Estimating ventilation time scales using overturning stream functions

Abstract

A simple method for estimating ventilation time scales from overturning stream functions is proposed. The stream function may be computed using either geometric coordinates or a generalized vertical coordinate, such as potential density (salinity in our study). The method is tested with a three-dimensional circulation model describing an idealized semi-enclosed ocean basin ventilated through a narrow strait over a sill, and the result is compared to age estimates obtained from a passive numerical age tracer. The best result is obtained when using the stream function in salinity coordinates. In this case, the reservoir-averaged advection time obtained from the overturning stream function in salinity coordinates agrees rather well with the mean age of the age tracer, and the corresponding maximum ages agree very well.

Appendix 1

1.1 Relation between mean age and turnover time

Bolin and Rodhe (1973) showed that the average transit time (i.e., the average age of the exiting fluid particles) is equal to the turnover time at a steady state. Björkström (1978) noted that the ratio between the average age and the average transit time of the fluid particles in a reservoir increases with increasing standard deviation in the transit-time distribution. From this, he was able to show that the average age of the fluid particles in the reservoir is at least half of the turnover time.

We will now show that the corresponding result holds also for an age tracer, which is affected by diffusion. The equation for the tracer distribution is obtained from Eq. (1) at a steady state,

$$ \nabla \cdot \left(\mathbf{u}\tau \right)=\nabla \cdot \left(\kappa \nabla \tau \right)+1. $$

(A1)

By integrating Eq. (A1) over the basin, we get

$$ {\displaystyle \underset{ strait}{\int}\tau \mathbf{u}\cdot d\mathbf{A}}={\displaystyle \underset{ strait}{\int}\kappa \mathit{\nabla \tau}\cdot d\kern-.1em \mathbf{A}}+{V}_0, $$

(A2)

where d A is the cross-sectional area in the strait. The only contribution to the integral on the left-hand side comes from the outflow, since τ = 0 in the incoming fluid. Thus, dividing Eq. (A2) by the total outflow F ₀ from the basin, we obtain

$$ {\tau}_t=\frac{1}{F_0}{\displaystyle \underset{ strait}{\int}\kappa \mathit{\nabla \tau}\cdot d\kern-.1em \mathbf{A}}+{\tau}_0, $$

(A3)

where ${\tau}_t=\frac{1}{F_0}{\displaystyle \underset{\mathrm{out}}{\int}\tau \mathbf{u}\cdot d\mathbf{A}}$ is the average transit time. The diffusive age flux through the strait, i.e., the first term on the right-hand side of Eq. (A3), is negative, since τ = 0 outside the domain, hence τ _t  < τ ₀. If we neglect this diffusive flux, we recover the result by Bolin and Rodhe (1973) that the mean transit time is equal to the turnover time, τ _t  = τ ₀.

We then multiply Eq. (A1) by τ and integrate. After a partial integration, we obtain

$$ {\displaystyle \underset{ out}{\int}\frac{\tau^2}{2}\mathbf{u}\cdot d\kern-.1em \mathbf{A}}={\displaystyle \underset{ out}{\int}\kappa \tau \mathit{\nabla \tau}\cdot d\kern-.1em \mathbf{A}}-{\displaystyle \int \kappa {\left|\nabla \tau \right|}^2 dV+{\displaystyle \int \tau dV}} $$

(A4)

The left-hand side can be rewritten as

$$ {\displaystyle \underset{ out}{\int}\frac{\tau^2}{2}\mathbf{u}\cdot d\mathbf{A}}=\frac{\tau_t^2}{2}{F}_0+{\displaystyle \underset{ out}{\int }{\tau}_t\Big(\tau -}{\tau}_t\left)\mathbf{u}\cdot d\mathbf{A}+\frac{1}{2}{\displaystyle \underset{ out}{\int}\Big(\tau -}{\tau}_t\right){}^2\mathbf{u}\cdot d\mathbf{A}. $$

(A5)

The second term on the right-hand side here vanishes identically. Thus, dividing Eq. (A4) by V ₀ and rearranging the terms, we obtain

$$ {\tau}_{\mathrm{mean}}=\frac{1}{2}\frac{{\tau_t}^2}{\tau_0}+\frac{1}{2{V}_0}{\displaystyle \underset{\mathrm{out}}{\int }{\left(\tau -{\tau}_t\right)}^2\mathbf{u}\cdot d\mathbf{A}}+\frac{1}{V_0}{\displaystyle \int \kappa {\left|\nabla \tau \right|}^2 dV-\frac{1}{V_0}{\displaystyle \underset{ out}{\int}\kappa \tau \mathit{\nabla \tau}\cdot d\kern-.1em \mathbf{A}}}. $$

(A6)

Again, neglecting the diffusive flux through the strait, we can neglect the last term in Eq. (A6) and use the relation τ _t  = τ ₀ from Eq. (A3), which gives

$$ {\tau}_{\mathrm{mean}}=\frac{1}{2}{\tau}_0+\frac{1}{2{V}_0}{\displaystyle \underset{ out}{\int }{\left(\tau -{\tau}_t\right)}^2\mathbf{u}\cdot d\kern-.1em \mathbf{A}}+\frac{1}{V_0}{\displaystyle \int \kappa {\left|\mathit{\nabla \tau}\right|}^2 dV} $$

(A7)

Hence,

$$ {\tau}_{\mathrm{mean}}\ge \frac{\tau_0}{2}. $$

Note that in order to derive this result, we neglected the diffusive flux through the strait, but not the diffusion and mixing in the interior of the basin, which gives rise to the last term in Eq. (A7).

If we set τ = 0 exactly at the boundary, ∇τ must be large at the outflow, and we cannot neglect the diffusive flux across the boundary. Near the boundary there is a thin boundary layer, with thickness of order κ/u, in which there is an advective–diffusive balance. The derivation above is valid if the age of the outflowing water is measured just inside this boundary layer.