---
title: "Conduit partly full with C = K R^n"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:conduit-partly-full-chezy-kr-n"
source: https://www.bctengineer.com/q/semester_4/hydraulics/conduit-partly-full-chezy-kr-n
site: BCT Engineering
---

# Conduit partly full with C = K R^n

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2075 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/conduit-partly-full-chezy-kr-n
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

A 900 mm diameter conduit 3600 m long is laid at a uniform slope of 1 in 1500 and connects two reservoirs. When the levels in the reservoirs are low, the conduit runs partly full and it is found that a normal depth of 600 mm gives a rate of flow of 0.322 m³/s. The Chezy coefficient C is given by $KR^n$ where K is a constant, R is the hydraulic radius and n = 1/6. Neglecting losses of head at entry and exit, obtain (i) the value of K, (ii) the discharge when the conduit is flowing full and the difference in level between the two reservoirs is 4.5 m.

## Short answer

From partly full geometry, Chezy C = 54.1 m¹/²/s, giving K = C/R¹/⁶ = 67.6. When full with a 4.5 m head difference, friction slope 0.00125 yields Q_full = 0.562 m³/s.

## Answer

#### Given data
$d=0.9$ m, $L=3600$ m, bed slope $S_0=1/1500$, normal depth $y=0.6$ m, $Q=0.322$ m³/s, $C=KR^{1/6}$.

#### (i) Value of $K$ (partly full)
Geometry for $y/d=0.6/0.9=0.667$. Angle subtended at the centre by the wetted arc:
$$
\cos\frac\theta2=1-\frac{2y}{d}=1-\frac{2\times0.6}{0.9}=-0.3333\;\Rightarrow\;\frac\theta2=109.47^\circ,\ \theta=218.94^\circ=3.8213\ \text{rad}
$$
$$
A=\frac{d^2}{8}(\theta-\sin\theta)=\frac{0.81}{8}(3.8213-\sin218.94^\circ)=0.4505\ \text{m}^2
$$
$$
P=\frac{d\theta}{2}=\frac{0.9\times3.8213}{2}=1.7196\ \text{m},\qquad R=\frac AP=0.2620\ \text{m}
$$
Chezy's equation $Q=CA\sqrt{RS_0}$ gives
$$
C=\frac{Q}{A\sqrt{RS_0}}=\frac{0.322}{0.4505\sqrt{0.2620/1500}}=54.08\ \text{m}^{1/2}/\text{s}
$$
$$
K=\frac{C}{R^{1/6}}=\frac{54.08}{0.2620^{1/6}}=67.60
$$

#### (ii) Discharge when flowing full
Full pipe: $A_f=\dfrac\pi4(0.9)^2=0.6362$ m², $R_f=\dfrac d4=0.225$ m.
$$
C_f=KR_f^{1/6}=67.60\times0.225^{1/6}=52.72
$$
When full and the reservoirs differ by 4.5 m (losses at entry and exit neglected), the energy (friction) slope is
$$
S_f=\frac{h_f}{L}=\frac{4.5}{3600}=0.00125
$$
$$
Q_f=C_fA_f\sqrt{R_fS_f}=52.72\times0.6362\times\sqrt{0.225\times0.00125}=0.5625\ \text{m}^3/\text{s}
$$
**Answer: (i) $K=67.6$ (with $C=54.1$ for the partly full flow); (ii) $Q_{full}=0.562$ m³/s.**

## Check yourself

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

1. How is the constant K in C = K R^(1/6) found?
   A. Use the full-pipe R = d/4 in the partly full flow data
   B. From C = Q/(A sqrt(R S0)) for the partly full flow using the geometry at y = 0.6 m, then K = C/R^(1/6)
   C. From the head difference of 4.5 m only
   D. By assuming C = 1 at R = 1
   Answer: B. The partly full data (y = 0.6 m, Q = 0.322) give A, R and hence C by Chezy's equation; K = C/R^(1/6) follows. (Key point: Method for finding K)

2. For full flow between the reservoirs with entry and exit losses neglected, which slope is used in Chezy's equation?
   A. The bed slope 1/1500
   B. S_f = 4.5/3600, since the whole level difference is lost to friction over the length
   C. S_f = 4.5/900
   D. The mean of 1/1500 and 4.5/3600
   Answer: B. With no minor losses, hf = 4.5 m over L = 3600 m, so the friction slope is 0.00125. (Key point: Energy slope for full flow)

3. Why is Chezy's C for full flow (about 52.7) different from the partly full value (about 54.1)?
   A. C = K R^(1/6) depends on R, and R_full = d/4 = 0.225 m is smaller than 0.262 m
   B. K changes when the pipe fills
   C. The pipe slope changes
   D. Entry losses reduce C
   Answer: A. K is a constant but C varies with R^(1/6); a smaller R gives a smaller C. (Key point: Hydraulic radius changes C)

4. For the full 0.9 m conduit, which values are used to get Q_full = C_f A_f sqrt(R_f S_f)?
   A. A = 0.4505 m2, R = 0.262 m
   B. A = 0.6362 m2, R = 0.225 m, giving about 0.56 m3/s
   C. A = 0.6362 m2, R = 0.45 m, giving about 0.8 m3/s
   D. A = 0.81 m2, R = 0.225 m
   Answer: B. A_full = pi d^2/4 = 0.6362 m2 and R_full = d/4 = 0.225 m, giving Q = 0.5625 m3/s. (Key point: Full-pipe geometry and result)

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