Thee following example illustrates the use of the equations, tables and concepts presented in the previous sections.

A small water-cooled liquid-fuel rocket engine is to be designed for a chamber pressure of 300 psi and a thrust of 20 lbs. The engine is to operate at sea level using gaseous oxygen and gasoline propellants.

Step 1

From Table I and Figures 3,4 and 5 we determine that the optimum O/F ratio is about 2.5 and that the ideal specific impulse will be about 260 sec. The total propellant flow rate is given by Equation (3)

wt = F/Isp = 20/260 = 0.077 lb/sec
Since the mixture ratio, r, is 2.5, we find from Equation (5)

wf = wt/(r + 1) = 0.077/3.5 = 0.022 lb/sec
From Equation (6) the oxygen flow rate is

wo = 0.077 - 0.022 = 0.055 lb/sec
As a check, we divide the oxygen flow rate by the fuel flow rate and the result is 2.5, as it should be.

Step 2

From Table I we note that the chamber gas temperature is 5742 degF or about 6202 degR.
From Equation (9) the gas temperature at the nozzle throat is

Tt = .909 (Tc) = .909 (6202) = 5650 R

Step 3

From Equation (12) the pressure at the nozzle throat is
Pt = .564 (Pc) = .564 (300) = 169 psi

Step 4

The nozzle throat area is given by Equation (7)

At = (w/Pt)(RTt/(gamma)gc)^(1/2)

At = (.077/169)(9500)1/2 = 0.0444 in2

Step 5

The nozzle throat diameter is given by Equation (17)

Dt = (4At/(pi))1/2 = (0.0566)1/2 = 0.238 in.

Step 6

From Table III we find that for a chamber pressure of 300 psi and a nozzle exit pressure of 14.7 psi (sea level)

Ae/At = 3.65 so that the nozzle exit area is, from Eq. (15)

Ae = 3.65 At = (3.655)(0.0444) = 0.162 in2

Step 7

The nozzle exit diameter is from Eq. (17)

De = (4Ae/(pi))^(1/2) = (.2065)^(1/2) = 0.4555 in.

Step 8

For this propellant combination we will assume a combustion chamber L* of 60 inches. The combustion chamber volume is given by Eq. (19)

Vc = L* At = (60)(.0444) = 2.67 in3

Step 9

The chamber length is found from Eq. (21)

Vc = (1.1) (Ac Lc)

However, we must first determine the chamber area or Ac. We do this by assuming that the chamber diameter is five times the nozzle throat diameter or Dc = 5Dt, therefore

Dc = 1.2 in. and Ac = 1.13 in2

Lc = Vc/(1.1)(1.13) = 2.67/1.245 = 2.15 in

Step 10

Copper will be used for the combustion chamber and nozzle wall. The chamber wall thickness, is given by Eq. (23)

tw = PD/16000 = (300)(1.2)/16000

tw = 0.0225 inch
To allow for additional stress and welding factors we shall set the wall thickness equal to 3/32 or 0.09375 inch and will assume that the nozzle wall has this thickness also.

Step 11

Previous experience with small watercooled rocket engines has shown that we ean expect the copper combustion chamber and nozzle to experience an average heat transfer rate, q, of ahout 3 Btu/in2-sec. The heat transfer area of the combustion chamber is the outer surfaee area of the chamber and nozzle. This surface area is given by

A = (pi)(Dc + 2tw)(Lc) + area of nozzle cone

A = 9.4 in2 + area of nozzle cone
The area of the nozzle cone up to the throat can be assumed to he ahout 10% of the chamber surface area so that

A = (1.1)(9.4) = 10.35 in2
The total heat transferred into the coolant is given by Eq. (24)

Q = q A = 3(10.35) = 31 Btu/sec

Step 12

The cooling water flow rate can be calculated by asssuming a desired temperature rise of the water. If this is 40 deg F then, from Eq. (24)

wv = Q/(deltaT), where cp for water = 1.0

w = 31/40 = 0.775 lb of water per sec.

Step 13

The annular flow passage between the combustion chamber wall and the outer jacket must be sized so that the flow velocity of the cooling water is at least 30 ft/sec. This veloeity is obtained when the flow passage has dimensions as determined below:

vw = ww/(rho)A
where vw = 30 ft/sec, ww = 0.775 lb/sec, (rho) = 62.4 lb/ft3, and A is the area of the annular flow passage, given by
A = ((pi)/4) (D22 - D12)
where D2 is the inner diameter of the outer jacket and D1 is the outer diameter of the combustion chamber, given by
D1 = Dc + 2tw
Substituting in the above equations

D2 = SQRT((4ww)/(vw(rho)(pi)) + D12)

D2 = (.0151)1/2 = .123 ft = 1.475 inch

D2 - D1 = 0.085 inch
The water flow gap is 0.0425 inch.

Step 14

The fuel injector for this small rocket engine will he a commercial spray nozzle with a 75 degree spray angle. The required capacitv of the nozzle is determined by the fuel flow rate
wf = 0.022 lb/sec = 1.32 lb/ minute,
Since there are six pounds of gasoline per gallon, the spray nozzle flow requirement is 0.22 gallon per minute (gpm). The spray nozzle can now be ordered from any of several suppliers (see List of Suppliers); nozzle material should be brass to ensure adequate injector heat transfer to the incoming propellant.

If an impinging jet injector had been chosen, the determination of the required injector hole number and size would have been as follows:

The flow area for fuel injection is given by Equation (25)

A = wf/(Cd) (2g(rho)(deltaP))^(1/2)
We will assume that Cd = 0.7 with a fuel injcction pressure drop of 100 psi. The density of gasoline is about 44.5 lb/ft3, so that

A = .022/(.7)(6430) = 0.0000049 ft2

A = 0.000706 in2

If only one injection hole is used (a poor practice which can lead to combustion instability) its diameter would be

D = (4A(pi))1/2 = (.0009)1/2 = 0.30 inch
A number 69 drill could be used for this hole.

If two injection holes are used, their diameter would be

D = (.00045)1/2 = 0.021 inch
A number 75 drill could be used for these holes.

Step 15

The injection holes for the gaseous oxygen will be simple drilled orifices. The size of these orifices should be such that a gas stream velocity or about 200 ft/sec is obtained at design oxygen flow rate. The holes must not be so small that sonic velocity is achieved in the orifice passages since this would result in a high upstream pressure requirement to drive the required amount of oxygen through the orifices.

If a spray nozzle fuel injector is used we will assume the use of four equally spaced oxygen injection ports parallel to the combustion chamber centerline around this nozzle. If we assume an injection pressure drop of 100 psi then the oxygen gas pressure at the entrance to the injection ports will be 400 psi (the chamber pressure plus the injection pressure drop). The density of gaseous oxygen at 400 psi and a temperature of 68 deg F is given by the perfect gas law (see Table II).

(rho2) = (rho1)(P2/P1) = 2.26 lb/ft3
Assuming, incompressibility, the injection flow area is given by

A = wo/(rho)vo

Since we know the oxygen flow rate and the desired injection velocity, we can easily find the total injection area

A = .055/(2.26)(200) = 0.0001217 ft2

A = 0.0175 in2
Since there are to he four holes, each hole has an area of 0.004375 in2 and the diameter of each hole is

D = (.00558)(1/2) = 0.0747 inch
A number 48 drill could be used for these holes.

These same size oxygen jets could also be used with two fuel jets in the impinging stream injector. The holes, oxygen and fuel, should be drilled at an angle of 45 (deg) with respect to the injector face with the intersection point of the streams about 1/4 inch inside the combustion chamber.


The foregoing design calculations provide the dimensions, thicknesses, and orifice sizes for the major components of our rocket engine. The actual design of the engine, however, requires engineering judgment and knowledge of machining, welding, and operational factors since these interact to determine the final configuration of the engine and its components. Perhaps the best way to accomplish the final design is to sit down with appropriate drafting materials and begin to draft a cross-section view of the engine. A scale of 2/1 (or twice actual size) is about right for these small engines and will enable the designer to better visualize the entire assembly.

Using the dimensions obtained in the example calculation, and the design technique described shove, the rocket engine assembly design shown in Figure 8 is obtained. The engine design features easy fabrication and assembly.