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Question

A 2.4-in-diameter soap bubble is to be enlarged by blowing air into it. Taking the surface tension of soap solution to be 0.0027 lbf/ft, determine the work input required to inflate the bubble to a diameter of 2.7 in.

Solution

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Answered 2 years ago
Answered 2 years ago
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Calculate the starting surface of the bubble,

As=4π(ds2)2\begin{align*} A_s = 4 \pi (\frac{d_s}{2})^2\\ \end{align*}

Insert the starting values for the area of the bubble ( convert inches to feet ),

As=4π(2.4122)2As=0.1256 ft2\begin{align*} &A_s = 4 \pi (\frac{\frac{2.4}{12}}{2})^2\\ &A_s = 0.1256 \text{ ft}^2\\ \end{align*}

Now calculate the final area of the bubble ( convert inches to feet ),

Af=4π(2.7122)2Af=0.159 ft2\begin{align*} &A_f = 4 \pi (\frac{\frac{2.7}{12}}{2})^2\\ &A_f = 0.159 \text{ ft}^2\\ \end{align*}

The change in the area of the bubble,

ΔA=AfAsΔA=0.1590.1256ΔA=0.03344 ft2\begin{align*} &\Delta A = A_f - A_s\\ &\Delta A = 0.159 - 0.1256\\ &\Delta A = 0.03344 \text{ ft}^2\\ \end{align*}

The expression to calculate the work input required to inflate the bubble (The 2×2 \times multiplication indicates that the bubble is hollow and needs to do work for the outer and inner layer of the bubble),

W=2σsΔA\begin{align*} W = 2 \sigma_s \cdot \Delta A\\ \end{align*}

Insert the values to calculate the work,

W=20.00270.03344W=1.80592×104 ft lbf\begin{align*} &W = 2 \cdot 0.0027 \cdot 0.03344\\ &W = 1.80592 \times 10^{-4} \text{ ft lbf}\\ \end{align*}

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