Lösung zu Widerstand einer Heizwasserleitung

1. Turbulenz

[math]\displaystyle{ I_{Vkrit} = \frac {R_V}{k} }[/math], [math]\displaystyle{ R_V = \frac {128 \eta l}{\pi d^4} = 7.53* 10^6 Pa/(m^3/s) }[/math], [math]\displaystyle{ k = \lambda \frac {8 \rho l}{\pi^2d^5}= 4.37* 10^{11} Pa/(m^3/s)^2 }[/math]

[math]\displaystyle{ I_{Vkrit} = 1.72* 10^{-5} m^3/s=0.0172 l/s }[/math]