= Solution
At a fixed height, first determine the stored quantities per unit vertical height and per unit length along the line source. With $\Delta\rho=\rho-\rho_0$, the <triangular-profile line plume> has the useful profile integral
$$
I_n=\int_{-b}^bf(x/b)^n\,dx=\frac{2b}{n+1},\qquad n\ge0.
$$
The geometric <volume> is $\mathcal A=2b$; the <mass> is $\mathcal D=\rho_0I_0+\Delta\rho I_1=b(\rho+\rho_0)$. The upward <momentum> is $\mathcal P=W(\rho_0I_1+\Delta\rho I_2)$, while the buoyancy force is $\mathcal B=-g\Delta\rho I_1$. Thus
$$
\boxed{\mathcal A=2b,\quad \mathcal D=b(\rho+\rho_0),\quad
\mathcal P=\frac{bW}{3}(\rho_0+2\rho),\quad \mathcal B=gb(\rho_0-\rho).}
$$
Here positive <buoyancy> corresponds to a rising light plume. If instead buoyancy is defined kinematically, divide $\mathcal B$ by $\rho_0$; the density-weighted convention here is the one matching the printed equations.
Multiply each local density by the vertical <velocity> to obtain the fluxes through a horizontal section, still per unit source length:
$$
\boxed{\begin{aligned}
V&=\int\widehat w\,dx=bW,\\
Q&=\int\widehat\rho\,\widehat w\,dx=\frac{bW}{3}(\rho_0+2\rho),\\
M&=\int\widehat\rho\,\widehat w^2\,dx=\frac{bW^2}{6}(\rho_0+3\rho),\\
F&=\int g(\rho_0-\widehat\rho)\widehat w\,dx=\frac23gbW(\rho_0-\rho).
\end{aligned}}
$$
In particular, the stored <momentum> equals the <mass flux> $Q$, and the <buoyancy flux> is $F=g(\rho_0V-Q)$. The different triangular-profile factors in storage and flux must not be replaced by top-hat coefficients.
Back to article page
