Answer:
pump head is 10.15 m
Step-by-step explanation:
given data
diameter suction = 400 mm
diameter discharge = 350 mm
discharge = 20000 l/m
to find out
pump head in meters
solution
we know discharge is 20000 l/m
so discharge =
m³/min
and
we know suction head hs is express as
hs = section gauge × specif gravity of Hg
hs = 0.127 × 13.6
hs = 1.7272 m of water
and
delivery head hd is express as
hd =
![(delivery gauge)/(density of water*g)](https://img.qammunity.org/2020/formulas/engineering/college/oa3gt3ga3jxthy1tixwk0i49hc2c766i9p.png)
hd =
![(75*10^3)/(1000*9.81)](https://img.qammunity.org/2020/formulas/engineering/college/ktx3qlpi7iqre2iyvryz6c42q16ykatfsf.png)
hd = 7.6453 m of water
and
we know distance between the gauge is here
Distance D = 0.075 + 0.45 = 0.525 m
so
discharge velocity vd is express as
Vd =
![(discharge)/(area)](https://img.qammunity.org/2020/formulas/engineering/college/rtdv17qz0jj6h34hz262k53dellmmlqe0p.png)
Vd =
![(1)/(3*(\pi )/(4) * 0.35^2)](https://img.qammunity.org/2020/formulas/engineering/college/ejjv22n5u23a8pf0nyioqwivsb9m33a5ez.png)
Vd = 3.465 m/s
and
suction velocity Vs is express as
Vs =
![(discharge)/(area)](https://img.qammunity.org/2020/formulas/engineering/college/rtdv17qz0jj6h34hz262k53dellmmlqe0p.png)
Vs =
![(1)/(3*(\pi )/(4) * 0.4^2)](https://img.qammunity.org/2020/formulas/engineering/college/2tzsyszymgf0ds9jop0pkb8xxnnp5huuaq.png)
Vs = 2.653 m/s
so
pump head will be here
pump head = hs + hd + D +
![(Vd^2 - Vs^2)/(2g)](https://img.qammunity.org/2020/formulas/engineering/college/nd2lv5dxbons554enuslxp18w3s3jf9oqq.png)
pump head = 1.7272 + 7.6453 + 0.525 +
![(3.465^2 - 2.653^2)/(2*9.81)](https://img.qammunity.org/2020/formulas/engineering/college/q5lwvrglj6qp53grh26jnoihbhfiyybkyg.png)
so pump head = 10.15 m