Circular Concrete Pipe Culvert Hydraulic Design Calculator

Culvert Hydraulic Design Calculator

MKS System | Circular Concrete Pipe | Inlet & Outlet Control Analysis

Based on FHWA HDS-5 & IRC:SP:13 Guidelines

Input Parameters

Stream Width

m

Design Flow (25-yr)

m³/s

Design Flow (50-yr)

m³/s

Culvert Length

m

Barrel Slope

m/m

Manning's n

Entrance Type Groove-end Projecting (kₑ=0.2) Square Edge Headwall (kₑ=0.5) Mitered (kₑ=0.7) Thin-edge Projecting (kₑ=0.9)

Allowable Headwater

m

Tailwater Depth (25-yr)

m

Tailwater Depth (50-yr)

m

Inlet Invert Level

m

Road Top Level (RTL)

m

Pipe Diameter

m

Wing Wall Angle

°

Culvert Design Calculator v1.0 | Based on Hydraulic Engineering Standards

Always verify results with professional engineering judgment and site-specific conditions

1. ⚙️ Input Parameters for Solved Example

Parameter	Symbol	Value	Unit	Description
Design Flow (50-yr)	$Q$	6.37	$$\text{m}^3/\text{s}$$	Critical design flow.
Pipe Diameter	$$D$$	1.40	$$\text{m}$$
Culvert Length	$$L$$	10.00	$$\text{m}$$
Barrel Slope	$$S$$	0.01	$$\text{m/m}$$
Manning's Roughness	$$n$$	0.012	-	Concrete pipe.
Entrance Loss Coeff.	$$k_e$$	0.50	-	Square Edge Headwall.
Tailwater Depth (50-yr)	$$TW$$	1.22	$$\text{m}$$	Water depth downstream.
Inlet Invert Level	$$\text{Invert}_{\text{In}}$$	100.00	$$\text{m}$$
Road Top Level (RTL)	$$\text{RTL}$$	103.40	$$\text{m}$$
Allowable Headwater	$$\text{HW}_{\text{Allow}}$$	3.05	$$\text{m}$$	Maximum acceptable $$\text{HW}$$.

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2. 🧮 Step-by-Step Calculation (50-Year Flow: $Q=6.37\ \text{m}^3/\text{s}$)

A. Preliminary Calculations

1. Full Pipe Area ($A$):

$$A = \frac{\pi D^2}{4} = \frac{\pi (1.40)^2}{4} = 1.54\ \text{m}^2$$

2. Full Pipe Velocity ($V$):

$$V = \frac{Q}{A} = \frac{6.37}{1.54} = 4.14\ \text{m/s}$$

3. Invert Drop ($\Delta Z$):

$$\Delta Z = S \times L = 0.01 \times 10.00 = 0.10\ \text{m}$$

Outlet Invert Level = $$100.00 - 0.10 = 99.90\ \text{m}$$

B. Inlet Control Analysis ($$\text{HW}_{\text{Inlet}}$$)

Inlet control headwater is determined by the control section at the entrance, acting as an orifice or weir. The calculation uses the submerged flow formula (FHWA HDS-5 simplified form) as the calculated Critical Depth is close to or greater than $D$.

1. Critical Depth ($$d_c$$): (Calculated iteratively)

$$d_c = 1.43\ \text{m}$$

2. Velocity Head ($$\frac{V^2}{2g}$$): ($$g = 9.81\ \text{m/s}^2$$)

$$\frac{V^2}{2g} = \frac{(4.14)^2}{2 \times 9.81} = 0.87\ \text{m}$$

3. Inlet Headwater ($$\text{HW}_{\text{Inlet}}$$):

$$\text{HW}_{\text{Inlet}} = d_c + (1 + k_e) \frac{V^2}{2g} = 1.43 + (1 + 0.50) \times 0.87 \approx \mathbf{2.80}\ \text{m}$$

C. Outlet Control Analysis ($$\text{HW}_{\text{Outlet}}$$)

Outlet control headwater is the sum of the outlet depth and all energy losses minus the invert drop.

$$ \text{HW}_{\text{Outlet}} = h_o + H_L - \Delta Z $$

1. Outlet Depth ($$h_o$$): The water depth at the outlet is $$\max(TW, D)$$.

$$h_o = \max(1.22\ \text{m}, 1.40\ \text{m}) = 1.40\ \text{m}$$

2. Entrance Loss ($$h_e$$):

$$h_e = k_e \frac{V^2}{2g} = 0.50 \times 0.87 = 0.44\ \text{m}$$

3. Friction Loss ($$h_f$$): ($$R = D/4 = 0.35\ \text{m}$$)

$$h_f = \frac{n^2 L V^2}{R^{4/3}} = \frac{(0.012)^2 \times 10.00 \times (4.14)^2}{(0.35)^{4/3}} = 0.09\ \text{m}$$

4. Exit Loss ($$h_{\text{exit}}$$):

$$h_{\text{exit}} = 1.0 \times \frac{V^2}{2g} = 1.0 \times 0.87 = 0.87\ \text{m}$$

5. Total Head Loss ($$H_L$$):

$$H_L = h_e + h_f + h_{\text{exit}} = 0.44 + 0.09 + 0.87 = 1.40\ \text{m}$$

6. Outlet Headwater ($$\text{HW}_{\text{Outlet}}$$):

$$\text{HW}_{\text{Outlet}} = 1.40 + 1.40 - 0.10 = \mathbf{2.70}\ \text{m}$$

D. Governing Headwater and Design Checks

1. Governing Headwater ($$\text{HW}_{\text{Gov}}$$):

$$\text{HW}_{\text{Gov}} = \max(\text{HW}_{\text{Inlet}}, \text{HW}_{\text{Outlet}}) = \max(2.80\ \text{m}, 2.70\ \text{m}) = \mathbf{2.80}\ \text{m}$$

Control Type: Inlet Control

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3. 📊 Solved Example Summary and Design Status

Design Check	Calculated Value	Limit/Target	Status
**Governing HW** ($$\text{HW}_{\text{Gov}}$$)	2.80 m	Allowable $$\text{HW}$$: 3.05 m	SAFE (2.80 m $$\le$$ 3.05 m)
Max HW Elevation	102.80 m	RTL: 103.40 m	-
**Freeboard** ($$\text{RTL} - \text{HW}_{\text{Elev}}$$)	0.60 m	Min Target: 0.30 m	SAFE (Sufficient Cushion)
**Outlet Velocity** ($$V_{\text{out}}$$)	5.17 m/s	Max Target: $$\approx$$ 5.0 m/s (Without protection)	HIGH (Requires Scour Protection)

💡 Design Recommendation

The culvert diameter of $$1.40\ \text{m}$$ provides adequate hydraulic capacity, as the resulting headwater is below the allowable level. However, the high exit velocity of $$\mathbf{5.17\ \text{m/s}}$$ necessitates the design and installation of a **rip-rap apron or other energy dissipator** at the culvert outlet to prevent downstream scour and erosion.