--- title: "Gauge-discharge relationship and discharge at 10.5 m" subject: "Engineering Hydrology" chapter: 4 key: "semester_5/engineering_hydrology:gauge-discharge-rating-7-5" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gauge-discharge-rating-7-5 site: BCT Engineering --- # Gauge-discharge relationship and discharge at 10.5 m Engineering Hydrology · Chapter 4: Surface Runoff - Asked in: 2075 Chaitra, 6+2 marks - Video: [Stage Discharge Rating Curve : Stream Flow Measurement : WRE](https://www.youtube.com/watch?v=RkorkZaEbL8) — The Classroom - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gauge-discharge-rating-7-5 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4 ## Question Following are the data of gauge and discharge collected at a particular section of the river by stream gauging operation. i) Develop a gauge-discharge relationship for this stream at this section for use in estimating the discharge for a known gauge reading. What is the coefficient of correlation of the derived relationship? Use a = 7.5 m for the gauge corresponding to zero discharge. ii) Estimate the discharge corresponding to a gauge reading of 10.5 m at this gauging station. | Gauge reading (m) | Discharge (m$^3$/s) | Gauge reading (m) | Discharge (m$^3$/s) | |---|---|---|---| | 7.65 | 15 | 8.48 | 170 | | 7.7 | 30 | 8.98 | 400 | | 7.77 | 57 | 9.30 | 600 | | 7.8 | 39 | 9.5 | 800 | | 7.9 | 60 | 10.5 | 1500 | | 7.91 | 100 | 11.1 | 2000 | | 8.08 | 150 | 11.7 | 2400 | ## Short answer Using log-log regression with a=7.5 m, the rating equation is Q=275.5(G–7.5)^1.456, correlation r=0.991. For G=10.5 m, Q≈1364 m³/s. ## Answer #### (i) Rating equation Use $Q = C(G-a)^b$ with $a = 7.5$ m. Taking logs: $\log Q = \log C + b\log(G-a)$, a straight line $y = \log C + b\,x$ with $x=\log(G-a)$, $y=\log Q$. | G (m) | G - a (m) | Q | x = log(G-a) | y = log Q | |---|---|---|---|---| | 7.65 | 0.15 | 15 | -0.8239 | 1.1761 | | 7.7 | 0.20 | 30 | -0.6990 | 1.4771 | | 7.77 | 0.27 | 57 | -0.5686 | 1.7559 | | 7.8 | 0.30 | 39 | -0.5229 | 1.5911 | | 7.9 | 0.40 | 60 | -0.3979 | 1.7782 | | 7.91 | 0.41 | 100 | -0.3872 | 2.0000 | | 8.08 | 0.58 | 150 | -0.2366 | 2.1761 | | 8.48 | 0.98 | 170 | -0.0088 | 2.2304 | | 8.98 | 1.48 | 400 | 0.1703 | 2.6021 | | 9.3 | 1.80 | 600 | 0.2553 | 2.7782 | | 9.5 | 2.00 | 800 | 0.3010 | 2.9031 | | 10.5 | 3.00 | 1500 | 0.4771 | 3.1761 | | 11.1 | 3.60 | 2000 | 0.5563 | 3.3010 | | 11.7 | 4.20 | 2400 | 0.6232 | 3.3802 | With $n=14$: $\sum x=-1.2617$, $\sum y=32.3255$, $\sum x^2=3.2388$, $\sum xy=1.6364$, $\sum y^2=81.3787$. $$ b=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}=1.4558 $$ $$ \log C=\frac{\sum y-b\sum x}{n}=2.4402\;\Rightarrow\; C=275.5 $$ **Rating equation:** $Q = 275.5\,(G-7.5)^{1.456}$ (m$^3$/s) **Coefficient of correlation** $$ r=\frac{n\sum xy-\sum x\sum y}{\sqrt{[n\sum x^2-(\sum x)^2][n\sum y^2-(\sum y)^2]}}=0.9913 $$ The high value shows a very good fit to the log-log straight line. #### (ii) Discharge at G = 10.5 m $$ Q = 275.5\times(10.5-7.5)^{1.456} = 275.5\times3^{1.456} = 1364\ \text{m}^3/\text{s} $$ **Answer: $Q=275.5(G-7.5)^{1.456}$, $r=0.991$; $Q(10.5\ \text{m})\approx 1364$ m$^3$/s** (the measured value is 1500 m$^3$/s, so the fit is about 9 % low at this stage). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What equation form is fitted, with a = 7.5 m? A. Q = C (G - a)^b B. Q = C + b(G - a) C. Q = C exp(G - a) D. Q = C / (G - a) Answer: A. Power form Q = C (G - a)^b is fitted. (Key point: Rating equation form) 2. How is the fit made linear? A. Plot Q vs G B. Take logs: log Q = log C + b log(G - a) and regress y = log Q on x = log(G - a) C. Plot G vs log Q only D. Plot Q vs 1/G Answer: B. The log-log form is a straight line with slope b and intercept log C. (Key point: Linearisation) 3. Using Q = 275.5 (G - 7.5)^1.456, what is Q at G = 10.5 m? A. About 1500 m3/s exactly B. About 826 m3/s C. About 1364 m3/s D. About 2000 m3/s Answer: C. 275.5 x 3^1.456 = 1364 m3/s, about 9 % below the measured 1500. (Key point: Result check) 4. What does r = 0.991 indicate? A. Poor fit B. That b equals 0.991 C. That a is wrong D. Very good fit to the log-log straight line Answer: D. r close to 1 means the data lie very near the straight line. 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