Sump pit calculations, sump pump flow rate calculation, sump pump duty

In this article I talk about sump pit calculations, sump pump flow rate calculation and sump pump duty. Let's start

Assumptions: 

• Rainwater coming from ramps is totally removed by trenches 
• Sprinkler discharge is the only source of water

Fire Data:

Parking Type of Hazard       Ordinary group I

Sprinkler protection Area (ft2) 130

Area of sprinkler Operation (ft2) 1500

Density (gpm/ft2)                         0,15

Flow from Sprinklers (gpm)  225

Flow rate Analysis

• For a 90 min operation of the fire pump the total flow from sprinklers is 20250 Gal = 76545 liters
• Adopting a flow of 225 gpm or 14.19 l/s for a sump pump means that:

1. All sprinklers are activated within 1500 ft2

2. The sump pump will remove all the water discharged instantaneously in 90 min

• If the total amount of 20250 gal are to be removed in 3 hours the pump flow rate would be 7.1 l/s
• If the total amount of 20250 gal are to be removed in 4 hours the pump flow rate would be 5,3 l/s
• The number of sprinkler in the operating area will be 11.54 let's say 12
• Depending on the layout this number could be up to14
• For a total flow rate of 225 gpm, the individual sprinkler flow rate is 18.75 gpm
• Assuming 3 sprinklers are first activated the total discharge flow is 56.25 gpm or 3.55 l/s

Trench Data :

• Trench Average Height (m) = 0.15
• Length of trench serving one pit (m) = 45
• Volume of water in trenches (m3) = 6,75
• Pump flow rate to remove water in trenches in 15 min = 7.5 l/s
• Pump flow rate to remove water in trenches in 30 min = 3.75 l/s

Pump Flow Rate

Design flow rate of the sump pump = 4 l/s

Selecting Duplex operation of the the sump pumps, the flow rate of each one to be selected is 2l/s

Thus when water reach a high level in the pit, the second pump operate and the total flow rate would be 4 l/s

Minimum Required Sump Volume:

Vreq = Tmin x Q / 4

Where:

Vreq = is the required sump volume between the start level and stop level of the pump

Tmin = is the minimum pump cycle time

Q = is the pump flow rate

Tmin = 600 sec = 10 minutes (The total # of starts & stops / hour is 6)

Q = 2 ltr/sec

Vreq = 300 liters = 0.3 m3

The Total Volume for the sump with 2 pumps

Vtot = Vreq + (n-1) ΔH х S

Where:

Vtot = Total sump volume between the start level and stop level of the pump

n = total number of pumps

ΔH = The constant height between the start up of two pumps (in meters)

S = Surface area of the sump (in sq meters)

Now,

n = 2

ΔH = 0.2 m

S = 1 m2

Vtot = 0.5 m3

Thus the pit is designed to have a sump volume of 0.5 m3