Question

A 2.2 kW pump operating at steady state draws in liquid water at 100 kPa, 20 C and delivers it at 500 kPa at an elevation 7 meters above the inlet. There is no significant change in velocity between the inlet and exit, and the local acceleration of gravity is 9.81 m/s2 . Would it be possible to pump 4000 L in 10 minutes or less? Explain using considerations of generalized reversible flow processes discussed in class

Answer 1

Maximum power required for the pump to do the job is higher than maximum power the pump is operating. In conclusion, it is not possible for the pump to pump 4000L in 10 minutes.

Step-by-step explanation:

Reversible processes is differentially removed from equilibrium with no (appreciable) internal temperature, pressure, and velocity changes.

•A reversible process can be reversed at any point by external conditions such as temperature

•Time-dependent reversible processes are invariant with regard to the algebraic sign of the time.

•Reversible processes traverse a succession of equilibrium states.

•A reversible process represents a limit to the performance of actual processes.

•For a given change of state the surroundings extract maximum possible work from the system if the process is carried out reversibly.

•If the surroundings are doing work on the system, for a given change of state, a minimum amount of work is needed for a reversible process.

P1= 100kpa

P2= 500kpa

Acceleration due to gravity is 9.81m/s^2

Temperature = 20 degree celsius

Elevation= 7meters

Wpump= 2.2kW

Possibility to pump 4000 L in 10 minutes is:

m=4000/10= 400l/min

Which is equal to 6.8 in kilogram per second.

So,

Wmaximum= M1V1(P2-P1) + mg × change in f

= 6.8 × 0.001 (500000-100000) + 6.8×9.81×7

W= 3186. 95 watt

Therefore, we can conclude that maximum power required for the pump to do the job is higher than maximum power the pump is operating. In conclusion, it is not possible for the pump to pump 4000L in 10 minutes.