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power is to be provided at the shaft. In general,
turbo-propeller aircraft provide one pound of thrust
for every 3.5 h,p. to 5 h.p.
Comparison between thrust and horse-power
14. Because the turbo-jet engine is rated in thrust
and the turbo-propeller engine in s.h.p., no direct
comparison between the two can be made without a
power conversion factor. However, since the turbopropeller
engine receives its thrust mainly from the
propeller, a comparison can be made by converting
the horse-power developed by the engine to thrust or
the thrust developed by the turbo-jet engine to t.h.p.;
that is, by converting work to force or force to work.
For this purpose, it is necessary to take into account
the speed of the aircraft.
Performance
217
g
Thrust (P P ) A W VJ
= − 0 ⋅ +
P0
30
0 273 T0
x 273 15
P
x 30
+
+
2.6
t.e.h.p. = s.h.p. + jet thrust lb.
15. The t.h.p. is expressed as
where F = lb. of thrust
V = aircraft speed (ft. per sec.)
Since one horse-power is equal to 550 ft.lb. per sec.
and 550 ft. per sec. is equivalent to 375 miles per
hour, it can be seen from the above formula that one
lb. of thrust equals one t.h.p. at 375 m.p.h. It is also
common to quote the speed in knots (nautical miles
per hour); one knot is equal to 1.1515 m.p.h, or one
pound of thrust is equal to one t.h.p. at 325 knots.
16. Thus if a turbo-jet engine produces 5,000 lb. of
net thrust at an aircraft speed of 600 m.p.h. the t.h.p.
would be
However, if the same thrust was being produced by
a turbo-propeller engine with a propeller efficiency of
55 per cent at the same flight speed of 600 m.p.h.,
then the t.h.p. would be
Thus at 600 m.p.h. one lb. of thrust is the equivalent
of about 3 t.h.p.
ENGINE THRUST IN FLIGHT
17. Since reference will be made to gross thrust,
momentum drag and net thrust, it will be helpful to
define these terms:
from Part 20, gross or total thrust is the product of the
mass of air passing through the engine and the jet
velocity at the propelling nozzle, expressed as:
The momentum drag is the drag due to the
momentum of the air passing into the engine relative
to the aircraft velocity, expressed as where
W = Mass flow in lb. per sec.
V = Velocity of aircraft in feet per sec.
g = Gravitational constant 32.2 ft. per sec. per
sec.
The net thrust or resultant force acting on the aircraft
in flight is the difference between the gross thrust
and the momentum drag.
18. From the definitions and formulae stated in
para, 17; under flight conditions, the net thrust of the
Performance
218
Fig. 21-2 The balance of forces and expression for thrust and momentum drag.
550 ft. per sec .
FV
8,000
375
5,000 x 600 =
14,545
55
8,000 x 100 =
g
(P P )A W vJ
− 0 +
g
WV
engine, simplifying, can be expressed as:
Fig. 21-2 provides a diagrammatic explanation.
Effect of forward speed
19. Since reference will be made to ’ram ratio’ and
Mach number, these terms are defined as follows:
Ram ratio is the ratio of the total air pressure at
the engine compressor entry to the static air
pressure at the air intake entry.
Mach number is an additional means of
measuring speed and is defined as the ratio of
the speed of a body to the local speed of sound.
Mach 1.0 therefore represents a speed equal to
the local speed of sound.
20. From the thrust equation in para. 18, it is
apparent that if the jet velocity remains constant,
independent of aircraft speed, then as the aircraft
speed increases the thrust would decrease in direct
proportion. However, due to the ’ram ratio’ effect from
the aircraft forward speed, extra air is taken into the
engine so that the mass airflow and also the jet
velocity increase with aircraft speed. The effect of
this tends to offset the extra intake momentum drag
due to the forward speed so that the resultant loss of
net thrust is partially recovered as the aircraft speed
increases. A typical curve illustrating this point is
shown in fig. 21-3. Obviously, the ’ram ratio’ effect, or
the return obtained in terms of pressure rise at entry
to the compressor in exchange for the unavoidable
intake drag, is of considerable importance to the
turbo-jet engine, especially at high speeds. Above
speeds of Mach 1.0, as a result of the formation of
shock waves at the air intake, this rate of pressure
rise will rapidly decrease unless a suitably designed
　
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