Question

1. A fluid with viscosity 1.2 Pa s moves through a 200-m-long pipe with a 20 cm diameter. What difference in pressure between the ends of the pipe is needed for a flow of 0.1 m^3/s?
1. 611 kPa

2. 380 kPa

3. 38 mPa

4. 0.61 Pa

2. A torpedo with characteristic length L=1.2 m is traveling through water (density 1000 kg/m^3, viscosity 1.0 mPa s) at speed 10 m/s. What is Reynold's number for this torpedo?

1. 1.2x10^4

2. 1.2

3. 1.2x10^7

4. 1.2x10^-3

3. A torpedo with characteristic length L=1.2 m is traveling through water (density 1000 kg/m^3, viscosity 1.0 mPa s) at speed 10 m/s. Is the flow of water around the torpedo laminar?

1. It is unstable, so it could be either laminar or turbulent.

2. It is neither because water isn't real.

3. No, it is turbulent.

4. Yes, it is laminar.

Answer 1

(1) The difference in pressure between the ends of the pipe is 611 kPa.

(2) The Reynold's number for this torpedo is 1.2  x 10⁷.

(3) The flow of water around the torpedo is turbulent.

Difference in pressure between the ends of the pipe

The difference in pressure is calculated by applying Poiseuille formula.

$V = \frac{\pi \Delta P r^4 t}{8\eta L} \\\\\frac{V}{t} = \frac{\pi \Delta P r^4 }{8\eta L}\\\\Q = \frac{\pi \Delta P r^4 }{8\eta L}$

where;

• Q is flow rate
• r is radius of the pipe
• L is length of the pipe
• η is viscosity
• ΔP is change in pressure

$\Delta P = \frac{8Q\eta L}{\pi r^4}$

$\Delta P = \frac{8(0.1)(1.2)(200)}{\pi (0.1)^4} \\\\\Delta P = 611,155 \ Pa\\\\\Delta P \approx 611 \ kPa$

Reynold's number for the torpedo

$Re = \frac{Vl \rho }{\mu} \\\\Re = \frac{(10)(1.2)(1000)}{1 \times 10^{-3}} \\\\Re = 1.2 \times 10^7$

Laminar or turbulent flow

• A flow is considered laminar if the Reynolds number is up to 2300.
• A flow is considered turbulent if the Reynolds number is greater than 3500.

The calculated Reynold's number (1.2 x 10⁷) is greater than 3500. Thus, the flow is turbulent.

Learn more about turbulent flow here: brainly.com/question/12081428

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