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Home » Umrechnung in PS: \( 1 \text HP = \frac{550 \text ft·lb/s}1 \). - Dyverse

Umrechnung in PS: \( 1 \text HP = \frac{550 \text ft·lb/s}1 \). - Dyverse

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 where $ p > q > 0 $ are hydraulic conductivity parameters. Find the minimum value of $ R $.

📌 Solution: Let $ a = rac{p + q}{p - q} $ and $ b = rac{p - q}{p + q} $. Note that $ b = rac{1}{a} $, and since $ p > q > 0 $, $ a > 1 $. Then

📌 R = a + rac{1}{a} - 2.

📌 Define $ f(a) = a + rac{1}{a} - 2 $ for $ a > 1 $. By AM-GM, $ a + rac{1}{a} \geq 2 $, with equality at $ a = 1 $. But $ a > 1 $, so $ f(a) > 0 $. However, as $ a o 1^+ $, $ f(a) o 0^+ $. The minimum value is not attained, but we seek the infimum. Since $ f(a) $ increases for $ a > 1 $, the expression approaches 0 as $ p o q^+ $, but never reaches it. However, we analyze the original expression algebraically:

📌 R = rac{(p+q)^2 + (p-q)^2}{(p-q)(p+q)} - 2 = rac{p^2 + 2pq + q^2 + p^2 - 2pq + q^2}{p^2 - q^2} - 2 = rac{2p^2 + 2q^2}{p^2 - q^2} - 2.

📌 R = 2\left( rac{p^2 + q^2}{p^2 - q^2} - 1

📌 ight) = 2\left( rac{2q^2}{p^2 - q^2}

📌 ight) = rac{4q^2}{p^2 - q^2}.

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