---
title: "Slope-area discharge, 10 km reach, 104.77 m"
subject: "Engineering Hydrology"
chapter: 4
key: "semester_5/engineering_hydrology:slope-area-104-77"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-104-77
site: BCT Engineering
---

# Slope-area discharge, 10 km reach, 104.77 m

Engineering Hydrology · Chapter 4: Surface Runoff

- Asked in: 2067 Mangsir (old course), 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-104-77
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4

## Question

Calculate the discharge of a stream having high water surface elevations noted at two sections A and B, 10 km apart. These elevations and other salient hydraulic properties are given below.

| Section | Water surface elevation (m) | Area of x-section (m$^2$) | Hydraulic radius (m) |
|---|---|---|---|
| A | 104.77 | 73.293 | 2.733 |
| B | 104.500 | 93.375 | 3.089 |

The eddy loss coefficient is 0.3 for gradual expansion, 0.1 for gradual contraction and Manning's roughness is 0.02.

## Short answer

Conveyances K₁=7163.5, K₂=9902.5, Kₘ=8424.4; starting with fall 0.27 m over 10 km, then iterating with velocity head and expansion loss (ke=0.3) gives discharge Q ≈ 44.2 m³/s.

## Answer

**Given:** $A_A=73.293$ m$^2$, $R_A=2.733$ m, $A_B=93.375$ m$^2$, $R_B=3.089$ m, $n=0.02$, $L=10$ km, fall $=104.77-104.50=0.27$ m. Velocity-head coefficients are taken as 1. Section B has the larger area, so the reach expands and $k_e=0.3$. The wetted perimeter is found from $P=A/R$ ($P_A=26.82$ m, $P_B=30.23$ m).

**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
$$
R_1=\frac{73.293}{26.8178}=2.7330\ \text{m},\qquad R_2=\frac{93.375}{30.2282}=3.0890\ \text{m}
$$
$$
K_1=\frac{1}{0.02}\times73.293\times2.7330^{2/3}=7163.5,\qquad K_2=\frac{1}{0.02}\times93.375\times3.0890^{2/3}=9902.5
$$
$$
K_m=\sqrt{K_1K_2}=8422.4
$$

#### Step 2: First trial (no velocity head)
$$
S_f=\frac{0.27}{10000},\qquad Q_1=K_m\sqrt{S_f}=43.764\ \text{m}^3/\text{s}
$$

#### 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 $v_B<v_A$ (expansion), so $h_e = 0.3(h_{vA}-h_{vB})$.

| 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 | 43.764 | 0.5971 | 0.4687 | 0.01817 | 0.01120 | 0.00209 | 0.27488 | 44.158 |
| 2 | 44.158 | 0.6025 | 0.4729 | 0.01850 | 0.01140 | 0.00213 | 0.27497 | 44.165 |
| 3 | 44.165 | 0.6026 | 0.4730 | 0.01851 | 0.01140 | 0.00213 | 0.27497 | 44.165 |

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. How is the mean conveyance Km obtained?
   A. Km = sqrt(K1 K2), with K = (1/n) A R^(2/3)
   B. Km = K1 + K2
   C. Km = (K1 - K2)/2
   D. Km = K1 x K2
   Answer: A. The geometric mean of the end-section conveyances is used. (Key point: Conveyance)

2. Why is ke = 0.3 used here?
   A. The reach is contracting
   B. Section B has the larger area, so the reach is expanding
   C. Manning's n is 0.02
   D. The reach is 10 km long
   Answer: B. Area increases from A (73.3) to B (93.4), so it is an expansion, giving he = 0.3 (hvA - hvB). (Key point: Eddy loss coefficient)

3. How does the trial procedure proceed and what is the result?
   A. Single step Q = Km sqrt(0.27/L) = 43.8 m^3/s with no correction
   B. Q = A x R x n = 4.5 m^3/s
   C. Start with Q = Km sqrt(0.27/L), correct hf with velocity head and eddy loss, iterate to Q about 44.2 m^3/s
   D. Iterate until hf equals the fall 0.27 m exactly
   Answer: C. hf rises slightly to 0.27497 m, giving Q converging to 44.165 m^3/s. (Key point: Trial method)

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