given by

(η_s)_c = \frac{(\dot{W}_c)_\text{isentropic}}{(\dot{W}_c)_\text{actua1}} = \frac{T_{4s} − T_3}{T_4 − T_3}

and, using k = 1.40 = \text{constant} for the cold ASC, we have

T_{4s} = T_3 (p_{4s}/p_3)^{(k−1)/k} = (520.) (200./14.7)^{(1.4−1)/1.4} = 1100  \text{R}= 640.°\text{F}

so that the compressor’s isentropic efficiency using constant specific heats is

(η_s)_{\substack{\text{compressor}\\\text{(constant}\\\text{specific heats)}}} = \frac{1100 − 520.}{1175 − 520.} = 0.886 = 88.6\%

Similarly, the turbine’s (prime mover) isentropic efficiency is given by

(η_s)_{pm} = \frac{(\dot{W}_{pm})_\text{actual}}{(\dot{W}_{pm})_\text{isentropic}} = \frac{T_{1} − T_2}{T_1 − T_{2s}}

where, using the constant specific heats, we obtain

T_{2s} = T_1 (p_{2s}/p_1)^{(k−1)/k} = (2060) (28.0/190.)^{(1.40−1)/1.40} = 1190  \text{R}= 730.°\text{F}

Then,

(η_s)_{\substack{pm\\\text{(constant}\\\text{specific heats)}}} = \frac{2060 − 1350}{2060 − 1190} = 0.816 = 81.6\%

3a. The Brayton cold ASC thermal efficiency is given by

(η_s)_{\substack{\text{Brayton}\\\text{cold ASC}\\}} = \frac{T_1 − T_{2s} – (T_{4s} − T_3)}{T_1 − T_{4s}}

=\frac{2060 − 1190 − (1100 − 520.)}{2060 − 1100} = 0.302 = 30.2\%

but the actual thermal efficiency of the engine, based on constant specific heats and the data provided in the schematic, is

(η_s)_{\substack{\text{Brayton}\\\text{(actual, constant}\\\text{specific heats)}}} = \frac{T_1 − T_2 − (T_4 − T_3)}{T_1 − T_4}

= \frac{2060 − 1350 − (1175 − 520.)}{2060 − 1175} = 0.062 = 6.2\%

2b. We can easily take into account the temperature-dependent specific heats by using Table C.16a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics. For the compressor, p_{r4} = (p_{4s}/p_3) =  (1.2147) (200./14.7) = 16.5 and, by interpolation in Table C.16a, we find that T_{4s} = 1084  \text{R} = 624°\text{F} . Then,

(η_s)_{\substack{c\\\text{(variable}\\\text{specific heats)}}} = \frac{T_{4s} − T_3}{T_4 − T_3} = \frac{1084 − 520.}{1175 − 520.} = 0.861 = 86.1\%

Similarly, for the turbine,

p_{r2} = p_{r1} (p_{2s}/p_1) = (196.16) (28.0/190.) = 28.9

and, by interpolation in Table C.16a, we find that T_{2s} = 1261  \text{R}= 801°\text{F} . Then,

(η_s)_{\substack{pm\\\text{(variable}\\\text{specific heats)}}} = \frac{T_{1} − T_2}{T_1 − T_{2s}} = \frac{2060 − 1350}{2060 − 1261 } = 0.889 = 88.9\%

3b. Finally, the Brayton hot ASC can be easily determined from

(η_T)_{\substack{\text{Brayton}\\\text{hot ASC }\\}} = \frac{h_1 − h_{2s} − (h_{4s} − h_3)}{h_1 − h_{4s}}

where, from Table C.16a,

h_3 = 124  \text{Btu/lbm (at 520. R)}

h_{4s} = 262  \text{Btu/lbm (by interpolation at 1084 R)}

h_1 = 521  \text{Btu/lbm (at 2060 R)}

h_{2s} = 307  \text{Btu/lbm (by interpolation at 1261 R)}

Then,

(η_T)_{\substack{\text{Brayton}\\\text{cold ASC}\\}} = \frac{521 − 307 − (262 − 124)}{521 − 262} = 0.293 = 29.3\%

and the engine’s actual thermal efficiency, based on temperature-dependent specific heats, is

(η_T)_{\substack{\text{Brayton}\\\text{(actual, variable}\\\text{specific heats)}}} = \frac{h_1 − h_2 − (h_4 − h_3)}{h_1 − h_4}

where h_4 = 284.9  \text{Btu/lbm (at 1175 R)} and h_2 = 329.9  \text{Btu/lbm (at 1350 R)} . Then,

(η_T)_{\substack{\text{Brayton}\\\text{(actual, variable}\\\text{specific heats)}}} = \frac{521 − 329.9 − (284.9 − 124)}{521 − 284.9} = 0.128 = 12.8\%

3c. The maximum work Brayton cold ASC thermal efficiency is given by Eq. (13.27) as

(η_T)_{\substack{\text{max work}\\\text{Brayton}\\\text{cold ASC}}} = 1 – \sqrt{\frac{T_3}{T_1}} = 1 – \sqrt{\frac{520.  \text{R}}{2060  \text{R}} } = 0.502 = 50.2\%

This is much greater than the actual thermal efficiency for this engine, because an aircraft engine need produce only enough work output to drive the engine’s auxiliary equipment (generator, fuel pump, etc.), and most of the engine’s energy output is in the kinetic energy of its exhaust (which produces thrust).