Question

Q3.1 (a) a pump forces 1 m3/min of water horizontally from an open well to a closed tank where the pressure is 0.9 mpa. Compute the work the pump must do upon the water in an hour just to force the water into the tank against the pressure. Sketch the system upon which the work is done before and after the process. (ans. 5400 kj/h) (b)if the work done as above upon the water had been used solely to raise the same amount of water vertically against gravity without change of pressure, how many meters would the water have been elevated

Answer 1

Answer:

Explanation:

(a)

Flow rate 1m3/hr.

Pressure of inlet water = 1 atm = 0.101325 MPa

Pressure of outlet water = 0.9 MPa

Power =Δ pv

(0.9- 0.101325 )$10^{3}$ kPa ×1 m^3/ 60 s

13.31 kJs

Answer 2

The work that the pump must do upon the water is 4800 KJ/ hr.

Given,

The pump forces 1 m³/ min of water horizontally.

∴ Flow rate of the water is = 1 m³/min.

Here, the pressure of water is = 0.9 MPa

∴ Pressure of outlet water (P$_o$) = 0.9 MPa

Pressure of inlet water (P$_i$) = 1 atm = 0.101325 MPa

We know,

The power of such an arrangement is calculated by the formula,

Power =ΔPV

            = (0.9- 0.101325 ) × 10³ Pa ×1 m³/ 60 s

            = 13.31 KJ/s

∴ The work done in 1 s is 13.31 KJ

⇒ The work done in 1hr = 3600 s will be = (13.31 × 3600) KJ

                                                                   = 47916 KJ

So, the work that the pump must do upon the water in an hour just to force the water into the tank against the pressure is ≈ 48000 KJ.