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but generally only rearward forces are produced and
these are due to the ’drag’ of the gas flow at the
propelling nozzle.
6. It will be seen that during the passage of the air
through the engine, changes in its velocity and
pressure occur (Part 2). For instance, where a
conversion from velocity (kinetic) energy to pressure
energy is required the passages are divergent in
shape, similar to that used in the compressor
diffuser. Conversely, where it is required to convert
the energy stored in the combustion gases to
velocity, a convergent passage or nozzle, similar to
that used in the turbine, is employed. Where the
conversion is to velocity energy, ’drag’ loads or
rearward forces are produced; where the conversion
is to pressure energy, forward forces are produced.
Part 2, fig. 2-3 illustrates velocity and pressure
changes at two points on the engine.
Thrust distribution
208
Fig. 20-1 Thrust distribution of a typical single-spool axial flow engine.
METHOD OF CALCULATING THE THRUST
FORCES
7. The thrust forces or gas loads can be calculated
for the engine, or for any flow section of the engine,
provided that the areas, pressures, velocities and
mass flow are known for both the inlet and outlet of
the particular flow section.
8. The distribution of thrust forces shown in fig. 20-
1 can be calculated by considering each component
in turn and applying some simple calculations. The
thrust produced by the engine is mainly the product
of the mass of air passing through the engine and the
velocity increase imparted to it (i.e. Newtons Second
Law of Motion), however, the pressure difference
between the inlet to and the outlet from the particular
flow section will have an effect on the overall thrust
of the engine and must be included in the calculation.
9. To calculate the resultant thrust for a particular
flow section it is necessary to calculate the total
thrust at both inlet and outlet, the resultant thrust
being the difference between the two values
obtained.
10. Calculation of the thrust is achieved using the
following formula:
Where A = Area of flow section in sq.in.
P = Pressure in lb. per sq.in.
W = Mass flow in lb. per sec.
vJ = Velocity of flow in feet per sec.
g = Gravitational constant 32.2 ft. per
sec. per sec.
CALCULATING THE THRUST OF THE ENGINE
11. When applying the above method to calculate
the individual thrust loads on the various components
it is assumed that the engine is static. The effect of
aircraft forward speed on the engine thrust will be
dealt with in Part 21. In the following calculations ’g’
is taken to be 32 for convenience. To assist in these
calculations the locations concerned are illustrated
by a number of small diagrams.
Compressor casing
12. To obtain the thrust on the compressor casing it
is necessary to calculate the conditions at the inlet to
the compressor and the conditions at the outlet from
the compressor. Since the pressure and the velocity
at the inlet to the compressor are zero, it is only
necessary to consider the force at the outlet from the
compressor. Therefore, given that the compressor-
OUTLET Area (A) = 182 sq.in.
Pressure (P) = 94 lb. per sq.in.
(gauge)
Velocity (vJ) = 406 ft. per sec.
Mass flow (W) = 153 lb. per sec.
The thrust
= 19,049 lb. of thrust in a forward direction.
Diffuser duct
13. The conditions at the diffuser duct inlet are the
same as the conditions at the compressor outlet, i.e.
19,049 lb.
Therefore, given that the diffuser--
OUTLET Area (A) = 205 sq.in.
Pressure (P) = 95 lb. per sq.in.
(gauge)
Velocity (vJ) = 368 ft. per sec.
Mass flow (W) = 153 lb. per sec.
The thrust
= 21,235 - 19,049
= 2,186 lb. of thrust in a forward direction.
Thrust distribution
209
g
Thrust (A x P) M VJ = +
0
32
= (182 x 94) + 153 x 406 −
19,049
32
= (205 x 95) + 153 x 368 −
19,049
g
(A x P) W VJ = + −
0
g
(A x P) M VJ = + −
Combustion chambers
14. The conditions at the combustion chamber inlet
are the same as the conditions at the diffuser outlet,
i.e. 21,235 lb. Therefore, given that the combustion
chamber-
OUTLET Area (A) = 580 sq.in.
Pressure (P) = 93 lb. per sq.in.
(gauge)
Velocity (vJ) = 309 ft. per sec.
Mass flow (W) = 153 lb. per sec.
The thrust
= 55,417 - 21,235
= 34,182 !b. of thrust in a forward direction.
Turbine assembly
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