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Elza [17]
3 years ago
13

Water flows in a tube that has a diameter of D= 0.1 m. Determine the Reynolds number if the average velocity is 10 diameters per

second. (b) Repeat the calculations if the tube is a nanoscale tube with a dimeter of D= 100 nm.
Engineering
1 answer:
Cloud [144]3 years ago
6 0

Answer:

a) Re_{D} = 111896.745, b) Re_{D} = 1.119\times 10^{-7}

Explanation:

a) The Reynolds number for the water flowing in a circular tube is:

Re_{D} = \frac{\rho\cdot v\cdot D}{\mu}

Let assume that density and dynamic viscosity at 25 °C are 997\,\frac{kg}{m^{3}} 0.891\times 10^{-3}\,\frac{kg}{m\cdot s}, respectively. Then:

Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (1\,\frac{m}{s} )\cdot (0.1\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }

Re_{D} = 111896.745

b) The result is:

Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (10^{-6}\,\frac{m}{s} )\cdot (10^{-7}\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }

Re_{D} = 1.119\times 10^{-7}

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The kinetic energy of A is twice the kinetic energy of B

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Engineers in Russia invented a new way to create colorful art with a __________.
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The Engineers in Russia invented a new way to create colorful art with constructivist art.

<h3>What is constructivist art?</h3>

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2 years ago
An escalator handles a steady load of 26 people per minute in elevating them from the first to the second floor through a vertic
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Answer:

\eta = 70.711\,\%

Explanation:

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\dot W  = \dot U_{g}

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\dot W = \left(26\,\frac{persons}{min}\right)\cdot (124\,lbm)\cdot \left(32.174\,\frac{ft}{s^{2}}\right)\cdot \left(\frac{1\,lbf}{32.174\,\frac{lbm\cdot ft}{s^{2}} } \right)\cdot (27.5\,ft)

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The mechanical efficiency of the escalator is:

\eta = \frac{2.687\,hp}{3.8\,hp}\times 100\,\%

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3 0
3 years ago
I will Brainlist<br> "Burning and Inch". Describe this measurement.
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Read 2 more answers
Applying the Entropy Balance: Closed Systems Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston–
Mrrafil [7]

Answer:

a) the amount of energy produced in kJ/K is 0.73145 kJ/K

b) the amount of energy produced in kJ/K is 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

Explanation:

Draw the T-s diagram.

a)

C_p = 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280

R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar

Δs = m[c_p ln(\frac{T_2}{T_1}) - Rln(\frac{P_2}{P_1})]

Substitute all parameters in the equation

Δs = 5[(0.939) ln(\frac{520}{280}) - (\frac{8.314}{44.01})ln(\frac{20}{2})]

Δs = 5 kg × 0.14629 kJ/kg.K

    = 0.73145 kJ/K

b)

Δs = m[\frac{s^0(T_2) - s^0(T_1)}{M} - Rln(\frac{P_2}{P_1})]

Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K

           T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K

R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar

Δs = 5[\frac{236.575 - 211.376}{44.01} - (\frac{8.314}{44.01})ln(\frac{20}{2})]

    = 5 kg (0.13795 kJ/kg.K)

    = 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

7 0
3 years ago
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