which may be seen on the diagram. As the gas flow passes through the exhaust system (Part 6), small forward forces may act on the inner cone or bullet, 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.
計算推力的方法
 
7．如果已經知道特定氣流段進口和出口的面積、壓力、速度和質量流量，就可計算發動機或發動機的任何氣流段的推力或氣體載荷。
9．為了計算一特定流段的總推力，必須計算進口和出口處的總的推力。總推力就是所得到的這兩個值之差。
8.圖20-1中表示的推力的分布可借助于依次考慮每一部件并采用一些簡單的計算方法來計算。發動機產生的推力主要是流過發動機的空氣的質量及其速度增量的乘積(即牛頓第二運動定律)，不過，某個特定流段的進口和出口之間的壓差對發動機的總推力具有影響，計算時必須加以考慮。
Thrust distribution

METHOD OF CALCULATING THE THRUST the compressor and the conditions at the outlet from FORCES the compressor. Since the pressure and the velocity at the inlet to the compressor are zero, it is only
7.
The thrust forces or gas loads can be calculated necessary to consider the force at the outlet from thefor the engine, or for any flow section of the engine, compressor. Therefore, given that the compressor-provided that the areas, pressures, velocities and mass flow are known for both the inlet and outlet of OUTLET Area (A) = 182 sq.in.

the particular flow section.  Pressure (P) = 94 lb. per sq.in. (gauge)

8.
The distribution of thrust forces shown in fig. 20-Velocity (vJ) = 406 ft. per sec. 1 can be calculated by considering each component Mass flow (W) = 153 lb. per sec.in turn and applying some simple calculations. The

The thrust

推力
thrust produced by the engine is mainly the product of the mass of air passing through the engine and the MVJ
= (A xP) +. 0velocity increase imparted to it (i.e. Newtons Second gLaw of Motion), however, the pressure difference 10．推力的計算采用下列公式：
式中
A=氣流段的面積，單位為平方英寸
P=壓力，單位為磅／平方英寸
W=質量流量，單位為磅／秒
vJ=氣流的速度，單位為英尺／秒
g=重力常數，32.2英尺／平方秒

153 x 406

between the inlet to and the outlet from the particular = (182 x 94) +. 0 flow section will have an effect on the overall thrust
32 of the engine and must be included in the calculation.
= 19,049 lb. of thrust in a forward direction.
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:

MVJThrust = (A xP) + g
Where  A = Area of flow section in sq.in.
P = Pressure in lb. per sq.in.

Diffuser duct
W = Mass flow in lb. per sec.

發動機推力計算

11．當采用上述方法計算不同部件上的單個推力載荷時，假設發動機是靜止狀態的。飛機前飛速度對發動機推力的影響將在第21章中討論。為方便起見，在下列計算中，“g”取32。為了幫助計算，有關的位置用許多小的簡圖來表示。
13. The conditions at the diffuser duct inlet are the vJ = Velocity of flow in feet per sec.
same as the conditions at the compressor outlet, i.e.
g = Gravitational constant 32.2 ft. per 19,049 lb.
sec. per sec.

Therefore, given that the diffuser--OUTLET Area (A) = 205 sq.in. Pressure (P) = 95 lb. per sq.in. CALCULATING THE THRUST OF THE ENGINE (gauge) Velocity (vJ) = 368 ft. per sec.
11. When applying the above method to calculate Mass flow (W) = 153 lb. per sec.the individual thrust loads on the various components it is assumed that the engine is static. The effect of The thrust aircraft forward speed on the engine thrust will be WVJ
= (A xP) +.19,049dealt with in Part 21. In the following calculations 'g' g

is taken to be 32 for convenience. To assist in these calculations the locations concerned are illustrated 153 x 368
= (205  x 95) +.19,049by a number of small diagrams.
32

Compressor casing = 21,235 - 19,049
12. To obtain the thrust on the compressor casing it
　
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