Question

) A pump can lift 160 kg of water per minute to a tank. If the potential energy of water is 21.6 kJ. what will be the height of water tank.( multiple can be accepted)​

Answer 1

Explanation:

If the potential energy of the water stored is 21.6 kJ, compute the height of the water tank. Take g= 10 m/s.

Answer 2

Answer:

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Explanation:

In any pumping system, the role of the pump is to

provide sufficient  pressure to overcome the

operating pressure of the system to move fluid at  

a required flow  rate.  The operating pressure of  

the system is  a function of the flow through the

system and the arrangement  of the system in  

terms  of the pipe length,  fittings,  pipe size,  the

change in liquid elevation, pressure on the liquid

surface, etc. To achieve a required flow through a

pumping system,  we need to calculate what  the

operating pressure of the system will be to select  

a suitable pump.

Figure 1: Typical Vertical Turbine Water Pumps  

MATHEMATICAL MODEL AND CALCULATIONS

Consider the pumping arrangement  shown in  

Figure 2 below:

Figure 2: Pumping Arrangement

Water is  pumped from the reservoir into a

receiving tank. This  kind of arrangement  is used

to lift water from a reservoir, or river, into a water

treatment  works  for treatment  before the water

goes into the supply network. The water level in  

the reservoir varies but the discharge level in the

receiving tanks remains  constant as the water is  

discharged from a point  above the water level.  

The pump is  required to pass  forward a flow  of  

2500 m 3 /hr to the receiving tank.  

The operating pressure of a pumped system is  

calculated in  the SI unit  of meters  (m).  To

maintain dimensional  consistency, any  pressure

values used within  the calculations are therefore

converted from kPa into m using the following

conversion;  

1 kPa =  0.102 m  

(as measured by a water filed U tube

manometer)

For the above system, the operating pressure or

the total system head, HTotal , is defined as:

( ) HTotal = H s + H D  + PRT  - PRES  … (1)

where, Hs =  Static head (m)

HD  =  Dynamic head (m)

PRT  =  

Pressure on the surface of the water in

the receiving tank (m)

PRES  =  

Pressure on the surface of the water in

the reservoir (m)

Although the atmospheric pressure changes with

height, the change in  pressure that  occurs  over

the pumping height is often so small that it can be

considered negligible.  In this  exemplar,  the

change in  pressure over the elevation from the

reservoir to the receiving tank  is  not  that  

significant  and hence is  negligible, i.e.,  PRT  - PRES  ª 0 .  Therefore, equation (1) becomes:

HTotal = H s + H D  … (2)