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Caitlin T.
Mechanical Engineering Grad Student and Tutor
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Physics (Fluid Mechanics)
TutorMe
Question:

Find the Reynolds number for the following pipe flow: Water flows through a pipe at 3.6 m/s and 273 K. Pipe diameter is 0.05 m. State whether this flow is laminar or turbulent.

Caitlin T.
Answer:

The equation for Reynolds number is as follows: $$ Re = \frac{\rho \cdot V\cdot D}{\mu } $$ Where $$ \rho $$ is the fluid density, D is the hydraulic diameter (diameter of the pipe in this case), V is the mean fluid velocity, and $$ \mu $$ is the kinematic viscosity. At 273 K, $$ \rho $$ = 999 $$ kg/m^{3} $$, $$ \mu $$ = 1.12 E-3$$N\cdot s / m^{2}$$ Since 1 N = $$ 1 \frac{kg\cdot m}{s^{2}} $$, 1.12 E-3 $$N\cdot s / m^{2}$$ = 1.12 E-3 $$ \frac{kg\cdot m}{s^{2}} \cdot s / m^{2} $$ = 1.12 E-3 $$ kg/(m \cdot s) $$ The Reynolds number would then be: $$ Re = \frac{999 kg/m^{3} \cdot 3.6 m/s \cdot 0.05 m}{1.12 E-3 kg/(m \cdot s) } $$ $$ Re = 160,554 $$ The flow is turbulent as the Reynolds number is well over 4000.

Basic Math
TutorMe
Question:

$​$ \frac{3}{5} - \frac{2}{7} = ?$​$

Caitlin T.
Answer:

In order to easily subtract fractions, one must find a common denominator. This can be done by finding a common multiple of the two denominators: $​$ 5*7 = 35 $​$ In order to have 35 be the denominator for both of the fractions, you must multiply the 3/5 fraction by 7 on both top and bottom, and the 2/7 fraction by 5 on both top and bottom: $​$ 3*7 = 21 $​$ $​$ 5*7 = 35 $​$ $​$ \frac{3}{5} = \frac{21}{35}$​$ $​$ 2*5 = 10 $​$ $​$ 7*5 = 35 $​$ $​$ \frac{2}{7} = \frac{10}{35}$​$ You obtain this new equation: $​$ \frac{21}{35} - \frac{10}{35} = ?$​$ Then, simply subtract the first numerator from the second numerator: $​$ 21-10 = 11 $​$ $​$ \frac{21}{35} - \frac{10}{35} = \frac{11}{35}$​$

Mechanical Engineering
TutorMe
Question:

Given a closed room with an open refrigerator and fan to disperse the cool air, what will happen to the temperature of the room over time?

Caitlin T.
Answer:

In this situation, energy is being added to the room in the form of electrical energy to power the refrigerator and fan. Although the interior of the refrigerator will be cool, the overall temperature of the room will rise as more energy is added. If there were an open window and heat exchange could occur, the situation may be different.

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