Question #eae28

1 Answer
A08
Aug 27, 2017

I get this.

Explanation:

The problem consists of sum of two pressures which the pump must create

  1. Pressure due to height of #40 m# of water column
  2. Pressure due to jet of water at a rate of #5ms^-1#

For 1. Assuming unit area, Force#=#Pressure

Pressure due #40m# water column #P_1=mg=rhoxx"volume"xxg#
#P_1=1000xx(40xx1)xx9.81=392400pa#

For 2. Assumed area of the jet be unit area.
Speed of water outlet from jet #=5 ms^-1#.
Total mass of water thrown out per sec#= rhoxx1xx5=5000kg#

Pump must exert pressure to support weight of this additional column of water of unit area#=5000xxg=49050N#

For unit area of jet pressure #P_2=49050pa#

Total pressure of the pump#P=P_1+P_2=392400+49050=441450pa#

Related questions
See all questions in Fluid Flow
Impact of this question
2074 views around the world
You can reuse this answer
Creative Commons License