A Merlin engine is bolted to a test stand and fired. The stand measures a push of \(845\ \mathrm{kN}\) (Falcon User’s Guide; see sources). Nine of them under a Falcon 9 give \(7605\ \mathrm{kN}\) on the pad, and a secondary compilation credits the same nine with \(8227\ \mathrm{kN}\) once the vehicle has climbed out of the atmosphere (see sources). Nothing about the engines has changed in between: the pumps turn at the same speed, the same propellant burns at the same rate. Yet the thrust has grown by 8 percent.

Two questions are hiding in that fact. The first is where the extra force comes from, and the answer will turn out to be that it was always there and the atmosphere was taking it away. The second is more basic. An engine is a pot with a hole in it. The gas leaves through the hole and touches nothing afterward. On what surface, exactly, does the force that lifts the vehicle act?

Chapter 1 obtained the thrust formula by following a fixed parcel of matter for an instant and promised a fuller treatment. This chapter supplies it. The tool is a way of applying Newton’s law to a region of space that fluid flows through, and once it is in hand the thrust equation, its dependence on altitude, and the place where the force is applied all follow from one balance. The chapter then splits the effective exhaust velocity into a factor that belongs to the combustion chamber and a factor that belongs to the nozzle, and ends by asking how much energy an exhaust jet carries. That last question is the bridge to Chapter 4.

3.1 Newton’s law for a region that gas flows through

Definition 1.1 gave the momentum principle for a closed system, a fixed collection of matter. An engine running on a stand is awkward to describe that way, because the matter keeps changing: every second a third of a tonne of new propellant enters and the same amount of gas leaves. What stays fixed is the hardware and the region of space it occupies. So fix the region instead.

A control volume is bookkeeping, not physics. The physics is still the law for a closed system, and the task is to translate it. Take the closed system to be the matter that is inside \(\mathcal V\) at time \(t\) , and follow it for a short time \(\mathrm dt\) . Its momentum at time \(t\) is the momentum inside the region, \(\int_{\mathcal V}\rho\,\mathbf u\,\mathrm d\mathcal V\) . At \(t+\mathrm dt\) the same matter has moved: part of it has passed out through the control surface, and other matter, which does not belong to the system, has come in. Through a patch of surface of area \(\mathrm dA\) , the volume of fluid that crosses in \(\mathrm dt\) is \((\mathbf u\cdot\hat{\mathbf n})\,\mathrm dA\,\mathrm dt\) , positive where the flow is outward and negative where it is inward. That fluid carries momentum \(\rho\,\mathbf u\) per unit volume. Hence

where the surface integral adds back what the system carried out and removes what outsiders carried in. Subtract \(\mathbf P_{\mathrm{sys}}(t)\) , divide by \(\mathrm dt\) , and use \(\mathrm d\mathbf P_{\mathrm{sys}}/\mathrm dt=\mathbf F\) for the closed system.

The second term is called the momentum flux. Its units are those of force: \((\mathrm{kg/m^3})(\mathrm{m/s})(\mathrm{m/s})(\mathrm{m^2})=\mathrm{kg\cdot m/s^2}=\mathrm N\) . The same counting applied to mass instead of momentum gives the companion statement that, in steady flow, the mass flow rates in and out balance: \(\oint\rho\,(\mathbf u\cdot\hat{\mathbf n})\,\mathrm dA=0\) .

The force \(\mathbf F\) on the contents has three kinds of contribution. Body forces act on every element of mass inside; gravity is the only one here. Surface forces are exerted across the control surface by whatever lies outside it. Where the surface passes through fluid, the outside fluid presses on the inside fluid with the local pressure \(p\) , perpendicular to the surface and directed inward, so the pressure force is \(-\oint p\,\hat{\mathbf n}\,\mathrm dA\) . Fluids also exert friction along a surface, but on the surfaces used below that contribution is small beside the pressure and is neglected. Finally, where the surface cuts through a solid member, such as a strut, the member transmits whatever force it carries.

One fact about pressure does a great deal of work in this chapter, and it is an immediate consequence of the divergence theorem. For a continuously differentiable scalar field \(p\) on a compact region with piecewise smooth boundary, \(\oint_{\partial\mathcal V}p\,\hat{\mathbf n}\,\mathrm dA=\int_{\mathcal V}\nabla p\,\mathrm d\mathcal V\) ; this is the divergence theorem applied to \(p\) times each constant coordinate vector, from Multivariable Calculus (math.multivariable-calculus.generalized-stokes-vector-calculus). If \(p\) has the same value \(p_a\) everywhere, its gradient is zero.

A sealed box in still air is pushed equally from all sides, whatever its shape. The lemma lets us subtract \(p_a\) from the pressure everywhere on a closed surface without changing the net force, and then only the places where the pressure differs from \(p_a\) contribute.

3.2 The engine on its stand

Now choose the region. Figure 3.1, on the left, shows an engine held by a strut on a test stand, firing horizontally so that gravity does not enter the balance along the axis. Let \(\hat{\mathbf t}\) be the unit vector along the engine axis pointing forward, the direction in which the engine pushes on the stand; the exhaust leaves in the direction \(-\hat{\mathbf t}\) .

Momentum flux. The only place where fluid crosses the surface with appreciable momentum is the exit plane. There the outward normal is \(\hat{\mathbf n}=-\hat{\mathbf t}\) and \(\mathbf u=-v_e\hat{\mathbf t}\) , so \(\mathbf u\cdot\hat{\mathbf n}=v_e\) and the mass flow rate is \(\dot m=\rho_ev_eA_e\) , with \(\rho_e\) the gas density at the exit. The momentum flux out of the control volume is

Pressure force. Write \(p=p_a+(p-p_a)\) on the whole surface. By Lemma 3.2 the uniform part contributes nothing. The remainder vanishes everywhere except on the exit plane, where it equals \(p_e-p_a\) , so

Force through the strut. Call the force that the engine exerts on the stand \(F\hat{\mathbf t}\) ; this is what a load cell in the strut reads, and it is the thrust. By Newton’s third law the stand pushes on the contents of the control volume with \(-F\hat{\mathbf t}\) .

Balance. In steady flow Theorem 3.1 says force on the contents equals momentum flux out. Along \(\hat{\mathbf t}\) ,

This is Definition 1.2, now derived instead of argued. The two derivations agree for a good reason. Chapter 1 followed the closed system of vehicle plus the parcel about to leave, and counted the pressure on a surface closed across the exit plane; the control surface here is that same surface, and the momentum carried away by the parcel in time \(\mathrm dt\) is the momentum flux times \(\mathrm dt\) . On a flying vehicle the tanks are inside the surface, the stand is replaced by the vehicle’s own inertia, and the same \(F\) appears in the equation of motion of Proposition 1.1.

How good are the assumptions? Assumption (iv) is easy to test with numbers. A flow of \(270\ \mathrm{kg/s}\) entering through pipes at \(10\ \mathrm{m/s}\) (illustrative values) carries momentum at a rate of \(2.7\ \mathrm{kN}\) , against a thrust of several hundred kilonewtons, and on a vehicle the pipes run from tanks that are inside the system anyway. Assumption (ii) is the weakest for real hardware: the gas at the rim of a nozzle leaves at an angle to the axis, and Chapter 5 puts a number on the loss. Assumption (i) fails during start-up and shutdown, when the momentum stored inside the engine is changing and the first term of Theorem 3.1 cannot be dropped. Assumption (iii) fails when the vehicle moves through the air, and by convention the resulting change in outside pressure is charged to aerodynamic drag, not to the engine.

3.3 Where the force is applied

The control surface of Section 3.2 never touched the gas inside the engine, which is why the derivation needed so little. It also means the derivation did not say where the force acts. For that, draw a second surface: one that contains the gas alone. It runs along the wetted inside walls, from the injector face at the front of the chamber, down the chamber wall, through the throat, along the inside of the nozzle bell, and closes across the exit plane (Figure 3.1, right).

Apply Theorem 3.1 again in steady flow. The momentum flux out through the exit plane is \(-\dot m v_e\hat{\mathbf t}\) as before. The forces on the gas are the pressure of the outside gas on the exit plane, \(-p_e\hat{\mathbf n}A_e=p_eA_e\hat{\mathbf t}\) , and the force exerted by the walls. So the walls push on the gas with

and by the third law the gas pushes on the walls with \((\dot m v_e+p_eA_e)\hat{\mathbf t}\) , forward. A fluid pushes on a wall through its pressure, so this force is also the integral of gas pressure over the wetted walls,

with \(\hat{\mathbf n}\) pointing out of the gas into the wall. Meanwhile the atmosphere presses on the outside of the hardware. That outside surface, together with the exit opening, would be a closed surface; by Lemma 3.2 a uniform \(p_a\) over the closed surface gives no force, so over the outside alone the atmosphere exerts \(-p_aA_e\hat{\mathbf t}\) , a rearward force equal to what is missing at the opening. Adding the two,

So the answer to the opening question is that thrust acts on the inside walls of the engine, as gas pressure. The three parts of the wall contribute differently. The injector face looks straight backward at the gas, and the pressure on it, close to the chamber pressure \(p_c\) , pushes the engine forward. The convergent wall that narrows toward the throat faces partly forward, and the gas pushes it backward. The bell beyond the throat faces partly backward again, and although the pressure there is far lower, it acts on a large area and pushes forward. If the engine were nothing but a chamber with a hole of the throat’s area \(A_t\) , and the gas inside were at rest at pressure \(p_c\) , the unbalanced force would be \(p_cA_t\) . Real engines deliver between about \(1.3\) and \(2\) times \(p_cA_t\) , numbers that Chapter 5 derives; the excess over \(1\) is what a shaped nozzle earns.

Nowhere in this account does the exhaust push on the air behind the engine. The atmosphere enters only through \(-p_aA_e\) , and it enters with a minus sign.

3.4 What the atmosphere takes away

The thrust equation contains the ambient pressure explicitly, and for a running engine nothing else in it depends on \(p_a\) . The mass flow, the exit speed, and the exit pressure are fixed by what happens inside, upstream of the exit plane; Chapter 5 shows that as long as the gas leaves faster than sound travels in it, no change outside can reach back in. Granting that for now:

Plotted against ambient pressure, thrust is a straight line with slope \(-A_e\) (Figure 3.2, left). Two published thrust figures at two known pressures therefore determine the exit area, and any third figure can be tested against the line.

Example 3.4: Thrust and specific impulse during a climb

Question. How do the thrust and specific impulse of the reference launcher’s first stage change during its burn, and what single effective exhaust velocity best represents the stage?

Thrust against altitude. From \(F=F_{\mathrm{vac}}-p_a(z)A_e\) with the atmosphere model above and the numbers of Example 3.2:

Altitude	Ambient pressure	Thrust	Specific impulse
\(0\ \mathrm{km}\)	\(101.4\ \mathrm{kPa}\)	\(2267\ \mathrm{kN}\)	\(285.0\ \mathrm s\)
\(5\ \mathrm{km}\)	\(54.1\ \mathrm{kPa}\)	\(2378\ \mathrm{kN}\)	\(299.0\ \mathrm s\)
\(10\ \mathrm{km}\)	\(26.5\ \mathrm{kPa}\)	\(2443\ \mathrm{kN}\)	\(307.1\ \mathrm s\)
\(20\ \mathrm{km}\)	\(5.5\ \mathrm{kPa}\)	\(2492\ \mathrm{kN}\)	\(313.4\ \mathrm s\)
\(30\ \mathrm{km}\)	\(1.2\ \mathrm{kPa}\)	\(2503\ \mathrm{kN}\)	\(314.7\ \mathrm s\)

Half of the gain arrives in the first \(5.5\ \mathrm{km}\) and nine tenths of it by \(16\ \mathrm{km}\) (Figure 3.2, right).

A rough altitude history. To turn altitude into time a trajectory is needed, and Chapter 7 will compute one. For an estimate, use the vertical climb of Section 1.7 with no drag and \(g=g_0\) , but with the thrust now depending on altitude:

\[ \frac{\mathrm dz}{\mathrm dt}=v,\qquad \frac{\mathrm dv}{\mathrm dt}=\frac{F_{\mathrm{vac}}-p_a(z)A_e}{m_{0,1}-\dot m_1t}-g_0,\qquad z(0)=0,\ v(0)=0 . \]

The right side is smooth except where the zones of the atmosphere model meet. There the model’s pressure jumps slightly, from \(22.707\) to \(22.718\ \mathrm{kPa}\) at \(11\ \mathrm{km}\) and from \(2.522\) to \(2.483\ \mathrm{kPa}\) at \(25\ \mathrm{km}\) , which changes the thrust by less than \(100\ \mathrm N\) in \(2.5\ \mathrm{MN}\) . The fourth-order Runge-Kutta scheme of Section 1.7 is used with \(200\) steps of \(0.845\ \mathrm s\) , and halving the step changes the results below by less than one part in a million. Because of the two jumps, the full fourth-order convergence of a smooth problem should not be expected.

Time	Altitude	Ambient pressure	Specific impulse
\(0\ \mathrm s\)	\(0\ \mathrm{km}\)	\(101.4\ \mathrm{kPa}\)	\(285.0\ \mathrm s\)
\(40\ \mathrm s\)	\(3.1\ \mathrm{km}\)	\(69.3\ \mathrm{kPa}\)	\(294.5\ \mathrm s\)
\(60\ \mathrm s\)	\(8.1\ \mathrm{km}\)	\(35.3\ \mathrm{kPa}\)	\(304.5\ \mathrm s\)
\(80\ \mathrm s\)	\(16.6\ \mathrm{km}\)	\(9.4\ \mathrm{kPa}\)	\(312.2\ \mathrm s\)
\(100\ \mathrm s\)	\(30.1\ \mathrm{km}\)	\(1.1\ \mathrm{kPa}\)	\(314.7\ \mathrm s\)

After \(80\ \mathrm s\) of a \(169\ \mathrm s\) burn the engine is within one percent of its vacuum performance.

The effective value. The ideal velocity increment is the integral of thrust acceleration (Section 1.5), which the same integration accumulates:

\[ \Delta v_1=\int_0^{t_{b,1}}\frac{F}{m}\,\mathrm dt=4467\ \mathrm{m/s}. \]

The mass ratio is still \(\mathcal R_1=4.359\) , so the single exhaust velocity that would give the same increment is \(4467/\ln4.359=3034\ \mathrm{m/s}\) , that is \(309.4\ \mathrm s\) . The plain time average of the specific impulse over the burn is lower, \(305.9\ \mathrm s\) ; the velocity-weighted value is higher because the late, high-performance part of the burn acts on a lighter vehicle (Exercise 17 makes this precise).

How far to trust it. A real vehicle turns toward the horizontal and is lower at a given time than this vertical climb. Repeating the calculation with the altitude fed to the pressure model halved gives \(307.0\ \mathrm s\) . So the stage’s effective specific impulse is between about \(307\) and \(309\ \mathrm s\) , well above the \(300\ \mathrm s\) assumed in Revision A and much closer to the vacuum figure than to the sea-level one. The stage spends most of its propellant above most of the atmosphere.

3.5 How far to expand: a first argument

The exit pressure \(p_e\) in Example 3.1 was below sea-level pressure, and the pressure term was negative there. Would the engine do better at sea level with a shorter nozzle, whose exit pressure is higher? Proposition 3.4 answers with almost no calculation.

Hold the chamber and the throat fixed, so the flow inside the existing nozzle does not change, and imagine extending the bell by a short ring of wall. The gas keeps expanding along the ring, so the pressure on the ring’s inside is a little below the old exit pressure \(p_e\) ; outside it is \(p_a\) . The ring faces partly backward, so its contribution to the thrust integral is its pressure excess, about \(p_e-p_a\) , times its projected area. If \(p_e\) exceeds \(p_a\) , the ring adds thrust: the nozzle was too short. If \(p_e\) is below \(p_a\) , the ring subtracts thrust: the outside air pushes it backward harder than the gas pushes it forward, and the nozzle was already too long. The best length is the one at which the gas reaches the exit at exactly the ambient pressure.

Chapter 5 turns this argument into a derivative, \(\mathrm dF/\mathrm dA_e=p_e-p_a\) , and computes the exit area that achieves it. Three warnings follow at once. First, the optimum is for one ambient pressure, and a first stage flies through all of them from \(101\ \mathrm{kPa}\) to nearly zero, so its nozzle is a compromise. A nozzle with \(p_e=p_a\) is called matched; one with \(p_e\) above \(p_a\) is underexpanded, and one with \(p_e\) below \(p_a\) is overexpanded. The engine of Example 3.1 is overexpanded on the pad and underexpanded for most of its burn. Second, in vacuum the proposition says the nozzle should be infinitely long, since any ring adds thrust when \(p_a=0\) . What stops an upper-stage nozzle is its size and mass, not the gas. Third, the thrust equation assumed that the gas fills the nozzle all the way to the exit. A nozzle whose exit pressure is far below ambient does not behave that way: the outside air forces its way in along the wall and the flow breaks away from it. Chapter 5 gives the limit; until then, treat the formula with suspicion whenever \(p_e\) is less than about half of \(p_a\) .

3.6 Total impulse

Thrust integrated over time is the total impulse of a burn,

in \(\mathrm{N\cdot s}\) . If \(c\) is constant it comes out of the integral and \(I_{\mathrm{tot}}=c\,m_p\) : the effective exhaust velocity is the impulse delivered per kilogram of propellant, which is how Section 1.4 read it. For the reference launcher’s first stage, \(137\,021\ \mathrm{kg}\) of propellant would deliver \(383\ \mathrm{MN\cdot s}\) at the sea-level exhaust velocity and \(423\ \mathrm{MN\cdot s}\) at the vacuum one; the climb of Example 3.4 delivers \(411\ \mathrm{MN\cdot s}\) .

Total impulse is the right measure for a motor considered apart from any vehicle, because it does not depend on what the motor is attached to. It is not the velocity the vehicle gains. Two motors with the same total impulse give the same \(\Delta v\) only to vehicles of the same mass history, and the rocket equation is what converts one into the other.

3.7 A chamber factor and a nozzle factor

An engine’s effective exhaust velocity depends on what burns in the chamber and on the shape of the nozzle, and a low value might be the fault of either. Engineers separate the two by measuring two more quantities on the test stand, the chamber pressure \(p_c\) and the throat area \(A_t\) , and forming two ratios.

Check the units of \(c^\ast\) : \(\mathrm{Pa\cdot m^2/(kg/s)}=\mathrm{N\cdot s/kg}=\mathrm{m/s}\) .

The force \(p_cA_t\) is the natural yardstick. It is the unbalanced pressure force of Section 3.3, the thrust of a chamber with a plain hole. The thrust coefficient says how many times that yardstick the real engine delivers, and so it measures the nozzle: its shape, its exit area, and the ambient pressure it works against. It changes with altitude along with \(F\) .

The characteristic velocity says how much chamber pressure a unit of mass flow sustains behind a given throat. A hotter or lighter gas needs less mass flow to hold the same pressure, and has a higher \(c^\ast\) . Nothing downstream of the throat enters its definition, and Chapter 5 shows that for an ideal gas it is a function of the chamber temperature and the gas properties alone. It measures the propellants and how completely they burn. For common liquid propellant pairs the theoretical values run from about \(1720\) to \(2420\ \mathrm{m/s}\) , the low end for storable propellants and the high end for hydrogen and oxygen, with kerosene and oxygen near \(1800\ \mathrm{m/s}\) (Wikipedia, “Liquid rocket propellant”, tabulating ideal calculations; see sources).

The factorization lets a test engineer assign blame. If a new injector lowers \(c^\ast\) while \(C_F\) holds, combustion got worse. If a nozzle change raises \(C_F\) at the same \(c^\ast\) , the chamber is untouched and the nozzle is extracting more.

3.8 The power in the jet

Force and momentum have carried the chapter this far. Energy has not appeared, and it is energy that the propellant supplies. How much is in the exhaust?

Sit on the engine. Each second, \(\dot m\) kilograms of gas that were at rest in the tanks leave at speed \(v_e\) , so the engine delivers kinetic energy to the jet at the rate \(\tfrac12\dot mv_e^2\) . For a matched nozzle, where \(c=v_e\) and \(F=\dot mc\) , this jet power is

in watts. The symbol \(\dot W\) denotes a power throughout these chapters.

Now stand on the ground and watch the vehicle pass at speed \(v\) . The thrust does work on the vehicle at the rate \(Fv=\dot m\,c\,v\) . The exhaust, which leaves at \(c\) backward relative to the vehicle, moves at \(v-c\) over the ground and is left carrying kinetic energy at the rate \(\tfrac12\dot m(c-v)^2\) . That residual energy does nothing for the vehicle. The fraction of the total that reaches the vehicle is the propulsive efficiency,

The denominator, \(\tfrac12\dot m(c^2+v^2)\) , is the jet power plus the kinetic energy that the propellant already had from riding on the vehicle. The subscripted symbol \(\eta_{\mathrm{prop}}\) is unrelated to the staging multiplier \(\eta\) of Chapter 2.

The efficiency is zero at \(v=0\) : on the pad all of the jet power stays in the exhaust. It rises to exactly \(1\) when \(v=c\) , the speed at which the exhaust is left standing still over the ground with no energy at all, and falls again beyond (Figure 3.3). Differentiating with respect to the ratio \(X=v/c\) gives \(2(1-X^2)/(1+X^2)^2\) , which confirms the maximum at \(X=1\) .

3.9 The first stage after this chapter

The reference launcher’s masses are unchanged from Revision A, and everything in this table still belongs to that revision; only the first stage’s thrust model has been refined. Its first-stage propulsion numbers are now these.

Two consequences are left for later chapters. With the vacuum thrust acting on the nearly empty stage, the sensed acceleration at burnout becomes \(F_{\mathrm{vac}}/(m_{f,1}g_0)=6.26\,g_0\) instead of \(5.67\,g_0\) , which is more than most payloads are asked to bear; Chapter 7 follows the acceleration through the ascent and puts a price on limiting it. And an effective specific impulse seven to nine seconds above the assumed one is worth, by the exchange ratio of Example 2.6, some \(240\) to \(310\ \mathrm{kg}\) of payload, which the launcher will bank when Chapter 9 redoes the budget and issues Revision B. The second stage works only in vacuum and keeps its single value of \(345\ \mathrm s\) for now.

The specific impulses themselves are still assumptions. Why \(285\) and \(315\) , and not \(250\) or \(400\) ? The next two chapters answer from the inside of the engine: Chapter 4 finds the exit velocity from the state of the gas in the chamber, and Chapter 5 finds the mass flow, the exit pressure, and the thrust coefficient from the shape of the nozzle.

Exercises

1. State the momentum theorem for a fixed control volume in steady flow. For the engine on a test stand, name the control surface used to derive the thrust equation, say what crosses it and where, list the five assumptions, and state which assumption fails first during the half second in which the engine starts.

2. An upper-stage engine (illustrative values) passes \(\dot m=85.0\ \mathrm{kg/s}\) and has \(v_e=3300\ \mathrm{m/s}\) , \(p_e=4.0\ \mathrm{kPa}\) , and \(A_e=2.10\ \mathrm{m^2}\) . (a) Find its thrust, effective exhaust velocity, and specific impulse in vacuum. (b) What fraction of the vacuum thrust is pressure thrust? (c) Find the attached-flow thrust the formula predicts at sea level and say why the engine could not be tested there with this nozzle.

   Answer

   (a) Momentum thrust \((85.0)(3300)=280.5\ \mathrm{kN}\) ; pressure thrust \((4.0\ \mathrm{kPa})(2.10\ \mathrm{m^2})=8.4\ \mathrm{kN}\) ; \(F_{\mathrm{vac}}=288.9\ \mathrm{kN}\) , \(c_{\mathrm{vac}}=3399\ \mathrm{m/s}\) , \(I_{sp}=346.6\ \mathrm s\) . (b) \(8.4/288.9=2.9\) percent. (c) The formula gives \(280.5+(4.0-101.325)(2.10)=76.1\ \mathrm{kN}\) , barely a quarter of the vacuum thrust. The exit pressure is \(4\) percent of ambient, far below the level at which the flow stays attached to the wall, so the gas would break away inside the bell and the formula would not describe what happens.

3. The reference launcher’s first stage produces \(2267\ \mathrm{kN}\) at sea level and has a total exit area of \(2.355\ \mathrm{m^2}\) . (a) Find its thrust where the ambient pressure is \(30\ \mathrm{kPa}\) and in vacuum. (b) Find the ambient pressure at which the thrust has risen halfway from its sea-level to its vacuum value. (c) Using the atmosphere model of Section 3.4, find the altitude at which that pressure occurs.

   Solution

   The mass flow and the exit state of a running engine do not depend on the ambient pressure, so Corollary 3.5 applies: thrust is a linear function of \(p_a\) with slope \(-A_e\) .

   (a) Start from the sea-level value and add back the pressure that is no longer subtracted:

   \[ F(30\ \mathrm{kPa})=2266.9\ \mathrm{kN}+(101.325-30)\ \mathrm{kPa}\times2.355\ \mathrm{m^2}=2435\ \mathrm{kN}, \]

   \[ F_{\mathrm{vac}}=2266.9+(101.325)(2.355)=2505\ \mathrm{kN}. \]

   (b) Because the relation is linear, the thrust is halfway between its end values when the pressure is halfway between its end values: \(p_a=101.325/2=50.66\ \mathrm{kPa}\) .

   (c) That pressure lies in the lowest zone of the model, where \(p=101.29\,[(288.14-0.00649z)/288.08]^{5.256}\) in kilopascals. Solving,

   \[ 288.14-0.00649\,z=288.08\left(\frac{50.66}{101.29}\right)^{1/5.256}=252.50,\qquad z=5491\ \mathrm m . \]

   Half of the thrust gain is in hand by \(5.5\ \mathrm{km}\) , which the stage of Example 3.4 passes about \(50\ \mathrm s\) after liftoff.

4. The F-1 engine of the Saturn V first stage is listed with thrust \(6770\ \mathrm{kN}\) at sea level and \(7770\ \mathrm{kN}\) in vacuum, specific impulse \(263\) and \(304\ \mathrm s\) , chamber pressure \(7.0\ \mathrm{MPa}\) , propellant flow \(2578\ \mathrm{kg/s}\) , and nozzle area ratio \(16\) (Wikipedia, “Rocketdyne F-1”; see sources). (a) Find the exit area and exit diameter. (b) Test the thrust ratio against the specific-impulse ratio and the listed flow against thrust over effective exhaust velocity. (c) Find the throat area, \(c^\ast\) , and \(C_F\) at sea level and in vacuum. (d) Comment on which results you trust least.

   Solution

   (a) By Corollary 3.5 the difference of the two thrusts is \(p_{\mathrm{SL}}A_e\) :

   \[ A_e=\frac{(7770-6770)\times10^{3}\ \mathrm N}{101\,325\ \mathrm{N/m^2}}=9.869\ \mathrm{m^2},\qquad d_e=\sqrt{4A_e/\pi}=3.54\ \mathrm m . \]

   (b) Thrust ratio \(6770/7770=0.8713\) ; specific-impulse ratio \(263/304=0.8651\) . They differ by \(0.7\) percent. The flows implied by thrust over effective exhaust velocity are \(6.770\times10^{6}/\bigl((263)(9.80665)\bigr)=2625\ \mathrm{kg/s}\) and \(2606\ \mathrm{kg/s}\) , against the listed \(2578\ \mathrm{kg/s}\) : one to two percent high. The figures are mutually consistent at the level of a couple of percent, and no better.

   (c) With area ratio \(16\) , \(A_t=9.869/16=0.6168\ \mathrm{m^2}\) and \(p_cA_t=(7.0\times10^{6})(0.6168)=4318\ \mathrm{kN}\) . Using the listed flow,

   \[ c^\ast=\frac{4.318\times10^{6}\ \mathrm N}{2578\ \mathrm{kg/s}}=1675\ \mathrm{m/s},\qquad C_{F,\mathrm{SL}}=\frac{6770}{4318}=1.568,\qquad C_{F,\mathrm{vac}}=\frac{7770}{4318}=1.800 . \]

   (d) The exit area comes from a difference of two rounded thrusts, so a rounding of either by \(20\ \mathrm{kN}\) moves it by \(2\) percent, and the throat area and both derived quantities inherit that error together with any error in the quoted area ratio. The characteristic velocity is about \(7\) percent below the theoretical \(1800\ \mathrm{m/s}\) for these propellants. Part of that is real, since this engine burned a fuel-rich mixture at moderate pressure, and part is the uncertainty of the chain. The thrust coefficients are of the expected size.

5. A magazine explains that “a rocket works by pushing its exhaust against the air, which is why engines are weaker in the thin air at altitude and need special nozzles in space.” Identify the two false statements, cite the term of the thrust equation that contradicts each, and compute the thrust of the engine in Example 3.1 at an altitude where the pressure is \(10\ \mathrm{kPa}\) .

   Answer

   False statement one: the exhaust pushes on the air. In the thrust equation the atmosphere appears only in \(-p_aA_e\) , a term that reduces thrust. False statement two: engines are weaker at altitude. Since \(F=F_{\mathrm{vac}}-p_aA_e\) , thrust rises as \(p_a\) falls. Large nozzles are used in vacuum because any added expansion gains thrust there, not because there is nothing to push on. For Example 3.1 at \(10\ \mathrm{kPa}\) : \(F=787.05+(60-10)(0.785)=826.3\ \mathrm{kN}\) .

6. A combustor is modeled as a straight tube of constant cross-section \(A\) , closed at the front by a flat wall and fully open at the back. Propellant enters through the front wall with negligible momentum and leaves the open end uniformly at speed \(v_e\) , density \(\rho_e\) , and pressure \(p_e\) . Neglect friction. (a) Use a control volume containing the gas alone to show that the pressure on the front wall is \(p_e+\rho_ev_e^2\) . (b) Show that the thrust in surroundings at \(p_a\) is \((p_e+\rho_ev_e^2-p_a)A\) and that this agrees with Theorem 3.3. (c) Explain why the pressure must fall along the tube even though its walls are parallel.

   Solution

   The system is a control volume containing the gas in the tube: bounded by the front wall, the side wall, and the open exit plane. The flow is steady, friction on the side wall is neglected, and the side wall is parallel to the axis, so pressure on it has no axial component.

   (a) Let \(\hat{\mathbf t}\) point forward. The momentum flux out through the exit plane is \(-\dot mv_e\hat{\mathbf t}\) with \(\dot m=\rho_ev_eA\) , that is \(-\rho_ev_e^2A\,\hat{\mathbf t}\) ; the entering propellant contributes nothing. The axial forces on the gas are the pressure of the outside gas on the exit plane, \(+p_eA\,\hat{\mathbf t}\) , and the push of the front wall, which acts rearward on the gas with magnitude \(p_{\mathrm{front}}A\) . Theorem 3.1 in steady flow gives

   \[ p_eA-p_{\mathrm{front}}A=-\rho_ev_e^2A\qquad\Longrightarrow\qquad p_{\mathrm{front}}=p_e+\rho_ev_e^2 . \]

   (b) The gas pushes forward on the front wall with \(p_{\mathrm{front}}A\) and the atmosphere pushes back on its outside with \(p_aA\) ; the side wall carries no axial force. So \(F=(p_e+\rho_ev_e^2-p_a)A=\dot mv_e+(p_e-p_a)A\) , which is Theorem 3.3 with \(A_e=A\) .

   (c) The gas starts with no axial momentum and leaves with \(\rho_ev_e^2\) per unit area per unit time. Something must push it rearward, and with parallel frictionless walls the only axial force available is a pressure difference between the two ends. The pressure therefore drops along the tube by exactly the momentum flux the gas acquires. This is Proposition 3.4 in its simplest geometry: all of the thrust is pressure excess on one flat wall.

7. An engine is fired in an altitude test cell at two cell pressures (illustrative values): it gives \(1.900\ \mathrm{MN}\) at \(80\ \mathrm{kPa}\) and \(2.050\ \mathrm{MN}\) at \(20\ \mathrm{kPa}\) . (a) Describe the graph of thrust against cell pressure and what its slope and intercept mean. (b) Find the exit area, the vacuum thrust, and the thrust at standard sea-level pressure. (c) A third firing gives \(1.975\ \mathrm{MN}\) at \(60\ \mathrm{kPa}\) ; is it consistent with the first two?

   Solution

   (a) By Corollary 3.5 the graph is a straight line falling to the right. Its slope is \(-A_e\) , and its intercept on the thrust axis, at zero pressure, is the vacuum thrust.

   (b) From the two firings,

   \[ A_e=\frac{(2.050-1.900)\times10^{6}\ \mathrm N}{(80-20)\times10^{3}\ \mathrm{Pa}}=2.50\ \mathrm{m^2}, \]

   \[ F_{\mathrm{vac}}=2.050\ \mathrm{MN}+(20\ \mathrm{kPa})(2.50\ \mathrm{m^2})=2.100\ \mathrm{MN},\qquad F_{\mathrm{SL}}=2.100-(101.325)(2.50)\times10^{-3}=1.847\ \mathrm{MN}. \]

   (c) The line predicts \(2.100-(60)(2.50)\times10^{-3}=1.950\ \mathrm{MN}\) at \(60\ \mathrm{kPa}\) . The third firing shows \(1.975\ \mathrm{MN}\) , which is \(25\ \mathrm{kN}\) or \(1.3\) percent above the line. Three points that should be collinear are not, so either a measurement is off or the engine did not run at the same chamber conditions in all three firings; the mass flow and chamber pressure records would show which.

8. Use the added-ring argument of Section 3.5. (a) Three nozzles on identical chambers have exit pressures of \(150\) , \(101\) , and \(40\ \mathrm{kPa}\) . For operation at sea level, say for each whether a short extension would raise or lower the thrust. (b) A ring added to the third nozzle has projected area \(0.10\ \mathrm{m^2}\) facing rearward and an average inside pressure of \(35\ \mathrm{kPa}\) . Find the change in thrust at sea level and in vacuum. (c) Explain why no finite nozzle is optimal in vacuum and what limits its size in practice.

   Answer

   (a) At sea level, \(101\ \mathrm{kPa}\) : the nozzle with \(p_e=150\ \mathrm{kPa}\) gains from an extension, the one with \(p_e=101\ \mathrm{kPa}\) is at the optimum and neither gains nor loses to first order, and the one with \(p_e=40\ \mathrm{kPa}\) loses. (b) \(\Delta F=(35-101.325)(0.10)=-6.6\ \mathrm{kN}\) at sea level and \(+3.5\ \mathrm{kN}\) in vacuum. (c) In vacuum every added ring has positive pressure excess, so thrust rises without limit in the model; in practice each ring adds less than the one before while its mass and length keep growing, and the stage has to fit inside the vehicle.

9. A solid motor (illustrative values) has a thrust history that rises linearly from zero to \(400\ \mathrm{kN}\) in \(2\ \mathrm s\) , holds that value until \(50\ \mathrm s\) , and falls linearly to zero at \(60\ \mathrm s\) . It burns \(8200\ \mathrm{kg}\) of propellant. Find the total impulse, the average effective exhaust velocity, the specific impulse, and the average thrust, and explain why the total impulse alone does not fix the velocity change the motor gives a vehicle.

   Solution

   Total impulse is the area under the thrust history, which here is a trapezoid made of a rising triangle, a rectangle, and a falling triangle:

   \[ I_{\mathrm{tot}}=\tfrac12(2)(400)+(48)(400)+\tfrac12(10)(400)=400+19\,200+2000=21\,600\ \mathrm{kN\cdot s}. \]

   The average effective exhaust velocity is the impulse per unit propellant mass, \(c=21.6\times10^{6}\ \mathrm{N\cdot s}/8200\ \mathrm{kg}=2634\ \mathrm{m/s}\) , so \(I_{sp}=2634/9.80665=268.6\ \mathrm s\) . The average thrust is \(21\,600/60=360\ \mathrm{kN}\) .

   Total impulse does not fix the velocity change because it takes no account of the mass being accelerated. The same \(21.6\ \mathrm{MN\cdot s}\) applied to a light vehicle gives a large \(\Delta v\) and to a heavy one a small \(\Delta v\) ; the rocket equation with this \(c\) and the vehicle’s actual mass ratio is what converts the motor’s impulse into velocity.

10. Two firings of one chamber with two different nozzles give the same \(c^\ast=1750\ \mathrm{m/s}\) and thrust coefficients of \(1.45\) and \(1.60\) . (a) What changed between the firings and what did not? (b) Compute the effective exhaust velocity and specific impulse of each. (c) A third firing uses the first nozzle and shows \(c^\ast\) lower by \(3\) percent at the same \(C_F\) ; give two physical causes and the new specific impulse.

    Solution

    (a) The characteristic velocity is unchanged, so the chamber did the same thing in both firings: the same combustion, and the same mass flow per unit of chamber pressure and throat area. The thrust coefficient rose from \(1.45\) to \(1.60\) , about \(10\) percent, so the second nozzle extracts more thrust from the same gas. It expands the gas further, or to a pressure closer to the ambient pressure, or it was fired at lower ambient pressure.

    (b) From \(c=c^\ast C_F\) :

    \[ c_1=(1750)(1.45)=2538\ \mathrm{m/s}\ (258.8\ \mathrm s),\qquad c_2=(1750)(1.60)=2800\ \mathrm{m/s}\ (285.5\ \mathrm s). \]

    The ratio \(c_2/c_1=1.103\) equals the ratio of thrust coefficients.

    (c) A lower \(c^\ast\) at the same nozzle means the chamber is delivering less pressure per unit of mass flow. Two physical causes are incomplete combustion, for example from poorer mixing at the injector, and an off-design mixture ratio that gives a cooler or heavier gas. The specific impulse falls in proportion: \((0.97)(1750)(1.45)=2461\ \mathrm{m/s}\) , or \(251.0\ \mathrm s\) .

11. A data sheet lists an engine’s chamber pressure as \(1410\ \mathrm{psi}\) , its sea-level thrust as \(845\ \mathrm{kN}\) , and its sea-level specific impulse as \(282\ \mathrm s\) (figures quoted for the Merlin 1D; secondary compilation; see sources). Take \(1\ \mathrm{psi}=6894.8\ \mathrm{Pa}\) and assume \(c^\ast=1760\ \mathrm{m/s}\) (illustrative). (a) Convert the chamber pressure to megapascals. (b) Find the mass flow, the throat area, and the throat diameter. (c) Find \(C_F\) at sea level. (d) What would your value of \(C_F\) have been had you used the pressure in psi as if it were kilopascals, and how would you have noticed?

    Answer

    (a) \(p_c=(1410)(6894.8)=9.72\ \mathrm{MPa}\) . (b) \(\dot m=845\times10^{3}/\bigl((282)(9.80665)\bigr)=305.6\ \mathrm{kg/s}\) ; \(A_t=\dot mc^\ast/p_c=(305.6)(1760)/(9.72\times10^{6})=0.0553\ \mathrm{m^2}\) ; diameter \(0.265\ \mathrm m\) . (c) \(C_F=845\times10^{3}/\bigl((9.72\times10^{6})(0.0553)\bigr)=1.571\) . (d) Using \(1410\ \mathrm{kPa}\) with the correct throat area gives \(C_F=10.8\) . A thrust coefficient lies between about \(1.3\) and \(2\) ; a value of ten is a units error, and the dimensionless check exposes it at once.

12. (Computation.) Write a program that evaluates the three-zone atmosphere model of Section 3.4 and integrates the vertical climb of Example 3.4 for the reference launcher’s first stage with a fourth-order Runge-Kutta scheme. Report the ideal velocity increment and the effective specific impulse for \(100\) , \(200\) , and \(400\) steps, then repeat with the altitude used in the pressure model multiplied by \(0.5\) and by \(0.25\) , and tabulate the effective specific impulse for the three cases.

    Answer

    With the true altitude: \(\Delta v_1=4467.20\ \mathrm{m/s}\) and \(309.41\ \mathrm s\) at each of \(100\) , \(200\) , and \(400\) steps, the three agreeing to within \(0.001\ \mathrm{m/s}\) . With the altitude in the pressure model multiplied by \(0.5\) : \(4433.0\ \mathrm{m/s}\) and \(307.0\ \mathrm s\) . Multiplied by \(0.25\) : \(4382.2\ \mathrm{m/s}\) and \(303.5\ \mathrm s\) . Even a path that stays four times lower than the vertical climb keeps the effective value well above the \(300\ \mathrm s\) of Revision A.

13. The Merlin 1D gives \(845\ \mathrm{kN}\) at sea level with a specific impulse of \(282\ \mathrm s\) (Falcon User’s Guide and a secondary compilation; see sources). (a) Estimate the jet power of one engine and of the nine on a Falcon 9 first stage, treating \(c\) as the jet speed. (b) Express the nine-engine figure in units of a \(1\ \mathrm{GW}\) power station. (c) Explain why this estimate is slightly off for an engine whose exit pressure differs from ambient.

    Answer

    (a) \(c=(282)(9.80665)=2765.5\ \mathrm{m/s}\) ; one engine \(\tfrac12(845\times10^{3})(2765.5)=1.17\ \mathrm{GW}\) ; nine engines \(10.5\ \mathrm{GW}\) . (b) About ten and a half such stations. (c) Jet power is \(\tfrac12\dot mv_e^2\) , and \(c\) equals \(v_e\) only for a matched nozzle. At sea level this engine’s \(c\) is below \(v_e\) if its exit pressure is below ambient, so \(\tfrac12Fc\) somewhat understates the kinetic power actually in the jet.

14. For the propulsive efficiency \(\eta_{\mathrm{prop}}(X)=2X/(1+X^2)\) with \(X=v/c\) : (a) show that \(\eta_{\mathrm{prop}}(X)=\eta_{\mathrm{prop}}(1/X)\) ; (b) find the two speed ratios at which the efficiency is \(0.9\) ; (c) evaluate it for a second stage with \(c=3383\ \mathrm{m/s}\) at vehicle speeds of \(2.5\) and \(7.7\ \mathrm{km/s}\) ; (d) explain in a sentence what becomes of the energy that is not counted as useful.

    Solution

    (a) Replace \(X\) by \(1/X\) and multiply numerator and denominator by \(X^2\) :

    \[ \eta_{\mathrm{prop}}(1/X)=\frac{2/X}{1+1/X^2}=\frac{2X}{X^2+1}=\eta_{\mathrm{prop}}(X). \]

    The curve is symmetric on a logarithmic speed axis about \(X=1\) .

    (b) Setting \(2X/(1+X^2)=0.9\) gives \(0.9X^2-2X+0.9=0\) , so

    \[ X=\frac{2\pm\sqrt{4-3.24}}{1.8}=0.627\ \text{or}\ 1.595 , \]

    a reciprocal pair, as part (a) requires.

    (c) At \(2.5\ \mathrm{km/s}\) , \(X=2500/3383=0.739\) and \(\eta_{\mathrm{prop}}=0.956\) . At \(7.7\ \mathrm{km/s}\) , \(X=2.276\) and \(\eta_{\mathrm{prop}}=0.737\) .

    (d) The energy not delivered to the vehicle remains as kinetic energy of the exhaust over the ground: below \(X=1\) the exhaust is left moving backward, above it the exhaust is left moving forward, and in both cases it carries energy that did no work on the vehicle.

15. An engineer proposes to “fix the poor propulsive efficiency at liftoff” by fitting a first stage with an engine of half the exhaust velocity, \(c=1500\ \mathrm{m/s}\) instead of \(3000\ \mathrm{m/s}\) , arguing that the efficiency at low speed is then higher. For a required ideal increment of \(3000\ \mathrm{m/s}\) , compute the mass ratio and propellant fraction with each engine, and explain what the proposal has optimized and why that is the wrong thing.

    Hint

    Compute \(\mathcal R=e^{\Delta v/c}\) for both engines: \(e^{1}=2.72\) (propellant fraction \(63\) percent) against \(e^{2}=7.39\) (\(86\) percent). The proposal raises the fraction of jet energy that reaches the vehicle and more than doubles the mass that must be lifted to do it. Ask which is scarce on a launcher: energy per kilogram of propellant, or kilograms.

16. Suppose a propellant combination releases \(10\ \mathrm{MJ}\) per kilogram and every joule became kinetic energy of the jet. (a) Find the exhaust velocity and compare it with \(3.0\) to \(4.5\ \mathrm{km/s}\) for real engines. (b) A common slip gives \(10\ \mathrm{km/s}\) ; identify the two errors behind it. (c) Find the energy per kilogram that an exhaust velocity of \(30\ \mathrm{km/s}\) requires and compare it with the \(13.3\ \mathrm{MJ/kg}\) of hydrogen and oxygen.

    Answer

    (a) \(c=\sqrt{2\times1.0\times10^{7}}=4472\ \mathrm{m/s}\) , at or just above the top of the range for real engines, as expected for a figure that assumes perfect conversion. (b) Reading megajoules per kilogram as if they were kilometres per second drops both the factor of one half and the square root; \(1\ \mathrm{J/kg}=1\ \mathrm{m^2/s^2}\) . (c) \(\tfrac12(3.0\times10^{4})^2=450\ \mathrm{MJ/kg}\) , about \(34\) times what hydrogen and oxygen release. No chemical reaction comes close, so such exhaust velocities need an energy source separate from the propellant.

17. A stage burns at constant mass flow \(\dot m\) while its effective exhaust velocity \(c(t)\) varies. (a) Show that its ideal velocity increment is \(\Delta v=\int_0^{t_b}c(t)\,\dot m/m(t)\,\mathrm dt\) . (b) Define \(c_{\mathrm{eff}}=\Delta v/\ln\mathcal R\) and show that it is a weighted average of \(c(t)\) with weight proportional to \(1/m(t)\) . (c) Deduce that \(c_{\mathrm{eff}}\) lies between the smallest and largest values of \(c\) , and that if \(c\) increases with time then \(c_{\mathrm{eff}}\) exceeds the plain time average of \(c\) . (d) Check both statements against the numbers of Example 3.4.

    Solution

    (a) With no external force, Proposition 1.1 in the form used in Section 1.4 reads \(m\,\mathrm dv/\mathrm dt=F=\dot m\,c(t)\) . Divide by \(m\) and integrate over the burn:

    \[ \Delta v=\int_0^{t_b}\frac{c(t)\,\dot m}{m(t)}\,\mathrm dt . \]

    (b) With constant flow, \(m(t)=m_0-\dot mt\) and \(\int_0^{t_b}\dot m/m\,\mathrm dt=\ln(m_0/m_f)=\ln\mathcal R\) . Therefore

    \[ c_{\mathrm{eff}}=\frac{\Delta v}{\ln\mathcal R}=\frac{\int_0^{t_b}c(t)\,\bigl[1/m(t)\bigr]\,\mathrm dt}{\int_0^{t_b}\bigl[1/m(t)\bigr]\,\mathrm dt}, \]

    a weighted average of \(c\) with the positive weight \(1/m(t)\) .

    (c) A weighted average with positive weights lies between the least and greatest values of the quantity averaged. For the comparison with the time average, note that \(1/m(t)\) increases with time. If \(c(t)\) also increases, the weight is larger where \(c\) is larger, so the weighted average exceeds the unweighted one. Formally, for two increasing functions the average of the product is at least the product of the averages (Chebyshev’s sum inequality), which gives \(\int c\,(1/m)\,\mathrm dt\ge\frac1{t_b}\int c\,\mathrm dt\int(1/m)\,\mathrm dt\) .

    (d) In Example 3.4, \(c_{\mathrm{eff}}\) corresponds to \(309.4\ \mathrm s\) , between \(285\) and \(315\ \mathrm s\) , and above the time average of \(305.9\ \mathrm s\) . Both statements hold.

18. This exercise uses Chapters 1 and 2. (a) With the exchange ratio of Example 2.6, estimate the payload gained if the first stage’s effective specific impulse is \(308\ \mathrm s\) instead of \(300\ \mathrm s\) . (b) The stage’s burn time fell from \(177.8\) to \(168.9\ \mathrm s\) ; if the burn were vertical with \(g=g_0\) , by how much would the gravity loss change? (c) Explain why the second effect is a consequence of holding liftoff thrust fixed while lowering the sea-level specific impulse.

    Solution

    (a) Example 2.6 gives \(34.2\ \mathrm{kg}\) of payload per second of first-stage specific impulse, with tanks and requirement fixed. Eight more seconds are worth about \((8)(34.2)=274\ \mathrm{kg}\) , a gain of \(5.5\) percent on the \(5000\ \mathrm{kg}\) payload. The exchange ratio is a derivative, so for a change this large the figure is an estimate good to a few percent.

    (b) A vertical burn in uniform gravity loses \(g_0t_b\) . The change is \((9.80665\ \mathrm{m/s^2})(177.8-168.9)\ \mathrm s=87\ \mathrm{m/s}\) less loss, from \(1744\ \mathrm{m/s}\) to \(1656\ \mathrm{m/s}\) .

    (c) Thrust is \(\dot mc\) . Liftoff thrust was held at \(2267\ \mathrm{kN}\) because the thrust-to-weight ratio at ignition is the design choice, and the exhaust velocity available at liftoff fell from that of \(300\ \mathrm s\) to that of \(285\ \mathrm s\) . The only way to keep the product is a larger flow, \(811.1\) instead of \(770.5\ \mathrm{kg/s}\) , and a larger flow empties the same tanks sooner. The stage then also has more thrust at altitude than Revision A assumed, which is the same fact seen from the other end.

19. A test gives an engine’s mass flow, exit area, sea-level thrust, and vacuum thrust. (a) Show that these four measurements cannot determine \(v_e\) and \(p_e\) separately, and identify the one combination of them that is determined. (b) Name one additional measurement that would separate them. (c) For the engine of Example 3.1, list two different pairs \((v_e,p_e)\) that give the same thrust at every ambient pressure.

    Hint

    Write the two thrust measurements as \(F_{\mathrm{vac}}=\dot mv_e+p_eA_e\) and \(F_{\mathrm{SL}}=F_{\mathrm{vac}}-p_{\mathrm{SL}}A_e\) . The second equation contains neither unknown. For part (c), any pair with \(\dot mv_e+p_eA_e=834.15\ \mathrm{kN}\) will do; try \(p_e=40\ \mathrm{kPa}\) . For part (b), think of a pressure tap in the wall at the nozzle lip.

20. For the reference launcher after this chapter: (a) compute the sensed acceleration at first-stage burnout with the vacuum thrust; (b) find the time after liftoff at which the sensed acceleration reaches \(5\,g_0\) , assuming vacuum thrust by then; (c) find the thrust-to-weight ratio at liftoff if one of the three engines fails to start, and the corresponding ratio if the same total thrust were divided among five engines; (d) say what this suggests about the choice of engine count.

    Solution

    (a) At burnout the stack mass is \(m_{f,1}=40\,791\ \mathrm{kg}\) and the vehicle is far above the atmosphere, so the thrust is the vacuum value:

    \[ \frac{F_{\mathrm{vac}}}{m_{f,1}g_0}=\frac{2.5055\times10^{6}\ \mathrm N}{(40\,791\ \mathrm{kg})(9.80665\ \mathrm{m/s^2})}=6.26 , \]

    so the sensed acceleration is \(6.26\,g_0\) .

    (b) The sensed acceleration is \(5\,g_0\) when \(m=F_{\mathrm{vac}}/(5g_0)=51\,097\ \mathrm{kg}\) . The stage has then burned \(177\,812-51\,097=126\,714\ \mathrm{kg}\) , which at \(811.1\ \mathrm{kg/s}\) takes \(156.2\ \mathrm s\) . By Example 3.4 the vehicle is far above \(30\ \mathrm{km}\) at that time, so using the vacuum thrust was justified. The limit is exceeded for the last \(12.7\ \mathrm s\) of the burn.

    (c) With one of three engines out, the thrust is two thirds and \(\psi_0=(2/3)(1.30)=0.87\) : the vehicle cannot leave the pad. With five engines and one out, \(\psi_0=(4/5)(1.30)=1.04\) : it lifts off, barely.

    (d) More, smaller engines make the vehicle tolerant of losing one, and also allow the acceleration limit of part (b) to be met by shutting engines down. The cost is more parts and more plumbing. With three engines the reference launcher has no engine-out capability at liftoff, a limitation to carry into the design review.