Question

how do you convert an absolute pressure of 5 bar to a pressure head (m water) assuming that the density of water is 1000kg/m-3

Answer 1

$$1bar = 10^{5} Pa$$ $$P = \rho * g * h (Pa) = 10^{-5} * \rho * g * h (bar)$$ $$h = 10^{5} \frac{P}{\rho * g}$$ $$h(5bar) = 10^{5} \frac{5}{1000 * 9.8} = 51.02 \, m$$

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