Question

During a big fire in the Unita Mountains, firefighters decide to use a large lake as a source of water (20°C, density of 998.2 kg/m^3). The firefighters place a 3-inch diameter hose at the bottom of a 10-foot deep lake and pump water to a cabin that is on fire (the entire hose is 3 inches in diameter). The outlet of the hose is 30 feet higher than the lake surface. The work performed by the pump is 200 J per kg of water that is pumped. Neglecting frictional losses, what is the volumetric flow rate of water exiting the hose (in gallons per minute)? If the pump uses 16kW of power during operation what is its efficiency?

Answer 1

diameter=3 inches = 0.0762 m

height = 9.144 m

depth = 10 foot = 3.048 m

now balancing energy,

200*1000 = 0.5*1000*v^2 + 1000*9.81*(9.144+3.048)

or v=12.68 m/s

so volume flow rate = A*v

=pi*(0.0762/2)^2*12.68

=0.0578 m^3/s

power output = 200*0.0578

=11.56 kW

so efficiency = 11.56*100/16

=72.25 %