Loading...

Messages

Proposals

Guarantee Your Academic Success!

50% Off On All Assignments For New Clients. Use Coupon Code "DISCOUNT21"

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

Web access to pump curve and pump selection resource, pump-flo.com.

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.  


Get Rewriting & Paraphrasing Help!

We have more than 1500 academic writers and we promise 0% plagiarism in your paper.

Our Top Online Writers.

Discuss your homework for free! Start chat

Top Grade Tutor

ONLINE

Top Grade Tutor

Singapore

I have more than 10 years of experience in ACADEMIC WRITING and STUDIES. I can …

Top Academic Guru

ONLINE

Top Academic Guru

Canada

I have more than 12 years of writing experience as I started my online writing …

A-Grade Writer

ONLINE

A-Grade Writer

Kenya

I am a full-time writer who has great command of Academic and Professional Writing because …

Order Your Homework Today!

We have over 1500 academic writers ready and waiting to help you achieve academic success

Private and Confidential

Yours all information is private and confidential; it is not shared with any other party. So, no one will know that you have taken help for your Academic paper from us.