An 16 An electric motor operates at 1000 rpm and displaces a piston of a hydraulic pump by $$ 200 \mathrm{~cm}^{3} $$ per revolution. Oil then starts flowing from a hydraulic pump through a pipe of diameter 25 mm . Oil has a kinematic viscosity of 70 cSt at $$ 40^{\circ} \mathrm{C} $$. a) Calculate the flow rate $$ Q $$ in $$ \mathrm{m}^{3} / \mathrm{s} $$. ( 1 mark) b) Calculate the oil velocity ( $$ \mathrm{m} / \mathrm{s} $$ ) in the pipe and decide where it is flowing (intake/return/ pressure). ( 2 marks) c) Calculate the Reynold's number and decide the type of flow in the pipe. ( 2 marks)

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### Problem a)

**Calculate the flow rate $$ Q $$ in $$ \mathrm{m}^{3} / \mathrm{s} $$.**

**Given:**
- Displacement per revolution = $$ 200 \, \mathrm{cm}^{3} $$
- Motor speed = $$ 1000 \, \mathrm{rpm} $$

**Calculation:**
1. **Total displacement per minute:**
   $$
   Q_{\text{per minute}} = 200 \, \mathrm{cm}^{3/rev} \times 1000 \, \mathrm{rev/min} = 200{,}000 \, \mathrm{cm}^{3/min}
   $$
2. **Convert $$ \mathrm{cm}^{3} $$ to $$ \mathrm{m}^{3} $$:**
   $$
   200{,}000 \, \mathrm{cm}^{3} = 200{,}000 \times 10^{-6} \, \mathrm{m}^{3} = 0.2 \, \mathrm{m}^{3}
   $$
3. **Convert minutes to seconds:**
   $$
   Q = \frac{0.2 \, \mathrm{m}^{3}}{60 \, \mathrm{s}} = \frac{1}{300} \, \mathrm{m}^{3}/\mathrm{s} \approx 3.33 \times 10^{-3} \, \mathrm{m}^{3}/\mathrm{s}
   $$

**Answer:**
$$
Q = 3.33 \times 10^{-3} \, \mathrm{m}^{3}/\mathrm{s}
$$
The flow rate $$ Q $$ is approximately $$ 3.33 \times 10^{-3} \, \mathrm{m}^{3}/\mathrm{s} $$. ### Problem b) **Calculate the oil velocity $$ v $$ in the pipe and determine where it is flowing (intake/return/pressure).** **Given:** - Pipe diameter $$ d = 25 \, \mathrm{mm} = 0.025 \, \mathrm{m} $$ - Flow rate $$ Q = 3.33 \times 10^{-3} \, \mathrm{m}^{3}/\mathrm{s} $$ **Calculation:** 1. **Calculate the cross-sectional area $$ A $$ of the pipe:** $$ A = \frac{\pi d^{2}}{4} = \frac{\pi (0.025)^{2}}{4} \approx 4.9087 \times 10^{-4} \, \mathrm{m}^{2} $$ 2. **Calculate the velocity $$ v $$:** $$ v = \frac{Q}{A} = \frac{3.33 \times 10^{-3}}{4.9087 \times 10^{-4}} \approx 6.78 \, \mathrm{m}/\mathrm{s} $$ **Determine the flow location:** - **Intake:** Typically has higher velocity and is where the oil enters the system. - **Return:** Usually has lower velocity as the oil returns to the reservoir. - **Pressure:** The pressure area maintains the flow but doesn't directly affect velocity. Given the calculated velocity of approximately $$ 6.78 \, \mathrm{m}/\mathrm{s} $$, it is likely flowing through the **intake** section of the hydraulic system, where higher velocities are common due to the pump's operation. **Answer:** $$ v \approx 6.78 \, \mathrm{m}/\mathrm{s} $$ The oil is flowing through the **intake** section of the hydraulic system. ### Problem c) **Calculate the Reynolds number and determine the type of flow in the pipe.** **Given:** - Kinematic viscosity $$ \nu = 70 \, \mathrm{cSt} = 70 \times 10^{-6} \, \mathrm{m}^{2}/\mathrm{s} $$ - Velocity $$ v = 6.78 \, \mathrm{m}/\mathrm{s} $$ - Pipe diameter $$ d = 0.025 \, \mathrm{m} $$ **Calculation:** 1. **Calculate the Reynolds number $$ Re $$:** $$ Re = \frac{v \times d}{\nu} = \frac{6.78 \times 0.025}{70 \times 10^{-6}} \approx \frac{0.1695}{70 \times 10^{-6}} \approx 2421.43 $$ 2. **Determine the type of flow:** - **Laminar Flow:** $$ Re < 2300 $$ - **Turbulent Flow:** $$ Re > 4000 $$ - **Transition Flow:** $$ 2300 \leq Re \leq 4000 $$ Given $$ Re \approx 2421.43 $$, the flow is in the **transition** regime, indicating a mix of laminar and turbulent characteristics. **Answer:** $$ Re \approx 2421.43 $$ The flow in the pipe is in the **transition** regime, exhibiting both laminar and turbulent characteristics.

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