Intro

The worked examples in the section below provide hand calculations of the Engelund manhole loss approach applied to three simple examples covering:

Example 1: Single Incoming/Outgoing Pipe with No Incoming Angle or Drop

A simple TUFLOW/ESTRY model was set up as shown in Figure 1 with the parameters for the manhole and the links upstream/downstream of the manhole shown in Table 1.

Figure 1: Example Model 1 Network Schematic

Figure 2: Example Model 1 Long Section

Parameter	Upstream Conduit	Downstream Conduit	Manhole
ID	UC-2	DC-1	Manhole
Type	R	R	R
Invert Level (m AD)	N/A	N/A	99.6
Upstream Invert Level (m AD)	99.8	99.7	N/A
Downstream Invert Level (m AD)	99.7	99.6	N/A
Width(m)	1	N/A	1
Form Loss	0	1	N/A
Flow Width(m)	N/A	N/A	1.8
K_Fixed	N/A	N/A	2
Km	N/A	N/A	1

Table 1: Relevant Model Parameters for Hand Calculations

The model was run with a steady inflow of 2m³s^-1. The resulting losses from the 1 hour steady inflow simulation are presented in the following figure.

Figure 3: Example Model 1 Results-Loss Coefficients

Example 1 - Entry Losses

Entry losses are applied as an exit loss on the incoming conduit and are calculated as follows.
Where:
With other parameters defined in the section above. The calculated flow area in the manhole is 3.6m² and 1m² in the adjacent culvert whilst the flow is 2m³/s for both Q_p and Q_om. Therefore, V_m equals:
And V_p:
Using these in the K_entry equation provides a loss coefficient of:
The value matches the downstream loss coefficient for the upstream conduit shown in Figure 3. The loss values are also shown in the *_TSL_P.shp file with the third value for the upstream conduit providing the entrance loss value.

Example 1 - Exit Losses

Exit losses are applied to the upstream end of the outgoing conduit and are calculated as follows:
From table 1, we can see that K_m is set to 1. A_p and A’_m are 1m² and 3.636m² respectively. Therefore, K_exit is:

Figure 3 shows that the calculated value matches those provided in the TUFLOW results. The value is also reported as the first value for the downstream conduit in the *_TSL_P.shp layer.

Example 1 - Outgoing Conduit Losses

The loss coefficient for the outgoing pipe represents the losses due to the incoming angle of the upstream conduit, any drops in inverts levels between the incoming and outgoing conduits, bend losses and any additional form losses. It is calculated as follows:

The default value of K_{Bend_max} is set to 4 but can be changed via the 1d_mh K_Bend_Max attribute. As shown above we have no angle for the incoming pipe and no drop in invert levels. Table 1 shows the K_Fixed is equal to 2 and the outgoing pipe has a form loss coefficient, to represent bend losses of 1. Therefore K_outletpipe is:

This matches the value within Figure 3 and the middle value for the downstream conduit in the *_TSL_P.shp, representing the outlet conduit losses, of 3.

Example 2: Single Incoming/Outgoing Conduit with Incoming Bend and Drop in Invert Levels

In this example, we will use the same simple setup but this time there is a 90 degree angle between the incoming and outgoing pipe as shown in the plan view in the figure below.

Figure 4: Example Model 2 Network Schematic

There is also a drop between the invert elevation of the incoming and outgoing conduits as shown in the difference in invert levels in table 2 below and the long profile in figure 5 . All other model parameters were kept the same including boundary inflows.

Parameter	Upstream Conduit	Downstream Conduit	Manhole
ID	UC-2	DC-1	Manhole
Invert Level (m AD)	N/A	N/A	99.4
Upstream Invert Level (m AD)	99.8	99.5	N/A
Downstream Invert Level (m AD)	99.7	99.4	N/A

Table 2: Updated Model 2 Parameters for Hand Calculations

Figure 5: Example Model 2 Long Profile

The resulting simulated losses are shown in the below figure 6.

Figure 6: Example Model 2 Results-Loss Coefficients

Example 2 - Entry Losses

In this example, the flow area in the manhole is 4.12m² and 1m² in the adjacent culvert whilst the flow is 2m³/s for both Q_p and Q_om. Therefore, Vm equals:
And V_p:
From this:

Which matches the value in figure 6.

Example 2 - Exit Losses

In this example, A'_m is 4.12m², therefore K_exit:

Which matches the result in figure 6.

Example 2 - Outgoing Conduit Losses

On this occasion, we have an 90 degree angle between the incoming and outgoing conduit. Therefore, we need to calculate Kθ using the following equation:
Q_f is calculated as:
Q_p and Q_om are both 2m³s^-1 and therefore Q_f is equal to 1. The incoming conduit angle is 90 degrees which is cancelled out by the denominator. K_θ is therefore:

We also have a drop between the incoming and outgoing pipes. K_drop which represents losses due to this drop is calculated as follows:

Which is:

Therefore K_outletpipe is:

This matches the simulated value in figure 6.

Example 3: Multiple Incoming Conduits with Incoming Bend and Drop in Invert Levels

In this example we have multiple incoming conduit. Each upstream conduit has a flow of 1m³s^-1 applied to it.

Figure 7: Example Model 3 Network Schematic
The conduits are the same shape and the same dimensions but have different invert levels as specified in the table below. All parameters for the manhole and other parts of the network are identical.

Parameter	Upstream Conduit 1	Upstream Conduit 2
ID	UC-2	UC-31
Upstream Invert Level (m AD)	99.8	100
Downstream Invert Level (m AD)	99.7	99.8

The resulting simulated loss values are presented in Figure 8.

Figure 8: Example Model 3 Results-loss Coefficients

Example 3 - Entry Losses

This time the flow area in the manhole is 3.57m² and 1m² in the adjacent culvert whilst the flow is 1m³/s for Q_p and 2m³s^-1 for Q_om. Therefore, V_m equals:
And V_p:
From this:

This matches the losses for both incoming pipes as shown in figure 8.

Example 3 - Exit Losses

A'_m is 3.57m² respectively. Therefore, K^exit:

The calculated losses match the losses calculated by the TUFLOW presented in Figure 8.

Example 3 - Outgoing Conduit Losses

In this model we have two incoming pipes, one with an 90 degree angle between the incoming and outgoing conduit and one with zero. Recall that:

With Qf calculated as:
Q_p is set to 1m³s^-1 and Q_om to 2m³s^-1. Q_f is equal to 0.5m³s^-1. For the incoming conduit with an angle of 90 degrees, K_θ is therefore: