Wastewater Treatment System Design Calculations

Treatment Design Calculations Group 1 Scheme Design 2014

Department of Civil, Construction & Mineral Engineering

Project

Course & Year

Wastewater Treatment Plant Design

Civil Engineering Year 5

Part of Structure

Calculation Sheet No.

Horizontal Flow Tank (Primary Settlement)

1

Drawing Ref:

Scheme Design Reference

of

Calculations By:

Checked By:

Date:

Maurice Healy

Jason Shields

22/03/2014

Calculations

2

Output

Minimum side wall depth = 2m Q = 59.4m3/day 60% reduction of TSS Peak flow factor = 2.5 Surface Overflow Rate = SOR = 185 - (2.5xTSS removal) = 185 - (2.5x60) = 35 m3/m2/day Step 1 - Calculate surface area (A) of tank required:

Surface Area=1.697m2 Step 2 – Calculate tank dimensions: Assume width/length ratio of 3:1 Hence: Length=2.256m

Width=0.752m

Step 3 - Calculate tank volume: Assume minimum tank depth of 2m.

Volume = V = 2mx2.256mx0.752m = 3.393m3

Volume=3.393m3

Department of Civil, Construction & Mineral Engineering

Project

Course & Year

Wastewater Treatment Plant Design

Civil Engineering Year 5

Part of Structure

Calculation Sheet No.

Horizontal Flow Tank (Primary Settlement)

2

Drawing Ref:

Scheme Design Reference

of

Calculations By:

Checked By:

Date:

Maurice Healy

Jason Shields

22/03/2014

Calculations

Step 4 – Hydraulic Retention Time (HRT)

Step 5 – Peak flow SOR

2

Output

HRT=1.37hours

Peak flow SOR=87.5m3/m2/day

Step 6 - Primary Sludge Production Weight of Dry solids = Wds = Q x TSS x F Assume F = 0.6 Q = 59.4m3/day Assume TSS = 300mg/l Wds = 59.4m3/d x 300mg/l x 0.6 x (103/106) Wds = 10.6922 kg/day Wds = 320 kg/day

Wds=320kg/day

Note: External Dimensions of tank are as follows: Length = 2.856m

Length=2.856m

Width = 1.352m

Width=1.352m

Depth = 2.1m

Depth=2.1m

Department of Civil, Construction & Mineral Engineering

Project

Course & Year

Wastewater Treatment Plant Design

Civil Engineering Year 5

Part of Structure

Calculation Sheet No.

Horizontal Flow Tank (Secondary Settlement)

1

Drawing Ref:

Scheme Design Reference

of

Calculations By:

Checked By:

Date:

Maurice Healy

Jason Shields

22/03/2014

Calculations

2

Output

Minimum side wall depth = 2m Q = 59.4m3/day 60% reduction of TSS Peak flow factor = 2.5 Surface Overflow Rate = SOR = 185 - (2.5xTSS removal) = 185 - (2.5x60) = 35 m3/m2/day

SOR=35m3/m2/day

Step 1 - Calculate surface area (A) of tank required:

Surface Area=1.697m2 Step 2 – Calculate tank dimensions: Assume width/length ratio of 3:1 Hence: Length=2.256m

Width=0.752m Step 3 - Calculate tank volume: Assume minimum tank depth of 2m.

Volume = V = 2mx2.256mx0.752m = 3.393m3

Volume=3.393m3

Department of Civil, Construction & Mineral Engineering

Project

Course & Year

Wastewater Treatment Plant Design

Civil Engineering Year 5

Part of Structure

Calculation Sheet No.

Horizontal Flow Tank (Secondary Settlement)

2

Drawing Ref:

Scheme Design Reference

of

Calculations By:

Checked By:

Date:

Maurice Healy

Jason Shields

22/03/2014

Calculations

2

Output

Step 4 – Hydraulic Retention Time (HRT) HRT=1.37 hours

Step 5 – Peak flow SOR

Peak flow SOR=87.5m3/m2/day

Step 6 - Primary Sludge Production Weight of Dry solids = Wds = Q x TSS x F Assume F = 0.6 Q = 59.4m3/day Assume TSS = 300mg/l Wds = 59.4m3/d x 300mg/l x 0.6 x (103/106) Wds = 10.6922 kg/day Wds = 320 kg/day

Wds=320kg/day

Note: External Dimensions of tank are as follows: Length = 2.856m

Length=2.856m

Width = 1.352m

Width=1.352m

Depth = 2.1m

Depth=2.1m

Department of Civil, Construction & Mineral Engineering

Project

Course & Year

Wastewater Treatment Plant Design

Civil Engineering Year 5

Part of Structure

Calculation Sheet No.

Extended Aeration (Secondary Treatment)

1

Drawing Ref:

Scheme Design Reference

of

Calculations By:

Checked By:

Date:

Maurice Healy

Jason Shields

22/03/2014

Calculations

3

Output

Population Equivalent = 220x1.5=330 Water use = 180l/day Influent Flow (Qo) = 330 p.e x 180l/day = 59.4m3/day Influent BOD (So) = 300mg/l Effluent BOD (S) = 20mg/l Assume Yield coefficient (Y) = 0.6 Assume decay rate (kd) = 0.06day-1 Assume MLSS (X) = 3500mg/l Assume WAS MLSS (Xw) = 14000mg/l Assume MCRT (ɸc) = 10day-1 Step 1 – Calculate the aeration tank volume (V):

V= 17.82m

3

Volume=17.82m3

Assume 2m depth.

Hence: Area = 17.82m3/2m2 = 8.91m2

Area=8.91m2

Width = 2.985m

Width=2.985m

Length = 2.985m

Length=2.985m

Department of Civil, Construction & Mineral Engineering

Project

Course & Year

Wastewater Treatment Plant Design

Civil Engineering Year 5

Part of Structure

Calculation Sheet No.

Extended Aeration (Secondary Treatment)

2

Drawing Ref:

Scheme Design Reference

of

Calculations By:

Checked By:

Date:

Maurice Healy

Jason Shields

22/03/2014

Calculations

3

Output

Step 2 – Calculate the Hydraulic Retention Time (HRT)

Step 3 – Volume of wasted sludge per day (Qw):

Qw=0.4455m3/day

Step 4 – Mass of sludge wasted each day (QwXw): QwXw = 0.4455 x 14 = 6.237kg/day

QwXw=6.237kg/day

Step 5 – Fraction of sludge recycled (Qr/Q0)

Qr/Qo=32.3%

Department of Civil, Construction & Mineral Engineering

Project

Course & Year

Wastewater Treatment Plant Design

Civil Engineering Year 5

Part of Structure

Calculation Sheet No.

Extended Aeration (Secondary Treatment)

3

Drawing Ref:

Scheme Design Reference

of

Calculations By:

Checked By:

Date:

Maurice Healy

Jason Shields

22/03/2014

Calculations

3

Output

Step 6 – Calculate f/m ratio f/m=0.29

Note: External Dimensions of tank are as follows: Length = 3.585m Width = 3.585m Depth = 2.1m

Length=3.585m Width=3.585m Depth=2.1m