Question

As shown in the figure below, air flows in parallel to a freshly painted solid surface of width 10 m, along the z-direction. The equilibrium vapor concentration of the volatile component A in the paint, at the air-paint interface, is \( C_{A,i} \). The concentration \( C_A \) decreases linearly from this value to zero along the y-direction over a distance \( \delta \) of 0.1 m in the air phase. Over this distance, the average velocity of the air stream is 0.033 m s\(^{-1}\) and its velocity profile \( v_z(y) \) is given by \[ v_z(y) = 10 y^2 \] where \( y \) is in meter. Let \( C_{A,m} \) represent the flow averaged concentration. The ratio of \( C_{A,m} \) to \( C_{A,i} \) is \(\underline{\hspace{1cm}}\) (round off to 2 decimal places). 

Hint:

The flow-averaged concentration can be calculated by integrating the concentration profile weighted by the velocity.

Answer 1

Correct Answer: 0.24

The flow-averaged concentration \( C_{A,m} \) is given by the following equation:
\[ C_{A,m} = \frac{\int_0^\delta C_A v_z(y) dy}{\int_0^\delta v_z(y) dy} \]
Since \( C_A \) decreases linearly with \( y \), we can express it as:
\[ C_A = C_{A,i} \left( 1 - \frac{y}{\delta} \right) \]
Thus, the numerator of the flow-averaged concentration is:
\[ \int_0^\delta C_A v_z(y) dy = \int_0^\delta C_{A,i} \left( 1 - \frac{y}{\delta} \right) 10 y^2 dy \]
And the denominator is:
\[ \int_0^\delta v_z(y) dy = \int_0^\delta 10 y^2 dy = 10 \times \frac{\delta^3}{3} = \frac{10 \delta^3}{3} \]
Calculating the numerator:
\[ C_{A,i} \times \int_0^\delta \left( 1 - \frac{y}{\delta} \right) 10 y^2 dy = C_{A,i} \times \frac{10 \delta^3}{6} \]
Thus:
\[ C_{A,m} = \frac{\frac{10 C_{A,i} \delta^3}{6}}{\frac{10 \delta^3}{3}} = \frac{1}{2} C_{A,i} \]
Therefore, the ratio of \( C_{A,m} \) to \( C_{A,i} \) is:
\[ \boxed{0.5} \]