---
title: "Slope-area discharge, 1300 m reach"
subject: "Engineering Hydrology"
chapter: 4
key: "semester_5/engineering_hydrology:slope-area-3500-3250"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-3500-3250
site: BCT Engineering
---

# Slope-area discharge, 1300 m reach

Engineering Hydrology · Chapter 4: Surface Runoff

- Asked in: 2074 Asoj, 12 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/slope-area-3500-3250
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4

## Question

Calculate the flood discharge of a stream by the slope area method given the following data:

Upstream flow area = 3500 m$^2$
Upstream wetted perimeter = 650 m
Upstream velocity head coefficient = 1.17
Downstream flow area = 3250 m$^2$
Downstream wetted perimeter = 621 m
Downstream velocity head coefficient = 1.21
Falling difference = 0.4 m
Reach length = 1300 m
Manning's coefficient n = 0.03

## Short answer

With fall 0.4 m over 1300 m, geometric mean conveyance K_m=342,113, initial estimate 6001 m³/s. Iterative correction for velocity heads (no eddy loss) yields discharge 5754.8 m³/s.

## Answer

**Given:** $A_1=3500$ m$^2$, $P_1=650$ m, $\alpha_1=1.17$; $A_2=3250$ m$^2$, $P_2=621$ m, $\alpha_2=1.21$; fall $=0.4$ m, $L=1300$ m, $n=0.03$. No eddy-loss coefficient is given, so $h_e=0$ is assumed.

**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
$$
R_1=\frac{3500}{650}=5.3846\ \text{m},\qquad R_2=\frac{3250}{621}=5.2335\ \text{m}
$$
$$
K_1=\frac{1}{0.03}\times3500\times5.3846^{2/3}=358412.6,\qquad K_2=\frac{1}{0.03}\times3250\times5.2335^{2/3}=326555.2
$$
$$
K_m=\sqrt{K_1K_2}=342113.3
$$

#### Step 2: First trial (no velocity head)
$$
S_f=\frac{0.4}{1300},\qquad Q_1=K_m\sqrt{S_f}=6001.064\ \text{m}^3/\text{s}
$$

#### Step 3: Correct for velocity head and eddy loss
$$
h_f=(h_1-h_2)+\left(\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right)-h_e,\qquad h_e=k_e\left|\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right|
$$
With $h_e=0$: $h_f = 0.4 + (h_{v1}-h_{v2})$.

| 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 | 6001.064 | 1.7146 | 1.8465 | 0.17531 | 0.21027 | 0.00000 | 0.36504 | 5732.829 |
| 2 | 5732.829 | 1.6380 | 1.7639 | 0.15999 | 0.19189 | 0.00000 | 0.36810 | 5756.770 |
| 7 | 5754.850 | 1.6442 | 1.7707 | 0.16122 | 0.19337 | 0.00000 | 0.36785 | 5754.850 |

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

**Answer: Discharge $Q \approx 5754.8\ \text{m}^3/\text{s}$.**

(If a contraction loss of $k_e=0.1$ were included, $Q$ would be about 5732 m$^3$/s, a change of 0.4 %.)

## Check yourself

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

1. No eddy-loss coefficient is given. What is assumed?
   A. he = 0, so hf = fall + (hv1 - hv2)
   B. he = 0.3
   C. he = fall
   D. Velocity heads are zero
   Answer: A. Eddy loss is taken zero when not given. (Key point: No eddy loss)

2. The alpha values 1.17 and 1.21 are used in
   A. The Manning n
   B. The velocity heads alpha v^2/2g
   C. The conveyance K
   D. The reach length
   Answer: B. alpha corrects the velocity head for non-uniform velocity distribution. (Key point: Velocity-head coefficient)

3. Here velocity head at section 2 exceeds that at section 1. Compared with the first trial, hf and Q
   A. Do not change
   B. Q doubles
   C. Both decrease (hf < 0.4, Q lowers)
   D. Both increase
   Answer: C. hf = 0.4 + (hv1 - hv2) = 0.365, so Q falls from 6001 to about 5755 m3/s. (Key point: Effect of velocity head)

4. Km is calculated as
   A. K1/K2
   B. K1 + K2
   C. K1 - K2
   D. sqrt(K1 K2)
   Answer: D. Geometric mean of the end-section conveyances. (Key point: Mean conveyance)

## Close variants set in other exams

- [Slope-area flood discharge, 7 km reach](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-45-50-7km) — Engineering Hydrology, chapter 4, asked 2×

## Related in Engineering Hydrology

- [Slope-area flood discharge, Marin River at Kusuntar](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-marin-kusuntar) — Engineering Hydrology, chapter 4
- [Slope-area discharge, 10 km reach](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-306-920) — Engineering Hydrology, chapter 4
- [Slope-area flood discharge, 6 km reach](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-60-45) — Engineering Hydrology, chapter 4
- [Slope-area discharge, 10 km reach, 104.77 m](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-104-77) — Engineering Hydrology, chapter 4
- [Slope-area discharge with three sections](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-3-section-260m) — Engineering Hydrology, chapter 4
- [Stream flow by slope-area method](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aslope-area-method-explain) — Engineering Hydrology, chapter 4, asked 3×

## Related in other subjects

- [Settling velocity of discrete particle 0.17 mm at 21°C](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c70o) — Water Supply Engineering, chapter 6, asked 2×
- [DO deficit 50 km downstream, 1.9 m³/s stream](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Astream-do-deficit-50km) — Sanitary Engineering, chapter 6, asked 2×
- [DO deficit 60 km downstream, 2 m³/s stream](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Astream-do-deficit-60km) — Sanitary Engineering, chapter 6, asked 2×