Please explain conceptually how they can assume the pressure

26 Solutlons to Ideal-Incompressible Pipe Flow 2. But p-P-atmospheric (or P,-P2-0 psig): hence the first term drops 3. But the veloci ty at the surface of any reservoir or large tank is negligible compared to the velocity in a pipe; hence the second term above drops out. 4. Since local g-standard g, g/&-1 5. There remains Z)ideal- Zi+(Won) where the work term is obtained from the given horsepower as follows: work(ft-lbf) ft-lbf 1bm Power time(sec) -w.xm, × where and Horsepower-wonXH ) (Won),,2= hp × 550 (Z)ideal-20+ 704 -20+77.12+97.12 ft. Horsepower-von×WX(10 sechp 550 ft-lbf Therefore 6. Thus × 550 62.4 × 8 7. The pressure at A (in Figure 2.3) can be estimated by applying the ideal Bernoulli equation (2.1) between 1 and A as follows: p 28e p 2ge 0+0+20-PA+ 18 +10. p 2ge

Solution

The free surface of the water in any reservoir is exposed to the atmosphere. Hence, the pressure at the free surface will be same as the atmospheric pressure.

In the above situation, since both the reservoirs ( 1 & 2) are open to the atmosphere, their pressures will be same as the atmospheric pressure. This is the reason p1 is equal to p2.

Further, as per gage pressure, atmospheric pressure is considered as zero. Hence, p1=p2=0.psig