---
title: "Khosla: residual pressures and HGL, pond 260 m"
subject: "Irrigation and Drainage Engineering"
chapter: 5
key: "semester_6/irrigation_and_drainage_engineering:khosla-weir-profile-pond260"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/khosla-weir-profile-pond260
site: BCT Engineering
---

# Khosla: residual pressures and HGL, pond 260 m

Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks

- Asked in: 2063 Baisakh (old course), 16 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/khosla-weir-profile-pond260
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5

## Question

Using Khosla's method, obtain the residual seepage pressures at the 'key' points for the weir profile shown below. Also calculate the value of the exit gradient. Consider the case of no flow at pond level. Also draw the subsoil HGL.

[Figure (not to scale): pond level 260 m, crest 258.5 m; upstream floor 256 m with 1 m thick floor and 0.5 m pile, upstream 2:1 (H:V) slope, upstream pile bottom 250 m; distances 3 m, 5 m and 2 m; downstream 3:1 (H:V) slope, downstream floor level 255 m, 1.2 m thickness, downstream steps with 2 m, 10.5 m, 15 m, 0.5 m and 1.2 m dimensions, pile bottoms 248 m and 248 m]

## Short answer

Residual pressure heads (m): E1=5.00, D1=3.80, C1=3.66; E2=3.46, D2=3.12, C2=2.80; E3=1.60, D3=1.29, C3=0. Exit gradient = 0.1266, safe (<0.2).

## Answer

#### Data and assumptions

Reading of the figure (not to scale): pond level 260 m; no flow, so the d/s water level is the d/s floor level 255 m and $H = 5$ m. U/s pile at 0 m: floor 256 m (1 m thick), bottom 250 m ($d = 6$ m). Then 3 m of level floor, the u/s 2:1 slope from 256 m up to the crest 258.5 m (5 m long) and 2 m to the intermediate pile at 10 m: floor 256 m, bottom 248 m ($d = 8$ m). Downstream of it: 2 m, the d/s 3:1 glacis from 258.5 m down to 255 m (10.5 m long) and 15 m of d/s floor to the d/s pile at 37.5 m: floor 255 m (1.2 m thick), bottom 248 m ($d = 7$ m). Both slopes end or start next to the intermediate pile, so their corrections are applied there, using $b'$ = distance between the two piles enclosing the slope: u/s slope (up, 6.5%): $-6.5\times 5/10$ at $E_2$; d/s glacis (down, 4.5%): $+4.5\times 10.5/27.5$ at $C_2$. Floor thickness correction uses 1 m (u/s) and 1.2 m (d/s); the thickness at the intermediate pile is neglected. Safe exit gradient (not given) assumed 0.2.

#### Khosla's constants

End piles: $\alpha = b/d$, $\lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-2}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-1}{\lambda}$ (in percent), $b = 37.5$ m; for the u/s pile $\phi_{D1} = 100-\phi_D$, $\phi_{C1} = 100-\phi_E$.

Intermediate pile ($b_1$ from the u/s end, $b_2$ from the d/s end): $\lambda_1 = \dfrac{\sqrt{1+\alpha_1^2}-\sqrt{1+\alpha_2^2}}{2}$, $\lambda = \dfrac{\sqrt{1+\alpha_1^2}+\sqrt{1+\alpha_2^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1-1}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1}{\lambda}$, $\phi_C = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda_1+1}{\lambda}$.

| Pile | d (m) | α | λ | φE (%) | φD (%) |
|---|---|---|---|---|---|
| U/s pile | 6.00 | 6.250 | 3.665 | 34.99 | 24.09 |
| D/s pile | 7.00 | 5.357 | 3.225 | 37.60 | 25.77 |

| Intermediate pile | d (m) | α1 | α2 | λ1 | λ | φE (%) | φD (%) | φC (%) |
|---|---|---|---|---|---|---|---|---|
| at 10 m | 8.00 | 1.250 | 3.438 | -0.990 | 2.590 | 77.88 | 62.48 | 49.87 |

#### Pressures at the key points

Pressure is expressed as a percentage of the head $H = 5.00$ m (no flow).

| Point | Simple φ (%) | Corrections (%) | Corrected φ (%) | Pressure head (m) |
|---|---|---|---|---|
| E1 (top of u/s pile) | 100.00 | 0 | 100.00 | 5.00 |
| D1 | 75.91 | - | 75.91 | 3.80 |
| C1 | 65.01 | thk +1.82, int +6.34 | 73.17 | 3.66 |
| E2 | 77.88 | int -5.49, slope -3.25 | 69.13 | 3.46 |
| D2 | 62.48 | - | 62.48 | 3.12 |
| C2 | 49.87 | int +4.37, slope +1.72 | 55.96 | 2.80 |
| E3 | 37.60 | thk -2.03, int -3.58 | 31.99 | 1.60 |
| D3 | 25.77 | - | 25.77 | 1.29 |
| C3 (d/s end) | 0 | 0 | 0 | 0 |

Corrections: thk = floor thickness, $+\dfrac{\phi_D-\phi_C}{d}t$ at C and $-\dfrac{\phi_E-\phi_D}{d}t$ at E; int = mutual interference, $19\sqrt{D/b'}\,\dfrac{d+D}{b}$, positive at C of the u/s pile and negative at E of the d/s pile of each pair ($D$ = depth of the neighbouring pile below the floor level at the pile being corrected, $b'$ = distance between the two piles); slope = slope correction. The D points are not corrected.

#### Exit gradient

$$
G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{5.00}{7.00}\times\frac{1}{\pi\sqrt{3.225}} = 0.1266\ (\approx 1/7.9)
$$

Permissible (assumed) $= 0.2000$. Safe (less than the permissible value).

#### Hydraulic gradient line

Bars show the corrected pressure (each # is 2.5%):
```text
 E1  100.0% |########################################
 D1   75.9% |##############################
 C1   73.2% |#############################
 E2   69.1% |############################
 D2   62.5% |#########################
 C2   56.0% |######################
 E3   32.0% |#############
 D3   25.8% |##########
 C3    0.0% |
```
The hydraulic gradient line (HGL) drops vertically across each pile (from E to D to C) and runs as a straight line along the floor between the pile junctions (C of one pile to E of the next). In levels (datum 255 m): $C_1$ 258.66 m, $E_2$ 258.46 m, $C_2$ 257.80 m, $E_3$ 256.60 m.

**Answer:** The residual seepage pressures at the key points are the corrected values (percent of $H$ and metres of water) in the table; the subsoil HGL is plotted above, and the exit gradient is about 1 in 7.9 (safe).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Pond level 260 m, d/s floor 255 m. What head H is used for the no flow case?
   A. 4 m
   B. 5 m
   C. 1.5 m
   D. 6 m
   Answer: B. H = 260 - 255 = 5 m. (Key point: Head with no flow)

2. Which sign applies to slope corrections at C1 (u/s slope) and E3 (d/s slope)?
   A. Negative at both
   B. Positive at C1, negative at E3
   C. Positive at both
   D. Negative at C1, positive at E3
   Answer: D. Slope corrections are negative at the u/s slope and positive at the d/s slope in this solution. (Key point: Slope correction sign)

3. Intermediate pile at 10 m and d/s pile at 37.5 m. What total length b is used for alpha = b/d?
   A. 37.5 m
   B. 10 m
   C. 47.5 m
   D. 27.5 m
   Answer: A. b is the total floor length from the u/s pile to the d/s pile, 37.5 m. (Key point: Floor length)

4. G_E = 0.1266 against a permissible 0.2. Conclusion?
   A. Safe
   B. Not safe
   C. Not safe since H/d is greater than 1
   D. Needs a deeper pile
   Answer: A. 0.1266 is less than 0.2, so the structure is safe against piping. (Key point: Exit gradient verdict)

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