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# Lab report on Pumping and Pumps

### Paper's Detail

 Category Engineering Paper Type Report Writing Reference Type APA Words 800

Introduction of Pumping and Pumps

PUMP-FLO connect is used as a web-based product in the PUMP-FLO family of products. This product is specially designed for the assistance of manufacturers in the process of pump selection. The connect provides complete process for the companies to list the digitized catalogs related to the performance curves for the hosted cloud environment that is convenient, affordable, and secure. The selection of PUMP-FLO is based on utilization protocol and gives access to manufacturers for the digitized and maintained pumping process. The present report estimates the requirements of the various system with the consideration of pump affinity and use of online resource.

Equipment of Pumping and Pumps

Objectives of Pumping and Pumps

Calculate pumping requirements for various systems.

Use pump affinity laws to adjust pump performance.

Select an appropriate pump for a given system using an online resource.

The report considers three systems that are categorized for various pumping process, affinity laws of pump to adjust the performance and pump selection through online resources.

System 1 of Pumping and Pumps:

new head =  30 psi  × 2.31  ( ft)/( psi)   =  69.3  ft

water power =  WHP  =   ( Q (( gal)/(〖 min〗⁡   ))  ×  head )/( 3960 gpm ( ft)/( HP))

water power  =   WHP =   (5 (( gal)/(〖 min〗⁡   ))  ×  69.3    )/(3960  gpm ( ft)/( HP))   =  ( 346.5)/( 3960)   =   0.087 hp

efficiency of pump =  E_( P)   =   50 %

brake power  =  BHP =   ?

BHP =   ( WHP)/(〖 E〗_( P)  )    =   ( 0.087)/( 0.50)   =  0.175 hp

motor efficiency  =  〖E 〗_m   =   60 %  = 0.60

input power requirement of motor =  ?

input power requirement of motor =   ( BHP)/(  E_( m) )    =     ( 0.175)/( 0.60)

=   0.291  hp

input power in kW=  ( 0.291  hp)/( 0.746 ( kW)/( hp))    =  0.39 k W

electricity cost  = \$ 0.15 ( k W)/( hr)

cost to run the motor for 24 hours =  ?

cost  =   0.39  ×  \$ 0.15 kW-hr  × 24 h   = \$  1.404

System 2 of Pumping and Pumps:

center pivot=   600  ( gal)/〖 min〗⁡

pivot point pressure  =   20  psi

water surface below surrounding land = 10  feet

center pivot height = 10  feet

location of reservoir  = 1700 feet from the pivot

quality=  PVC class 160 IPS pipe

head loss =  ( 10 ft)/(  10 ft)  =  1

Hazen Williams formulas  =   〖h 〗_l   =  K_( p)  〖 Q〗^( 1.85)   =  〖600〗^( 1.85)   =  137904.9

minimum pipe size =   1 feet

total dynamic head =  TDH for system  =  dynamic suction head + dynamic discharge

=20 psi (2.31 ( ft)/( psi)  )   =  46.2

water power  =    ?

water power  =  WHP =   ( Q(( gal)/( min⁡   ))  ×   head )/( 3960 gpm ( ft)/( HP))

water power =   WHP  =   (600 (( gal)/(〖 min〗⁡   ))  ×  46.2    )/(3960  gpm ( ft)/( HP))   =  ( 27720)/( 3960)   =  7 hp

efficiency of pump =  〖E 〗_P   =   80 %

brake power =  BHP =  ?

BHP =   ( WHP)/(〖 E〗_( P)  )   =   ( 7)/( 0.80)    =  8.75 hp

motor efficiency  =  〖E 〗_m   =   80 %  = 0.80

input power requirement of motor =  ?

input power requirement of motor =   ( BHP)/E_( m)    =     ( 8.75)/( 0.80)

=   10.9 hp

input power in kW =  ( 10.9 h p)/( 0.746 ( kW)/( hp))    =   14.66 k W

electricity cost  = \$ 0.15  ( k W)/( hr)

cost to run the motor for 24 hours =  ?

cost  =   14.66  ×  \$  0.15 k W-hr × 24  h  =\$ 52.7

System 3 of Pumping and Pumps:

Closed pump

Open pump

The diameter available for the pump is ranging from 6.6 in to 7.14 in.

System 4 of Pumping and Pumps:

initial flow rate = Q_( 1)  =   200  ( gal)/( min)

final flow rate   = Q_( 2)   =   250  ( gal)/( min)

initial operating speed = ω_( 1)   =   1800 rpm

final operating speed   =〖 ω〗_( 2)    =    ?

Q_( 1)/〖Q 〗_2    =   ω_( 1)/ω_( 2)

ω_( 2)  =  ω_( 1)  (( 〖Q 〗_2)/Q_( 1) )   =   1800 ( ( 250)/( 300)  )   =   1500 rpm

Total head loss of Pumping and Pumps

height of the water  =   H_( 2)

√(H_( 2)/H_( 1) )  =  N_( 2)/N_( 1)

H_( 2)   =   (N_( 2)/N_( 1) )  ×  H_( 1)

=  1500/1800   × 100 ft =  83.33 ft

Power developed of Pumping and Pumps

P =  ρ Q g H

ρ  =  1000 kg/m^2

P =  ((( 1000 ( kg)/( m^( 2) ))   ( 250 (  gal)/(  min))  ( 9.8 )( 83.33)  ))/( 746 )   =  23 H

Conclusion of Pumping and Pumps

The present report considers different pump system and conducted analysis. It is concluded that by varying any parameter the value and specifications of pump changes. All the parameters are determining factors that measure the outcomes.

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