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Use the instantaneous recycling approximation with a well-mixed gas reservoir, constant stellar yield $p$, no gas inflow, and an irreversible galactic outflow. Let $M_s$ be the mass locked into long-lived stars and remnants after prompt recycling, and put $\psi (t)={\dot{M}}_s(t)\ge 0$. The yield $p$ is the newly synthesized mass of heavy elements returned to the gas per unit increase of $M_s$. Thus $M_h$ in $Z=M_h/M_g$ denotes the heavy-element mass in the gas, excluding metals already locked into stars.

Assume $M_s(0)=0$, $M_g(0)=M_0$, and $Z(0)=Z_0$; initially pristine gas means $Z_0=0$. Take the outflow rate to be $\alpha \psi$ with constant $\alpha \ge 0$, and assume its metallicity equals the current gas-phase metallicity $Z$. This is the well-mixed leaky-box model of galactic chemical evolution. The parameter $\alpha$ is a mass-loading factor relative to the net rate of locking mass into stars. If a gross star formation rate $\Psi$ and a prompt returned fraction $\mathcal{R}$ are used instead, $\psi =(1-\mathcal{R})\Psi$ and a wind rate $\eta \Psi$ corresponds to $\alpha =\eta /(1-\mathcal{R})$.

Conservation of mass givesand the baryonic mass remaining in the box is $M_g+M_s=M_0-\alpha M_s$. Newly made metals enter the gas at rate $p\psi$, while pre-existing metals are locked into stars at rate $Z\psi$ and leave in the wind at rate $\alpha Z\psi$. HenceUsing $M_h=Z{M}_g$ and the product rule,The cancellation expresses the fact that a well-mixed wind removes gas and metals in the same proportion, while fresh stellar production raises the abundance in the remaining gas.

Divide this equation by ${\dot{M}}_g=-(1+\alpha )\psi$ along the evolving reservoir, or integrate in time through intervals with no star formation. The resulting gas-phase metallicity isEquivalently, its fully explicit dependence on a prescribed net star formation rate isvalid while $M_g(t)>0$. With pre-existing stars, replace $M_s(t)$ in the gas-mass relation by $M_s(t)-M_s(0)$. Without specifying $\psi (t)$ or a gas-consumption law, the model fixes $Z$ as a function of gas mass but does not determine a unique function of time.

For the closed-box model of galactic chemical evolution, set $\alpha =0$. Then $M_g=M_0-M_s$ andAt the same remaining fraction $g=M_g/M_0$ of the initial gas reservoir, the leaky box has the logarithmic coefficient $p/(1+\alpha )$, lower than the true stellar yield by the wind factor. This coefficient is often called an effective yield, but the definition of the gas fraction must be specified.

In particular, the observable gas fraction of a galaxy is often $\mu =M_g/(M_g+M_s)$, using the mass still present, rather than $g$. In our initially star-free model,ThereforeFor $0<\mu <1$ and $\alpha >0$, the leaky-box enrichment is smaller at the same $\mu$: $\ln [1+\alpha (1-\mu )]\le \alpha (1-\mu )<\alpha \ln (1/\mu )$. If effective yield is defined observationally as $(Z-Z_0)/\ln (1/\mu )$, it is consequently not simply the constant $p/(1+\alpha )$ at all $\mu$.

As an explicit time example, choose a linear gas-consumption law $\psi =ϵM_g$, with constant $ϵ>0$. The mass and metallicity equations giveThe closed box with the same $ϵ$ has the same metallicity at a given time, while the leaky box exhausts its gas faster. Thus a comparison at fixed time depends on the chosen star formation rate, even though the comparison at fixed gas fraction above is unambiguous.

Finally, the assumption about wind metallicity is essential. If the expelled gas has abundance $Z_w$ instead, its metal-loss rate is $\alpha Z_w\psi$, and the same bookkeeping yieldsA wind enriched relative to the ambient gas lowers the enrichment rate; the logarithmic solution derived above applies to $Z_w=Z$.