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tangare [24]
3 years ago
6

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

d. Each packet is of length L and the link has transmission rate R. What is the average queuing delay for the N packets? (b) (5 points) Now suppose that one of such N packets arrives to the link every L/2R seconds. What is the average queuing delay of a packet?
Engineering
1 answer:
Bezzdna [24]3 years ago
4 0

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)

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A double-pane insulated window consists of two 1 cm thick pieces of glass separated by a 1.8 cm layer of air. The window measure
Elanso [62]

Answer:

(b). T = 22.55 ⁰C

(c). q = 557.8 W

Explanation:

we take follow a step by step process to solving this problem.

from the question, we have that

The two glass pieces is separated by a 1.8 cm distance layer of air.

the thickness of glass piece is 1 cm

width = 4 m

the height = 3 m

(a). the sketch of the thermal circuit is uploaded in the picture below.

(b).  the thermal resistance due to the conduction in the first glass plane is given thus;

R₁ = Lg / Kg A ................(1)

given that Kg rep. the thermal conductivity of the glass plane

A = conduction surface area

Lg = Thickness of glass plane4

taking the thermal conductivity of glass plane as Kg = 0.78 w/mk

inputting values into equation (1) we have,

R₁ = [1 (cm) ˣ 1 (m)/100 (cm)] / [(0.78 w/mk)(4m ˣ 3m)]

R₁ = 1.068 ˣ 10 ⁻³ k/w

Being that we have same thermal resistance in the first and second plane,

therefore R₁ = R₃ = 1.068 ˣ 10 ⁻³ k/w

⇒ Also the thermal resistance between air and glass as a result of the conduction by the layer is given thus

R₂ = La/KaA .....................(2)

given Ka = thermal conductivity of air

A = surface area

La = thickness of air

substituting values into the equation we have

R₂ = [1.8 (cm) ˣ 1 (m)/100 (cm)] / [(0.0262 w/mk)(4m ˣ 3m)]

R₂ = 5.73 ˣ 10⁻² k/w

Given the thermal resistance on the outer surface due to convection, we have

R₄ = 1/hA

inputting value gives R₄ = 1 / (12 w/m² ˣ 12m) = 6.94 ˣ 10⁻³k/w

R₄ = 6.94 ˣ 10⁻³k/w

Finally the sum total of thermal resistance = R₁ + R₂ + R₃ + R₄

R-total = 0.0663 kw

From this we can calculate the rate of heat loss

using  q = Ti - To / R-total ..............(3)

given Ti and To is the inside and outside temperature i.e. 27⁰C and -10⁰C

from equation (3),

q = 27- (-10) / 0.0063 = 557.8 W

q = 557.8 W  

⇒ Applying the heat transfer formula for inside surface glass temperature gives;

q = Ti - T₂ / R₃ + R₄

T₂ = Ti - q (R₃ + R₄)

T₂ = 27 - 557.8 (1.068ˣ10⁻³ + 6.94ˣ10⁻³ ) = 22.55°C

T₂ = 22.55°C

cheers i hope this helps

8 0
3 years ago
Koch traded Machine 1 for Machine 2 when the fair market value of both machines was $60,000. Koch originally purchased Machine 1
Mariana [72]

Answer:

Koch's adjusted basis in machine 2 after the exchange is $60,000

Explanation:

given data

fair market value = $60,000

originally purchased Machine 1 = $76,900

Machine 1 adjusted basis = $40,950

Machine 2 seller purchase = $64,050

Machine 2 adjusted basis = $55,950

solution

As he exchanged machine for another at $60,000

and this exchanged in fair market

so adjusted basis =  $50,000

Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell

Adjusted basis increases when a person deducts expenses from factor taxes and operating statements

so Koch's adjusted basis in machine 2 after the exchange is $60,000

3 0
3 years ago
If the density of states function in the conduction band of a particular semiconductor is a constant equal to K, derive the expr
s2008m [1.1K]

Answer:

full details of the answer is attached

5 0
3 years ago
Comparison of density values determines whether an item will float or sink in water. For each of the values below, determine the
geniusboy [140]

Answer:

a) the object floats

b) the object floats

c) the object sinks

Explanation:

when an object is less dense than in the fluid in which it is immersed, it will float due to its weight and volume characteristics, so to solve this problem we must find the mass and volume of each object in order to calculate the density and compare it with that of water

a)

volumen for a cube

V=L^3

L=1.53in=0.0388m

V=0.0388 ^3=5.8691x10^-5m^3=58.69ml

density=m/v

density=13.5g/58.69ml=0.23 g/ml

The wooden block floats  because it is less dense than water

b)

m=111mg=0.111g

density=m/v

density=0.111g/0.296ml=0.375g/ml

the metal paperclip floats   because it is less dense than water

c)

V=0.93cups=220.0271ml

m=0.88lb=399.1613g

Density=m/v

density=399.1613/220.027ml=1.8141g/ml

the apple sinks   because it is  denser than water

4 0
3 years ago
Q1. (20 marks) Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high- temper
Anvisha [2.4K]

The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

\frac{w}{Q1} = \frac{35}{100}

W = 1.2*\frac{35}{100}*1000kj/s

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

<h3>What is the workdone by this engine?</h3>

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ

<h3>Cournot efficiency = W/Q1</h3>

= 1200 - 369.6/1200

= 69.2 percent

change in s is zero for the reversible heat engine.

Read more on enthropy here: brainly.com/question/6364271

6 0
2 years ago
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