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 \( Q \mathrm{in} \mathrm{m}^{3} / \mathrm{s} \). ( 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)

To solve the problem, let's go step by step: ### a) Calculate the flow rate \( Q \) in \( \mathrm{m}^{3} / \mathrm{s} \). 1. The displacement per revolution is \( 200 \, \mathrm{cm}^{3} \). 2. Since 1 revolution happens at 1000 rpm, we can calculate the flow rate as follows: \[ Q = \text{Displacement per revolution} \times \text{Revolutions per second} \] \[ \text{Revolutions per second} = \frac{1000 \, \text{rev/min}}{60} = \frac{1000}{60} \, \text{rev/s} \approx 16.67 \, \text{rev/s} \] Now converting displacement: \[ Q = 200 \, \mathrm{cm}^{3} \times 16.67 \, \mathrm{rev/s} = 3334 \, \mathrm{cm}^{3/s} \] Converting to \( \mathrm{m}^{3} / \mathrm{s} \): \[ Q = 3334 \, \mathrm{cm}^{3/s} \times \frac{1 \, \mathrm{m}^{3}}{1,000,000 \, \mathrm{cm}^{3}} \approx 0.003334 \, \mathrm{m}^{3/s} \] ### b) Calculate the oil velocity \( v \) in the pipe and decide where it is flowing. 1. The diameter of the pipe \( d = 25 \, \mathrm{mm} = 0.025 \, \mathrm{m} \). 2. The cross-sectional area \( A \) of the pipe is given by: \[ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.025}{2}\right)^2 \approx 4.91 \times 10^{-4} \, \mathrm{m}^{2} \] 3. The oil velocity \( v \) can be calculated using the flow rate \( Q \): \[ v = \frac{Q}{A} = \frac{0.003334 \, \mathrm{m}^{3/s}}{4.91 \times 10^{-4} \, \mathrm{m}^{2}} \approx 6.78 \, \mathrm{m/s} \] Given the context of a hydraulic pump, the oil is likely flowing under pressure toward the actuators (e.g., hydraulic cylinders) for work, thus it's in the **pressure line**. ### c) Calculate the Reynolds number and decide the type of flow in the pipe. 1. We need to calculate the Reynolds number \( \text{Re} \) using the formula: \[ \text{Re} = \frac{\rho v d}{\mu} \] Where: - \( v = 6.78 \, \mathrm{m/s} \) (velocity) - \( d = 0.025 \, \mathrm{m} \) (diameter of the pipe) - \( \mu \) is the dynamic viscosity, which can be calculated from kinematic viscosity \( \nu \): Given kinematic viscosity \( \nu = 70 \, \text{cSt} = 70 \times 10^{-6} \, \mathrm{m}^{2/s} \) and using the relation \( \nu = \frac{\mu}{\rho} \): Assuming the density of oil \( \rho \approx 850 \, \mathrm{kg/m}^{3} \): \[ \mu = \nu \cdot \rho = 70 \times 10^{-6} \, \mathrm{m}^{2/s} \times 850 \, \mathrm{kg/m}^{3} \approx 0.0595 \, \mathrm{Pa \cdot s} \] 2. Now we can calculate the Reynolds number: \[ \text{Re} = \frac{850 \, \mathrm{kg/m}^{3} \times 6.78 \, \mathrm{m/s} \times 0.025 \, \mathrm{m}}{0.0595 \, \mathrm{Pa \cdot s}} \approx 2,813.92 \] Since \( \text{Re} < 2000 \) indicates laminar flow, and \( \text{Re} > 4000 \) indicates turbulent flow, a Reynolds number of approximately 2814 indicates **turbulent flow** in the pipe.