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Orlov [11]
3 years ago
7

Flow at the inlet and outlets can be assumed One-dimensional for the sake of simplicity. (a) Explain how this assumption simplif

ies the Reynolds Transport Theorem equation (provide complete, detailed answer) (b) This assumption introduces errors into the calculations, as it assumes flow properties do not vary across the inlet/outlet area. How this error changes with the area (does it increase or decrease)?
Engineering
1 answer:
Anna [14]3 years ago
5 0

Answer:

a) reduces the need to determine complicated flow properties.

b) The error decreases with the change in the area.

Explanation:

a) The mechanical properties of a fluid can sometimes be complicated. This results in an a need to calculate all the forces acting on the surface of the fluid particle. This then introduces the concept of the resolving the various forces. To reduce the error, we use the control volume. The  Reynolds Transport Theorem states that the rate of change of an extensive property N, for a system is equal to the time rate of change of N in the control volume and the net rate of flux of the property N through the control surface.

b) By making the assumptions, it decreases the likelihood of errors occurring in the equation.

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3. (a) (5 points) Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queue
Bezzdna [24]

Answer:

(N-1) × (L/2R) = (N-1)/2

Explanation:

let L is length of packet

R is rate

N is number of packets

then

first packet arrived with 0 delay

Second packet arrived at = L/R

Third packet arrived at = 2L/R

Nth packet arrived at = (n-1)L/R

Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R

Now

L / R = (1000) / (10^6 ) s = 1 ms

L/2R = 0.5 ms

average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2

the average queuing delay of a packet = 0 ( put N=1)

4 0
3 years ago
What tool should be used to loosen or tighten brake or fuel lines?
emmainna [20.7K]

The tool you would use are brake line wrenches.

4 0
3 years ago
A square aluminum plate 5 mm thick and 150 mm on a side is heated while vertically suspended in quiescent air at 75°c. determine
Doss [256]

By using the boundary layer equation, the average heat transfer coefficient for the plate is equal to 4.87 W/m²k.

<u>Given the following data:</u>

Surface temperature = 15°C

Bulk temperature = 75°C

Side length of plate = 150 mm to m = 0.15 meter.

<h3>How to calculate the average heat transfer coefficient.</h3>

Since we have a quiescent room air and a uniform pole surface temperature, the film temperature is given by:

T_f=\frac{T_{s} + T_{\infty} }{2} \\\\T_f=\frac{15 + 75 }{2} \\\\T_f = 45

Film temperature = 45°C to K = 273 + 45 = 318 K.

For the coefficient of thermal expansion, we have:

\beta =\frac{1}{T_f} \\\\\beta =\frac{1}{318}

From table A-9, the properties of air at a pressure of 1 atm and temperature of 45°C are:

  • Kinematic viscosity, v = 1.750 \times 10^{-5} m²/s.
  • Thermal conductivity, k = 0.02699 W/mk.
  • Thermal diffusivity, α = 2.416 \times 10^{-5} m²/s.
  • Prandtl number, Pr = 0.7241.

Next, we would solve for the Rayleigh number to enable us determine the heat transfer coefficient by using the boundary layer equations:

R_{aL}=\frac{g\beta \Delta T l^3}{v\alpha } \\\\R_{aL}=\frac{9.8 \;\times \;\frac{1}{318} \;\times \;(75-15) \;\times \;0.15^3 }{1.750 \times 10^{-5}\; \times \;2.416 \times 10^{-5} } \\\\R_{aL}=\frac{9.8\; \times 0.00315 \;\times \;60\; \times\; 0.003375 }{4.228 \times 10^{-10}  }\\\\R_{aL}=1.48 \times 10^{7}

Also take note, g(Pr) is given by this equation:

g(P_r)=\frac{0.75P_r}{[0.609 \;+\;1.221\sqrt{P_r}\; +\;1.238P_r]^\frac{1}{4} } \\\\g(P_r)=\frac{0.75(0.7241)}{[0.609 \;+\;1.221\sqrt{0.7241}\; +\;1.238(0.7241)]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{[0.609 \;+\;1.221\sqrt{0.7241}\; +\;1.238(0.7241)]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{[2.5444]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{1.2630 }

g(Pr) = 0.430

For GrL, we have:

G_{rL}=\frac{R_{aL}}{P_r} \\\\G_{rL}=\frac{1.48 \times 10^7}{0.7241} \\\\G_{rL}=1.99 \times 10^7

Since the Rayleigh number is less than 10⁹, the flow is laminar and the condition is given by:

N_{uL}=\frac{h_{L}L}{k} = \frac{4}{3} (\frac{G_{rL}}{4} )^\frac{1}{4} g(P_r)\\\\h_{L}=\frac{0.02699}{0.15} \times  [\frac{4}{3} \times  (\frac{1.99 \times 10^7}{4} )^\frac{1}{4} ]\times 0.430\\\\h_{L}= 0.1799 \times 62.9705 \times 0.430\\\\h_{L}=4.87\;W/m^2k

Based on empirical correlation method, the average heat transfer coefficient for the plate is given by this equation:

N_{uL}=\frac{h_{L}L}{k} =0.68 +  \frac{0.670 R_{aL}^\frac{1}{4}}{[1+(\frac{0.492}{P_r})^\frac{9}{16}]^\frac{4}{19}   } \\\\h_{L}=\frac{0.02699}{0.15} \times ( 0.68 +  \frac{0.670 (1.48 \times 10^7)^\frac{1}{4}}{[1+(\frac{0.492}{0.7241})^\frac{9}{16}]^\frac{4}{19}   })\\\\h_{L}=4.87\;W/m^2k

Read more on heat transfer here: brainly.com/question/10119413

3 0
2 years ago
What historical event allowed both aerospace fields to make enormous strides forward?
Lostsunrise [7]

Answer:

The world wars. Especially world War II

Explanation:

The world wars were the single most notable events that led to the leap in development of the aircraft from a simple apparatus capable of flight to a sophisticated flight machine. Airplane production was at its maximum during these periods (especially world war II) and has never seen that capacity of production again since. Countries focused on research that could add even the slightest enhancements to their aircraft in order to for it to surpass that of other countries.

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3 years ago
If the symbol is on the top and bottom of the reference line, we call this __________ sides.
sattari [20]
I think the correct answer is B
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