Question

What is the total static head pressure loss in psi from a water meter to the highest fixture, which is 2' above the top floor, in a 4 story building with floor-to-floor height of 12 feet

Answer 1

Answer:

The value is $P = 21.6 \  psi$

Explanation:

From the question we are told that

   The height of the highest fixture above the top floor is   $h  =  2 ft$

   The floor to floor height is  $d =  12 \  feet$

Generally given that the building is a four story building then that height of the building is

      $k =  4 *  12$

=>   $k =  48 \ ft$

Generally the height of the highest fixture is  

       $H  =  h  + k$

=>     $H  =  2  + 48$

=>     $H  = 50 \ ft$    

Generally the pressure loss is mathematically represented as

       $P=  \rho  *  g *  H$

Here $\rho$ is the density of water with value $\rho =  1.940 \  slugs/ft^3$

   and  g is the acceleration due to gravity with value  

        $g = 32.174 \  ft/s^2$

=>    $P=  1.940   *  32.174 *  50$

=>    $P=  3121 \ lb/ft^2$    

converting to psi

        $P = \frac{3121}{144}$

=>      $P = \frac{3121}{144}$

=>      $P = 21.6 \  psi$