Mitchell Community College - REVISED-6028-MCC Calc Book_Storm

ENGINEERING CALCULATIONS

FOR

MITCHELL COMMUNITY COLLEGE
HEALTH SCIENCES BUILDING

Prepared by:
LandDesign, Inc.
223 North Graham St.
Calculations by: GR
Checked by: MDM

Revised:

March 22, 2019

LandDesign PN: 1016028

Project Name:Mitchell Community CollegeProject Number:1016028Design Requirements:85% TSS RemovalGiven Information:Land Use: InstitutionalDrainage Area to Facility (Acres): 2.00 NCDENR (Maximum 5 Acres)Meck. County (Maximum 10 Ac for a Surface Filter)Meck. County (Maximum 2 Ac for a Perimeter Filter)Meck. County (Maximum 5 Ac for an Underground Filter)Impervious Area (Acres): 0.74Percent Impervious: 37.0%Soil Type: CgCSoil Group: BCurve Number (CN) Pre: 87Curve Number (CN) Post: 75Time of Concentration Pre (hour): 0.230Time of Concentration Post (hour): 0.083Design Storm Rainfall Depth (in): 1.0 (Typically 1")Detention Time (Day): 1.66 Meck. County:2 days - 85% TSS & 70% TP (Optimal Efficiency)1 day - 70% TSS & 35% TP (Standard Efficiency)2 days - 85% TSS & 0% TP (TSS - Only Efficiency)NCDENR1.66 daysCoefficient of Permeability for the Sand Filter Bed: 1.75 inches/hourWatershed:Yadkin-Southeast Catawba - 85% TSS & 70% TPEfficiency:NCDENR
Step 1: Compute the water quality volume (WQV).
WQV (Ft3) = (RV / 1) X (AD / 1) X (43,560 Ft2 / 1 Acre) X (RD / 1) X (1 Ft / 12 in)
WQVAdj (Ft3) = (0.75) WQV
Where:
WQV = Water Quality Volume (Ft 3)
WQVAdj = Adjusted Water Quality Volume (Ft3)
AD = Drainage Area to Sand Filter (acres). Max. 5 acres.
RV = Volumetric Runoff Coefficient = 0.05 + 0.9 (% Impervious)
RD = Design Storm Rainfall Depth (in). Typically 1.0".
Percent Impervious = 37%
AD =2.00 acres
RD = 1.0 in
RV = 0.38
WQV = 2,781 Ft3
WQVAdj =2,086 Ft3
Sand Filter Design (NCDENR - 2007)
Project Name: Mitchell Community College
Job Number: 1016028
Step 2: Determine the maximum head on the sand filter and sedimentation basin.
ha (ft) = hMaxFilter / 2
hMaxFilter (Ft) = (WQV Adj) / (AS + Af)
AS (Ft2) = -(Q0 / w) x (ln (1 - E))
Af (Ft2) = (WQV x dF) / (k x t x (hA + dF))
Q (Ft3/Sec) = k x (1 Day / 86,400 Sec) x (Af) x ((ha + dF) / dF)
Where:
ha = Average Head (Ft)
hMaxFilter = Maximum Head on the Sand Filter (Ft)
AS = Surface Area of the Sedimentation Basin (Ft2)
Af = Surface Area of the Sand Filter Bed (Ft2)
Q0 = Average Rate of Outflow From Sedimentation Chamber (Ft3 / sec.) = WQV / 86,400
E = Trap Efficiency of the Chamber = 0.9
w = Settling Velocity of Particle (Ft/sec) (Assume Particle Diameter of 20 microns = 0.0004 Ft/sec)
dF = Depth of Sand Filter Bed (Ft) (Minimum of 1.5 Ft)
k = Coefficient of Permeability for the Sand Filter Bed = 3.5 Ft/day
t = Time Required to Drain the WQV through the Sand Filter Bed (day).
Q = Sand Filter Infiltration Rate at hMaxFilter (Ft3/Sec). (Darcy's Law)
For maximum heads between 2 and 6 feet, the following combinations of variables will work:
(Ideal situation AS = Af)
hMaxFilter WQVAdj AS + Af
(Ft)(Ft3) (Ft
2)
2 2,086 1,043
3 2,086 695
4 2,086 522
5 2,086 417
6 2,086 348
Area available for sand filter = 695 Ft2
hMaxFilter =3 Ft
ha = 1.5 Ft
Minimum AS =185 Ft2
dF =2.0 Ft
k = 3.5 Ft/day
t = 1.66 day
Minimum Af =274 Ft2
Min. required area for sand filter = 459 Ft2 TRUE
Af =348 Ft2
AS =348 Ft2
hMaxFilter =3.0 Ft
ha = 1.5 Ft
Q = 0.025 Ft3/Sec
Step 3: Ensure that the Water Quality Volume is Contained.
(Af + AS) x hMaxFilter ≥WQVAdj
2,088 Ft3 ≥2,086 Ft3 TRUE
Step 4: Additional Design Requirements.
For underground sand filters, provide at least 5 feet of clearance between the surface of
the sand filter and the bottom of the roof of the underground structure to facilitate
cleaning and maintenance.
Step 5: Calculate Surface Area and Underdrain Requirements
Q (Ft3/Sec) = k x (1 Day / 86,400 Sec) x (Af) x ((ha + dF) / dF)
Where:
ha = Average Head (Ft)
hMaxFilter = Maximum Head on the Sand Filter (Ft) (Maximum of 2.0 Ft)
Af = Surface Area of the Sand Filter Bed (Ft2)
AS = Surface Area of the Sedimentation Basin (Ft2)
Cd = Sediment Chamber Depth (Ft)
dF = Depth of Sand Filter Bed (Ft) (Minimum of 2.0 Ft)
k = Coefficient of Permeability for the Sand Filter Bed = 3.5 Ft/day
QMaxFilter = Sand Filter Infiltration Rate at hMaxFilter (Ft3/Sec). (Darcy's Law)
Q0 = Sand Filter Infiltration Rate at Top of Sand (Ft3/Sec). (Darcy's Law)
Required Af =274 Ft2
Provided Af = 373 Ft2
Required As =185 Ft2
Provided As = 2400 Ft2
Required Cd =0.3 Ft
Provided Cd =3 Ft
QMaxFilter =0.026 Ft3/Sec
Q0 =0.000 Ft3/Sec
Number of Underdrain Rows Required = 1 (Maximum Spacing = 10 Ft)
Total Length of Underdrain = 47 Ft
Number of Underdrain Perforations = 368 Holes
50% of Perforations = 184 Holes
Area of One Perforation = 0.00077 Ft2
Capacity of One Hole = 0.0058 Ft3/Sec
Total Capacity of Perforations = 1.07 Ft3/Sec
Check:
1.07 Ft3/Sec > 0.026 Ft3/Sec :. GOOD
Capacity of 6" PVC Underdrain at 0.50% Slope = 0.39 Ft3/Sec
Assume 50% Clogging = 0.20 Ft3/Sec
Check:
0.20 Ft3/Sec > 0.026 Ft3/Sec :. GOOD