---
title: "Khosla: cistern floor thickness from corrected phi"
subject: "Irrigation and Drainage Engineering"
chapter: 5
key: "semester_6/irrigation_and_drainage_engineering:khosla-corrected-phi-barrage-cistern"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/khosla-corrected-phi-barrage-cistern
site: BCT Engineering
---

# Khosla: cistern floor thickness from corrected phi

Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks

- Asked in: 2078 Chaitra, 6 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-corrected-phi-barrage-cistern
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5

## Question

Following corrected pressure potentials were determined underneath a barrage floor by Khosla's theory: At the junction of upstream sheet pile with floor φc1 = 82%; At the junction of downstream sheet pile with floor φc2 = 35%. Calculate the minimum thickness of the cistern floor at the beginning and end of the cistern. The following data are given: Full reservoir level = 505 masl, River bed level = 500 masl, Cistern floor level = 499 masl, Total length of barrage between upstream and downstream sheet piles = 40 m, Length of cistern = 15 m. Assume tailwater depth is nil on the downstream side and specific gravity of concrete floor = 2.4. Also, calculate the exit gradient.

## Short answer

Using corrected φc1=82%, φc2=35%, head 6m, floor length 40m, cistern starts at 25m, linear interpolation gives uplift 52.6% at start, 35% at end. Thickness t = h/(2.4-1): 2.26m and 1.50m. Exit gradient from d/s pile depth 6.40m is 0.156 (1 in 6.4).

## Answer

#### Data and assumptions

Corrected pressures: $\phi_{C1} = 82\%$ (u/s pile junction), $\phi_{C2} = 35\%$ (d/s pile junction). Full reservoir level 505 m, cistern floor 499 m, no tail water, so the head is $H = 505 - 499 = 6$ m. Floor length between piles $b = 40$ m, cistern length 15 m at the d/s end, $G = 2.4$.

#### Pressure along the floor

The HGL is a straight line between the junctions. The cistern starts $40 - 15 = 25$ m from the u/s pile:

$$
\phi_{start} = 82 - (82-35)\times\frac{25}{40} = 52.62\%,\qquad \phi_{end} = 35\%
$$

Uplift heads above the cistern floor level:

$$
h_{start} = 52.62\%\times 6 = 3.157\ \text{m},\qquad h_{end} = 35\%\times 6 = 2.100\ \text{m}
$$

#### Thickness of the cistern floor

$$
t = \frac{h}{G-1}:\qquad t_{start} = \frac{3.157}{1.4} = 2.26\ \text{m},\qquad t_{end} = \frac{2.100}{1.4} = 1.50\ \text{m}
$$

#### Exit gradient

The depth of the d/s pile is found from $\phi_{C2} = \phi_E$ of the d/s pile (the junction pressure):

$$
\cos(\pi\times 0.35) = \frac{\lambda-2}{\lambda} \Rightarrow \lambda = \frac{2}{1-0.4540} = 3.663,\quad \alpha = \sqrt{(2\lambda-1)^2-1} = 6.246,\quad d = \frac{b}{\alpha} = 6.40\ \text{m}
$$

$$
G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{6}{6.40}\times\frac{1}{\pi\sqrt{3.663}} = 0.1558\ (\approx 1/6.4)
$$

**Answer:** Minimum cistern floor thickness $\approx 2.26$ m at the beginning (u/s end of the cistern) and $\approx 1.50$ m at the end; exit gradient $\approx 0.156$ (about 1 in 6.4), below the usual safe value of 1/6.

## Check yourself

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

1. With no tail water the head is:
   A. 505 - 499 = 6 m
   B. 505 - 500 = 5 m
   C. 6 + 40
   D. 499 - 500
   Answer: A. Head is full reservoir minus the cistern floor level. (Key point: Head)

2. Pressure phi at the start of the cistern (25 m from the u/s pile) is:
   A. 82 - (82-35) x 25/40 = 52.62%
   B. 82%
   C. 35%
   D. 58.5%
   Answer: A. The HGL is straight between the junctions, so interpolate over 25 of 40 m. (Key point: Pressure at cistern start)

3. Minimum thickness at the start of the cistern is:
   A. 3.157/1.4 = 2.26 m
   B. 3.157/2.4
   C. 3.157 x 1.4
   D. 2.10/1.4
   Answer: A. t = h/(G-1) with G = 2.4, giving 2.26 m. (Key point: Thickness)

4. Why is end thickness smaller (1.50 m)?
   A. Uplift head there is 35% x 6 = 2.1 m, lower
   B. Cistern ends are thinner by rule
   C. Because G changes
   D. The exit gradient is small
   Answer: A. Pressure decreases along the floor, so the required thickness decreases. (Key point: End thickness)

5. d/s pile depth is found by:
   A. Using phi_E = 35% to get lambda, alpha, then d = b/alpha = 6.4 m
   B. Taking d = 15 m
   C. d = H
   D. d = b/2
   Answer: A. Inverting phi_E gives lambda 3.663 and alpha 6.246, hence d = 6.40 m and G_E = 0.156. (Key point: Pile depth)

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