Question

What gauge pressure must a pump produce in order to pump water from the bottom of

Grand Canyon (730 m elevation) in Pascals and in atmosphere?

Answer 1

By the definition of the gauge pressure we have:

$$P_{gauge}=P_{abs}-P_{atm}=\rho_{water}g\Delta h,$$ $$P_{gauge}=\rho_{water}g(h_2-h_1).$$

From the definition of the question, we know $h_1=730\ m$. But, we don't know $h_2$. Let's suppose that $h_2=1200\ m$. Then, we get:

$$P_{gauge}=1000\ \dfrac{kg}{m^3}\cdot9.8\ \dfrac{m}{s^2}\cdot(1200\ m-730\ m)=4.61\cdot10^6\ Pa,$$ $$P_{gauge}=4.61\cdot10^6\ Pa\cdot\dfrac{1\ atm}{101325\ Pa}=45.5\ atm.$$