---
title: "Khosla: corrected pressures, slope correction 4%"
subject: "Irrigation and Drainage Engineering"
chapter: 5
key: "semester_6/irrigation_and_drainage_engineering:khosla-corrected-pressure-slope-4pct"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/khosla-corrected-pressure-slope-4pct
site: BCT Engineering
---

# Khosla: corrected pressures, slope correction 4%

Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks

- Asked in: 2065 Shrawan (old course), 10 marks
- Video: [khosla's theory- exit gradient and H.G.Line - problem 1 part 3](https://www.youtube.com/watch?v=ECOS1KrBrDg) — HASEENA SALAH
- 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-corrected-pressure-slope-4pct
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5

## Question

Find out the corrected pressure at upstream, downstream and intermediate key points of a hydraulic structure founded on permeable foundation as given below. Use factor for slope correction as 4%. Assume suitable data if necessary. Also calculate the exit gradient [?] and plot the hydraulic gradient line.

[Figure: pond level 157.00 m, crest 155.00 m; upstream floor 154.00 m (1 m thick) with 0.6 m pile projection, upstream sheet pile bottom 148.00 m, intermediate pile bottom 149.00 m; horizontal distances 3 m, 2 m and 15 m; downstream floor levels 152.00 m (1.3 m thick) and 150.00 m, steps of 10 m and 10 m, downstream sheet pile bottom 141.7 m (as printed) with 0.6 m projection; total distances 15.8 m and 40 m]

## Short answer

Corrected pressures: u/s C1 70.0%, intermediate E2 57.7% and C2 58.0%, d/s E3 36.6%. Exit gradient is 0.156, safe (<0.200).

## Answer

#### Data and assumptions

Reading of the figure: pond level 157.00 m; with no flow the d/s water level is the d/s floor level 150.00 m, so $H = 7$ m. U/s pile at 0 m: floor 154.00 m (1 m thick), bottom 148.00 m ($d = 6$ m). Intermediate pile at 15.8 m: floor 152.00 m (1.3 m thick), bottom 149.00 m ($d = 3$ m). D/s pile at 40 m: floor 150.00 m (1.3 m thick), bottom 141.70 m as printed ($d = 8.3$ m). The 4% slope correction is multiplied by (slope length 3 m / $b'$), with $b' = 15.8$ m between the u/s and intermediate piles, and is negative at $C_1$ (up slope). Mutual interference is included. The permissible exit gradient is not given; 0.2 is assumed.

#### 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), $b = 40$ 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 | 6.667 | 3.871 | 33.94 | 23.40 |
| D/s pile | 8.30 | 4.819 | 2.961 | 39.48 | 26.96 |

| Intermediate pile | d (m) | α1 | α2 | λ1 | λ | φE (%) | φD (%) | φC (%) |
|---|---|---|---|---|---|---|---|---|
| at 15.8 m | 3.00 | 5.267 | 8.067 | -1.384 | 6.745 | 61.50 | 56.58 | 51.81 |

#### Pressures at the key points

Pressure is expressed as a percentage of the head $H = 7.00$ m (no flow).

| Point | Simple φ (%) | Corrections (%) | Corrected φ (%) | Pressure head (m) |
|---|---|---|---|---|
| E1 (top of u/s pile) | 100.00 | 0 | 100.00 | 7.00 |
| D1 | 76.60 | - | 76.60 | 5.36 |
| C1 | 66.06 | thk +1.76, int +2.94, slope -0.76 | 69.99 | 4.90 |
| E2 | 61.50 | thk -2.13, int -1.67 | 57.69 | 4.04 |
| D2 | 56.58 | - | 56.58 | 3.96 |
| C2 | 51.81 | thk +2.06, int +4.12 | 58.00 | 4.06 |
| E3 | 39.48 | thk -1.96, int -0.90 | 36.62 | 2.56 |
| D3 | 26.96 | - | 26.96 | 1.89 |
| 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, e.g. at $C_2$: $D = 152-141.7 = 10.3$ m, $d = 3$ m, $b' = 24.2$ m); slope = slope correction. The D points are not corrected. (The corrected $C_2$ comes out slightly above $E_2$ because the shallow intermediate pile receives large thickness and interference corrections.)

#### Exit gradient

$$
G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{7.00}{8.30}\times\frac{1}{\pi\sqrt{2.961}} = 0.1560\ (\approx 1/6.4)
$$

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   76.6% |###############################
 C1   70.0% |############################
 E2   57.7% |#######################
 D2   56.6% |#######################
 C2   58.0% |#######################
 E3   36.6% |###############
 D3   27.0% |###########
 C3    0.0% |
```
The hydraulic gradient line (HGL) drops 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).

**Answer:** Corrected pressures: u/s pile $D_1 = 76.6\%$, $C_1 = 70.0\%$; intermediate pile $E_2 = 57.7\%$, $D_2 = 56.6\%$, $C_2 = 58.0\%$; d/s pile $E_3 = 36.6\%$, $D_3 = 27.0\%$ (pressure heads in the table). Exit gradient $\approx 0.156$ (1 in 6.4), which is safe.

## Check yourself

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

1. Pond level 157 m and d/s floor level 150 m. What head H is used?
   A. 5 m
   B. 2 m
   C. 7 m
   D. 9 m
   Answer: C. With no flow, the d/s water level equals the d/s floor level, so H = 157 - 150 = 7 m. (Key point: Head with no flow)

2. Which corrections are applied to the simple Khosla pressures?
   A. Only mutual interference
   B. Temperature and soil type
   C. Only slope
   D. Mutual interference of piles, floor thickness and slope
   Answer: D. Corrected pressure = simple phi plus interference, thickness and slope corrections. (Key point: Corrections applied)

3. How is the 4% slope correction applied at C1 (upstream slope)?
   A. Positive, in proportion to slope length / b
   B. Negative, in proportion to slope length / b
   C. Positive, the full 4%
   D. Negative, the full 4%
   Answer: B. The correction is 4% x 3/40 = 0.30%, negative at C1 for an up slope. (Key point: Slope correction sign)

4. G_E = 0.156 compared with a permissible value of 0.2. What follows?
   A. Cannot compare without floor thickness
   B. Unsafe because G_E is not zero
   C. Unsafe because H/d is large
   D. Safe against piping
   Answer: D. 0.156 is less than 0.2, so the structure is safe against piping. (Key point: Exit gradient result)

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