SINGLE-STAGE ROCKET ENGINES

D. P. Mishra

5.6 SINGLE-STAGE ROCKET ENGINES

Let us consider whether a single chemical rocket engine can provide suf- ficient velocity increment for various aerospace applications. Before initi- ating a meaningful discussion, we need to consider certain parameters in terms of mass of various components of the rocket engine which will be helpful in describing the performance of both single-stage and multistage rocket motors. As the main objective of the rocket engine is to place the payload, we need to consider the payload mass m_l. In a chemical rocket engine, the major portion of the mass is the propellant mass m_p. The struc- tural mass m_s, which consists of engine structure, supporting structure, tankages, valves, guidance, and control, must be reduced to have higher velocity increment. Then, the total initial mass of the rocket engine m₀ is the sum of all the quantities:

m m m m0= l+ p+ s (5.34) When the entire propellant is being burnt during its operation, the burn- out mass m_b as defined earlier is equal to the sum of the structural and payload masses:

m m mb = ₀- p =m ml+ s (5.35) Let us now define three mass ratios, namely, payload fraction LF, structural fraction SF, and propellant fraction PF, by dividing Equation 5.34 with the total initial mass m₀:

m m

m m

m

m LF SF PF

l s p

0 + 0 + 0 = + + =1 (5.36)

As mentioned earlier, the mass ratio MR can be defined as follows:

MR mm

m

m m LF SF

b l s

= =

+ =

0 0 +1 (5.37)

Similarly, this mass ratio MR can be expressed in terms of propellant frac- tion PF as follows:

MR mmb PF

= =

-

0 1

1 (5.38)

Let us define another important parameter known as payload coefficient:

l = - =

+ m

m m

m

l m m

l

l

p s

0 (5.39)

It indicates the mass of the payload that can be carried compared to the propellant and structural masses. Unfortunately, this payload coefficient happens to be small for aerospace applications. Hence, engineers strive to obtain a higher payload fraction for meaningful space applications although it is very difficult to achieve in reality. Another important param- eter known as structural coefficient ε is defined as the ratio of structural mass to the sum of structural and propellant masses as follows:

e = + = -

- m

m m

m m m m

s

p s

b l

o l (5.40)

Note that this expression is true only when the entire propellant is burnt out without any residual unburnt propellant mass. It is desirable to have a smaller value of structural coefficient for space applications because the smaller is its value, the lighter will be the vehicle. In other words, the struc- tural coefficient indicates how far the designer can manage to reduce the structural mass. When the structural coefficient is small and the rocket engine is quite huge, then the total engine mass including its structural mass will be dictated by the initial propellant mass. Hence, under this condition, the structural coefficient can be considered to remain constant, indicating that it would not be dependent on the vehicle size and, in turn,

velocity increment. By using the two mass ratios, namely, payload ratio λ and structural coefficient ε, we can express the mass ratio MR as follows:

MR mm

m

b m mo p

= =

- = +

0 0 1+l

e l (5.41)

For achieving higher attainable velocity increment as per Equation 5.17, with limited I_sp, higher MR must be used. In order to have a higher MR, lower m_b is to be used for a given initial engine mass. Thus, a very careful structural design is essential for a given payload. Generally, the payload capacity can be enhanced further for a given m_p and MR, by reducing structural mass (see Equation 5.37). Thus, it is advantageous to reduce the structural fraction SF as much as possible in consistence with the strength requirements of the vehicle. Let us understand further how payload mass can be related to velocity increment by considering Equations 5.17 and 5.37:

DV V MR V

LF SF

eq eq

=

( )

^{= æ} ₊

èç ö

ø÷

ln ln 1 (5.42)

The payload fraction can be expressed in terms of velocity increment and V_eq and SF:

LF e= ^-(^{DV V}/ ^eq)-SF

(5.43) Note that the payload fraction LF gets enhanced when higher V_eq is used for a lower value of velocity increment and structural fraction SF. In other words, when the velocity increment demand is high for a given value of structural mass fraction SF and V_eq, the payload fraction will be small.

The variation in LF is plotted in Figure 5.10 with velocity increment ratio ΔV/V_eq for three values of SF. It may be noted that a higher payload fraction can be achieved only when the vehicle has a higher jet velocity and lower SF for a certain demand of velocity increment ΔV. Hence, the designer of a rocket engine has to enhance jet velocity while decreasing the payload fraction SF.

Example 5.2

A single-stage chemical rocket with I_sp = 250 is designed to escape with the following mass: m_l = payload mass = 200 kg; m_s = structural

mass = 800 kg; m₀ = total mass = 30,000 kg. Determine the mass ratio, velocity increment, payload, and structural fraction for this rocket engine assuming there are no drag and gravity effects.

Solution

The burntout mass m_b can be determined as m_b =200 800 1000 kg+ =

By using Equation 5.16, the mass ratio MR can be determined as MR m m= ₀/ _b=30 000 1 000 30, / , =

Then the propellant fraction can be estimated using Equation 5.16:

PF=1 1- /MR=0 97.

This indicates that only 3% of the total mass can be used for structure and payload mass. In other words, a tiny payload can be carried into space by a monstrous rocket.

0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2

0.0 0.1 0.2 0.3 0.4 0.5 0.6

LF

ΔV/Veq

SF= 0.1 SF= 0.15

FIGURE 5.10 Variation in LF with velocity increment ratio ΔV/V_eq for two values of SF.

The payload fraction LF and structural fraction SF are determined as LF m= _l/m0=200 30 000 0 0066/ , = .

SF m m= s/ 0 =800 30 000 0 0266/ , = .

The velocity increment for an ideal case can be determined by using Equation 5.42:

DV =9 81 250. ´ ln

( )

30 8 34⁼ . km s/

The velocity can be attained ideally without taking into account drag and gravitational pull. In actual sense, it will be quite a low value.

In order to escape the earth’s gravity, velocity increment should be 11.2 km/s. Hence, it would not be possible to escape the earth’s grav- ity with this rocket engine. One solution would be to use a multistage engine, which is discussed in the following section.