Answer:
The pump operates satisfactorily.
Step-by-step explanation:
According to the NPSH available definition:
![NPSHa = (P_(a) )/(density*g) + (V^(2) )/(2g) - (P_(v))/(density*g)](https://img.qammunity.org/2020/formulas/engineering/college/y9afzpafdutxxfz2ei1st3xq7d0vb373as.png)
Where:
![P_(a) absolute pressure at the inlet of the pump](https://img.qammunity.org/2020/formulas/engineering/college/jmbmpbhgoarpdepp62549nx9iz5beh84ev.png)
![V velocity at the inlet of te pump = 4m/s](https://img.qammunity.org/2020/formulas/engineering/college/n7jsnzvmga25oupv6cer4stx34mkdeoyfh.png)
![g gravity acceleration = 9,8m/s^(2)](https://img.qammunity.org/2020/formulas/engineering/college/sl19usd1foumke3ndlg1dkzs49rcry5qtg.png)
![P_(v) vapor pressure of the liquid, for water at 65°C = 25042 Pa](https://img.qammunity.org/2020/formulas/engineering/college/js40fb1y248362fkmpikbcvtlts387d2y7.png)
The absolute pressure is the barometric pressure Pb minus the losses: Suction Lift PLift and pipe friction loss Ploss:
To convert the losses in head to pressure:
![P = density*g*H](https://img.qammunity.org/2020/formulas/engineering/college/b92620dksdmla3xhr7z0y8b5fo69myqm0n.png)
So:
![P_(b) = 760 mmHg = 101325 Pa](https://img.qammunity.org/2020/formulas/engineering/college/54pf8r45qe3gitpgfn0mov1gia4t5aw580.png)
![P_(lift) = 33634,58 Pa](https://img.qammunity.org/2020/formulas/engineering/college/3o6oibrq929y55l8itpxk9svp464cz7e6r.png)
![P_(loss) = 8648,89 Pa](https://img.qammunity.org/2020/formulas/engineering/college/4m64yvo3r1xk17dyolp2jksbkst8ktqs0b.png)
The absolute pressure:
![P_(a) = P_(b) - P_(lift) - P_(loss) = 59044,53 Pa](https://img.qammunity.org/2020/formulas/engineering/college/nvn8mqix7wndeo5gq4b33vx1ojccmdzpf8.png)
replacing on the NPSH available equiation:
![NPSHa = 6,14 m + 0,816 m - 2,6 m = 4,356 m](https://img.qammunity.org/2020/formulas/engineering/college/67r8mg21z933udiuo3m8k7ff2gfmz5extb.png)
As the NPSH availiable is higher than de required the pump should operate satisfactorily.