Problem:

In a given month, a watershed with an area of 1500 m² received 100 cm of precipitation. During the same month, the loss due to evaporation was 15 cm. Ignore losses due to transpiration and infiltration due to ground water. What would be the average rate of flow measured in a gage at the outlet of the watershed in m³/day?

Solution:

We are given the following values:

\begin{align*}
\text{Area}, \ A & =1050 \ \text{acres} \\
\text{Precipitation}, \ P & = 100 \ \text{cm} \\
\text{Evaporation}, E & = 15 \ \text{cm}
\end{align*}

In this case, the inflow is $\text{Precipitation}, P$. Our formula now becomes

\sum \text{Inflow}- \sum \text{Outflow}=\ \Delta S
\\
P-E=O

Therefore, the outflow is

\begin{align*}
O & = 100 \ \text{cm} - 15\ \text{cm} \\
O & = 85 \ \text{cm}
\end{align*}

Since this is measured in 1 month, the outflow is measured as

O = 85\ \text{cm} / \text{month}

Now, convert to m and multiply the area of the watershed

\begin{align*}
O & = \left( 85\ \frac{\text{cm}}{\text{month}} \times \frac{1\ \text{month}}{30\ \text{days}} \times \frac{1\ \text{m}}{100\ \text{cm}} \right) \left( 1500\ \text{m}^2\right) \\
O & = 42.5 \ \text{m}^3/\text{day}
\end{align*}

Therefore, the average rate of flow measured in the gage at the outlet is \boxed{42.5\ \text{m}^3/\text{day}}.