GATE 2018 IN Q.50: solution and answer

GATE 2018 · INQ.50 · 2 marks · NAT

The average velocity $v$ of flow of clear water in a 100 cm (inner) diameter tube is measured using the ultrasonic flow meter as shown in the figure. The angle $θ$ is $4 5^{\circ}$ . The measured transit times are $t_{1} = 0.9950$ ms and $t_{2} = 1.0000$ ms. The velocity $v$ (in m/s) in the pipe is (up to one decimal place) ___.

GATE 2018 IN question 50: A horizontal pipe drawn as two parallel horizontal walls, with the inner diameter marked 100 cm by a vertical… — Figure from the GATE 2018 IN paper. It is part of the question.

Answer and solution

ANSWER

The average flow velocity is $v \approx 5.0 m/s$ .

Official answer: 3.5 to 5.7Source: Official answer key, GATE 2018 IN (IIT Guwahati)

Transit-time (ultrasonic) flowmeter. The sound path crosses the pipe at $θ = 4 5^{\circ}$ to the axis, so its length is $L = \frac{D}{sin θ} = \frac{1 m}{sin 4 5 ^{\circ}} = 2 m$ . Here $D = 100 cm = 1 m$ .
The flow adds $v cos θ$ to the sound speed $c$ on the downstream trip and subtracts it on the upstream trip. The shorter time $t_{1} = 0.9950 ms$ is therefore the downstream trip and $t_{2} = 1.0000 ms$ is the upstream trip: $t_{1} = \frac{L}{c + v cos θ}$ and $t_{2} = \frac{L}{c - v cos θ}$ .
Subtract the reciprocals. This removes $c$ , which is not given: $\frac{1}{t _{1}} - \frac{1}{t _{2}} = \frac{2 v cos θ}{L}$ .
Solve for $v$ : $v = \frac{L}{2 cos θ} (\frac{1}{t _{1}} - \frac{1}{t _{2}}) = \frac{D}{2 sin θ cos θ} (\frac{1}{t _{1}} - \frac{1}{t _{2}}) = \frac{D}{sin 2 θ} (\frac{1}{t _{1}} - \frac{1}{t _{2}})$ .
For $θ = 4 5^{\circ}$ , $sin 2 θ = 1$ and $D = 1 m$ , so $v = (\frac{1}{t _{1}} - \frac{1}{t _{2}}) \times 1 m$ .
Numbers: $\frac{1}{0.9950 \times 1 0 ^{- 3} s} = 1005.03 s^{- 1}$ and $\frac{1}{1.0000 \times 1 0 ^{- 3} s} = 1000 s^{- 1}$ . Therefore $v = 5.03 m/s \approx 5.0 m/s$ .

KEY IDEA

In a transit-time ultrasonic flowmeter the difference of the reciprocal transit times, $\frac{1}{t _{1}} - \frac{1}{t _{2}} = \frac{2 v cos θ}{L}$ , gives the velocity without needing the sound speed. With $L = D / sin θ$ this becomes $v = \frac{D}{sin 2 θ} (\frac{1}{t _{1}} - \frac{1}{t _{2}})$ .

WATCH OUT FOR

Taking the path length as the pipe diameter $D$ instead of $D / sin θ$ , or using $Δ t / t^{2}$ with $L = D$ and so being off by the factor $sin θ cos θ$ or $cos θ$ . Another slip is using $2 D / sin 2 θ$ , or forgetting to convert ms to s.

From the GATE 2018 IN paper. Every question of this paper · Download the full paper · All 2 Pressure & flow questions, with notes