Question

Operating labor requirements L in the chemical process industry is described in terms of the plant capacity C (kg day\(^{-1}\)) over a wide range (10\(^3\) − 10\(^6\)) by a power law relationship: \[ L = \alpha C^\beta \] where \(\alpha\) and \(\beta\) are constants. It is known that \[ L = 60 \text{ when } C = 2 \times 10^4 \text{and} L = 70 \text{ when } C = 6 \times 10^4 \] The value of L when C = \(10^5\) kg day\(^{-1}\) is \(\underline{\hspace{1cm}}\) (round off to nearest integer).

Hint:

For power law relationships, use the known values to solve for \(\alpha\) and \(\beta\) and then calculate the desired value.

Answer 1

Correct Answer: 73

Using the power law relationship: \[ L = \alpha C^\beta \] We have two equations: 1) \(60 = \alpha (2 \times 10^4)^\beta\) 2) \(70 = \alpha (6 \times 10^4)^\beta\) By dividing these two equations: \[ \frac{70}{60} = \left(\frac{6 \times 10^4}{2 \times 10^4}\right)^\beta \] \[ \frac{7}{6} = 3^\beta \] Taking the natural log of both sides: \[ \ln \left(\frac{7}{6}\right) = \beta \ln 3 \] Solving for \(\beta\): \[ \beta = \frac{\ln \left(\frac{7}{6}\right)}{\ln 3} \approx 0.077 \] Now substitute \(\beta\) into either equation to solve for \(\alpha\): \[ 60 = \alpha (2 \times 10^4)^{0.077} \] \[ \alpha = \frac{60}{(2 \times 10^4)^{0.077}} \approx 56.8 \] Now, calculate \(L\) when \(C = 10^5\): \[ L = 56.8 (10^5)^{0.077} \approx 77 \] Final answer: 73 to 77.