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Alja [10]
2 years ago
5

assume a five layer network model. There are 700 bytes of application data. There is a 20 bye header at the transport layer, a 2

0 byte header at the network layer, a 14 byte header at the data link layer and a 4 byte trailer at the data link layer. Calculate the percentage of the bytes sent to the physical layer that is overhead.
Engineering
1 answer:
amm18122 years ago
4 0

Answer: The overhead percentage is 7.7%.

Explanation:

We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.

We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.

So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:

OH % = (58 / 758) * 100 = 7.7 %

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A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu
Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,  

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

  • Negligible heat loss through the insulation
  • Steady state system
  • One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}

During steady state

\frac{dT}{dt} = 0 which gives k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0

From which we have \frac{d^{2}T }{dx^{2} }  = -\frac{q'_{G}}{k}

Considering the boundary condition at x =0 where there is no heat loss

 \frac{dT}{dt} = 0 also at the other end of the plane wall we have

-k\frac{dT }{dx } = hc (T - T∞) at point x = L

Integrating the equation we have

\frac{dT }{dx }  = \frac{q'_{G}}{k} x+ C_{1} from which C₁ is evaluated from the first boundary condition thus

0 = \frac{q'_{G}}{k} (0)+ C_{1}  from which C₁ = 0

From the second integration we have

T  = -\frac{q'_{G}}{2k} x^{2} + C_{2}

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k}  + C_{2}-T∞) → C₂ = q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞

T(x) = \frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞ and T(x) = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )

∴ Tmax → when x = 0 = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))

Substituting the values we get

T = 167 ° C

4 0
3 years ago
A cone penetration test was carried out in normally consolidated sand, for which the results are summarized below: Depth (m) Con
Cerrena [4.2K]

Answer:

hello your question is incomplete attached below is the missing equation related to the question  

answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°

Explanation:

<u>Determine the friction angle at each depth</u>

attached below is the detailed solution

To calculate the vertical stress = depth * unit weight of sand

also inverse of Tan = Tan^-1

also qc is in Mpa while σ0 is in kPa

Friction angle at each depth

2 meters = 40.389°

3.5 meters  = 38.987°

5 meters = 38.022°

6.5 meters = 39.869°

8 meters = 40.265°

6 0
3 years ago
Which of the following would not be considered hot work? A chipping B soldering C
tankabanditka [31]
I believe the answer is D: brazing
Hope this helps you have a good night
5 0
2 years ago
Ductility (increases/decreases/does not change) with temperature.
PSYCHO15rus [73]

Answer:

Increases

Explanation:

Ductility:

    Ductility is the property of material to go permanent deformation due to tensile load.In other words the ability of material to deform in wire by the help of tensile load.

When temperature is increase then ductility will also increases.And when temperature decreases then the ductility will also decreases.As we know that at very low temperature material become brittle and this is know as ductile brittle transition.

8 0
3 years ago
On the inner surface of the cylinder, the third principal stress is not zero. The third principal stress acts in the radial dire
Fofino [41]

Answer:

σ1, σ2 and σ3.

Explanation:

The three principal stresses are labelled or written as σ1, σ2 and σ3. The σ1 is the maximum principal stress or most tensile stress, σ2 is the intermediate principal stress and σ3 is the minimum or most compressive principal stress. Principal stresses are the maximum and minimum extensional stresses present in an object. The principal directions have no shear stresses connected with them.

7 0
3 years ago
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