---
title: "Slope-area discharge, 10 km reach"
subject: "Engineering Hydrology"
chapter: 4
key: "semester_5/engineering_hydrology:slope-area-306-920"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-306-920
site: BCT Engineering
---

# Slope-area discharge, 10 km reach

Engineering Hydrology · Chapter 4: Surface Runoff

- Asked in: 2078 Kartik, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-306-920
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4

## Question

The high flow water surface elevations of a stream at two sections 10 km apart are 306.920 m and 306.650 m. The cross-sectional area and wetted perimeters are as follows:

| Section | Area (m$^2$) | Wetted perimeter (m) |
|---|---|---|
| A | 73.3 | 26.80 |
| B | 93.4 | 30.23 |

Assume n = 0.02. The eddy loss coefficient is 0.30 for gradual expansion and 0.10 for gradual contraction. Estimate the discharge in the stream. Section A is upstream of B.

## Short answer

Slope-area method with Manning’s conveyance and trial correction for velocity head and expansion eddy loss (ke=0.3) gives discharge ≈ 44.2 m³/s.

## Answer

**Given:** $A_1=73.3$ m$^2$, $P_1=26.80$ m, $A_2=93.4$ m$^2$, $P_2=30.23$ m, $n=0.02$, $L=10$ km, fall $h_1-h_2 = 306.920-306.650 = 0.27$ m. Velocity-head coefficients are taken as $\alpha_1=\alpha_2=1$. The area increases downstream, so the reach is an expansion and $k_e = 0.3$.

**Method:** Manning's equation gives conveyance $K=\dfrac{1}{n}AR^{2/3}$ and $Q = K\sqrt{S_f}$. The friction slope is found from the energy equation between the two sections, so the discharge is found by trial.

#### Step 1: Conveyance
$$
R_1=\frac{73.3}{26.8}=2.7351\ \text{m},\qquad R_2=\frac{93.4}{30.23}=3.0896\ \text{m}
$$
$$
K_1=\frac{1}{0.02}\times73.3\times2.7351^{2/3}=7167.8,\qquad K_2=\frac{1}{0.02}\times93.4\times3.0896^{2/3}=9906.6
$$
$$
K_m=\sqrt{K_1K_2}=8426.6
$$

#### Step 2: First trial (no velocity head)
$$
S_f=\frac{0.27}{10000},\qquad Q_1=K_m\sqrt{S_f}=43.786\ \text{m}^3/\text{s}
$$

#### Step 3: Correct for velocity head and eddy loss
$$
h_f=(h_1-h_2)+\left(\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right)-h_e,\qquad h_e=k_e\left|\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right|
$$
Here $v_2<v_1$ (expansion), so $h_e = 0.3(h_{v1}-h_{v2})$.

| Trial | $Q$ assumed | $v_1$ | $v_2$ | $\alpha_1 v_1^2/2g$ | $\alpha_2 v_2^2/2g$ | $h_e$ | $h_f$ | $Q=K_m\sqrt{h_f/L}$ |
|---|---|---|---|---|---|---|---|---|
| 1 | 43.786 | 0.5974 | 0.4688 | 0.01819 | 0.01120 | 0.00210 | 0.27489 | 44.181 |
| 2 | 44.181 | 0.6027 | 0.4730 | 0.01852 | 0.01140 | 0.00213 | 0.27498 | 44.188 |
| 3 | 44.188 | 0.6028 | 0.4731 | 0.01852 | 0.01141 | 0.00213 | 0.27498 | 44.188 |

The trials converge to $h_f$ and $Q$ shown in the last row.

**Answer: Discharge $Q \approx 44.2\ \text{m}^3/\text{s}$.**

## Check yourself

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

1. Area at A is 73.3 and at B is 93.4 m2 with A upstream. Which eddy-loss coefficient is used?
   A. 0.30 for expansion
   B. Average of 0.1 and 0.3
   C. 0.10 for contraction
   D. Zero because the reach is uniform
   Answer: A. Area increases downstream so the reach is an expansion and ke = 0.3. (Key point: Expansion vs contraction)

2. How is the mean conveyance Km of the reach found from K1 and K2?
   A. The smaller of K1 and K2
   B. Geometric mean sqrt(K1 K2)
   C. Arithmetic mean (K1 + K2)/2
   D. K1 + K2
   Answer: B. Km = sqrt(K1 K2) = 8426.6. (Key point: Mean conveyance)

3. Why must the discharge be found by trial?
   A. Manning's n is unknown
   B. The length of the reach is unknown
   C. Friction head loss depends on velocity heads, which depend on Q itself
   D. The fall is not measured
   Answer: C. hf includes velocity-head difference and eddy loss that depend on Q, so Q is assumed, hf computed, and Q updated until it converges. (Key point: Why trial is needed)

4. Which expression gives the friction head loss hf?
   A. hf = fall x he
   B. hf = fall - (hv1 - hv2) + he
   C. hf = fall + he only
   D. hf = fall + (hv1 - hv2) - he
   Answer: D. hf = (h1 - h2) + (alpha1 v1^2/2g - alpha2 v2^2/2g) - he. (Key point: Friction head)

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