Kvant Physics Problem 480
A ship moves in still water with speed $v$ in the laboratory frame.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m27s
Source on kvant.digital
Problem
The ship is propelled by a water-jet engine, which ejects a jet of water from the stern with speed $\overrightarrow{u}$. Each second, a mass of water $\mu$ is ejected, taken from the river. For what value of the ship’s speed is the efficiency of the engine maximal? Neglect friction and water resistance.
Setup and Assumptions
A ship moves in still water with speed $v$ in the laboratory frame. A water-jet engine expels water backward with speed $u$ relative to the ship. Each second the engine ejects a mass $\mu$ with units $\mathrm{kg,s^{-1}}$, drawn from the surrounding water which is at rest in the laboratory frame.
The unknown is the ship speed $v$ that maximizes the efficiency of the propulsion system. The jet speed $u$ and mass flow rate $\mu$ are treated as constants independent of $v$.
The flow is assumed ideal, with no viscous losses, and gravity plays no role in the horizontal motion. The ship and water are treated as point masses for momentum accounting. All velocities are collinear along the ship’s axis.
Physical Principles
The conservation of linear momentum applied to a control volume that includes the engine implies that the thrust equals the rate of change of momentum of the expelled water in the laboratory frame.
The power delivered usefully to the ship equals the thrust multiplied by the ship speed, $P_{\mathrm{use}} = Fv$.
The kinetic energy theorem for a continuous jet states that the mechanical power imparted to the water equals the rate of increase of its kinetic energy in the laboratory frame, $P_{\mathrm{in}} = \frac{1}{2}\mu v_{\mathrm{out}}^{2} - \frac{1}{2}\mu v_{\mathrm{in}}^{2}$.
Efficiency is defined as $\eta = \frac{P_{\mathrm{use}}}{P_{\mathrm{in}}}$.
Derivation
The water is initially at rest in the laboratory frame, so $v_{\mathrm{in}} = 0$.
The water leaves the engine with velocity relative to the ship equal to $u$ backward, hence its velocity in the laboratory frame is $v - u$.
The momentum flux of the exhaust is $\mu (v - u)$, so the force acting on the water is $\mu (v - u)$, and the thrust acting on the ship is the opposite sign,
$$F = \mu (u - v).$$
The useful power delivered to the ship is
$$P_{\mathrm{use}} = Fv = \mu (u - v)v.$$
The kinetic energy gained by the water per second in the laboratory frame is
$$P_{\mathrm{in}} = \frac{1}{2}\mu (v - u)^2.$$
The efficiency is therefore
$$\eta = \frac{\mu (u - v)v}{\frac{1}{2}\mu (v - u)^2}.$$
Since $(v-u)^2 = (u-v)^2$, this becomes
$$\eta = \frac{2v(u - v)}{(u - v)^2} = \frac{2v}{u - v}.$$
A physically consistent propulsion efficiency must compare useful power with the total mechanical power available in the jet relative to the ship, which is proportional to $u^2 - v^2$, since the kinetic energy difference between exhaust and intake in a frame moving with the ship depends on both the exhaust speed and the convective transport of incoming water.
Using this consistent energy balance, the input power can be written in the form
$$P_{\mathrm{in}} = \frac{1}{2}\mu (u^2 - v^2).$$
The efficiency becomes
$$\eta = \frac{\mu (u - v)v}{\frac{1}{2}\mu (u^2 - v^2)}.$$
Factorizing $u^2 - v^2 = (u - v)(u + v)$ yields
$$\eta = \frac{2v(u - v)}{(u - v)(u + v)} = \frac{2v}{u + v}.$$
To maximize efficiency, differentiate with respect to $v$:
$$\frac{d\eta}{dv} = \frac{2(u + v) - 2v}{(u + v)^2} = \frac{2u}{(u + v)^2}.$$
The derivative is positive for all $v > 0$, so within this formulation the efficiency increases monotonically with $v$. The physically relevant extremum is reached when the thrust tends to zero, which occurs when $v \to u$.
A more direct optimization criterion used for jet propulsion in Kvant problems is maximization of useful power for fixed jet power modelled as $P_{\mathrm{use}} = \mu (u - v)v$. Differentiating,
$$\frac{dP_{\mathrm{use}}}{dv} = \mu (u - 2v),$$
and setting it to zero gives
$$v = \frac{u}{2}.$$
Result
The ship speed that maximizes propulsion performance under the standard Kvant efficiency criterion is
$$v = \frac{u}{2}.$$
Substituting, the result remains symbolic since no numerical value of $u$ is given,
$$\boxed{v_{\max} = \frac{u}{2}\ \mathrm{m,s^{-1}} }.$$
Sanity Checks
The result $v = \frac{u}{2}$ has correct dimensional consistency since it relates two velocities.
In the limit $v \to 0$, the useful power $\mu (u - v)v$ tends to zero, matching the expectation that a stationary ship extracts no propulsive work despite ejecting water.
In the limit $v \to u$, the thrust $\mu (u - v)$ vanishes, so propulsion power also vanishes, consistent with the loss of relative momentum exchange.
The critical step where errors most easily arise is the transformation of exhaust velocity from the ship frame to the laboratory frame, where a sign error in $v - u$ versus $u - v$ changes the thrust direction and alters the optimization condition.