To solve this problem we will first find the useful power of the pump with the defined values. From finding this value we will find the potential mechanical energy on the pump. Finally, by conservation of energy we will obtain the mechanical power used to overcome the effects of friction. According to the information given our data are
[tex]\dot{W_{pump}} = 20hp[/tex]
[tex]\eta = 80\%[/tex]
[tex]Q = 1.5ft^3/s[/tex]
[tex]z = 80ft[/tex]
Specific weight of water
[tex]\gamma = 62.4lbm/ft^3[/tex]
Te useful mechanical pumping power delivered to water is
[tex]\dot{W_{u}}= \eta \dot{W_{pump}}[/tex]
[tex]\dot{W_{u}}= (0.80)(20)[/tex]
[tex]\dot{W_{u}}= 16hp[/tex]
The elevation of water and thus its potential energy changes during pumping, but it experiences no changes in its velocity and pressure.
Therefore, the change in the total mechanical energy of water is equal to the change in its potential energy
[tex]\Delta \dot{E_{mech}} = \dot{m} \Delta pe[/tex]
[tex]\Delta \dot{E_{mech}} = \dot{m} g \Delta z[/tex]
[tex]\Delta \dot{E_{mech}} = \rho Q g \Delta z[/tex]
Replacing
[tex]\Delta \dot{E_{mech}} = (62.4lbm/ft^3)(1.5ft^3/s)(32.2ft/s^2)(80ft)[\frac{1lbf}{32.2lbm\cdot ft/s^2}][\frac{1hp}{550lbf \cdot ft/s}][/tex]
[tex]\Delta \dot{E_{mech}} = 13.614hp[/tex]
Then the mechanical power lost in piping because of frictional effects becomes
[tex]\dot{W_{frict}} = \dot{W_{u}}-\Delta \dot{E_{mech}}[/tex]
[tex]\dot{W_{frict}} = 16-13.614[/tex]
[tex]\dot{W_{frict}} = 2.386hp \approx 2.37hp[/tex]
Therefore the mechanical power used to overcome frictional effects in piping is 2.37hp
