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Keith_Richards [23]
3 years ago
8

Compare and contrast ""centralized"" and ""decentralized"" routing algorithms. (What are the advantages and disadvantages of eac

h?) Why do we prefer a ""decentralized"" algorithm for routing messages through the internet?
Engineering
1 answer:
pantera1 [17]3 years ago
5 0

Answer:

Comparison between centralized and decentralized routing algorithms:

The major similarity between both centralized and decentralized routing algorithms is that they are both communication serving systems. They both utilize node system(e.g, computer) and are both a communication liking system( e.g, Cable).

Contrasting between centralized and decentralized routing algorithms:

There are few differences between these two type of communication link or path way system but to name a couple of them,

For centralized, there is a single client server distribution node which simply means that one or more client server system are connected to a central processing server.

This also means that if the central processing server or pathway fails, it leads to the failure of the entire system. That is, there is no sending, responding or general processing of any form of requests. Example of a system that uses the centralized routing algorithm is the google search engine.

WHILE:

for the decentralized routing algorithms, there are multiple client server distribution pathway and each server makes its own decision. Here,there is no single entity that receives and responds to the request therefore,failure of any form of central path way processing node does not lead to the failure of the whole system unlike for the centralized system.

Advantages of centralized routing algorithm:

It can be easily protected or secured due to the nature of the system. If the central node is been secured, it generally translate to the different client node being secured.

It is easy to disconnect a connected client node from the central pathway node or central server as the case may be.

Disadvantages of centralized routing algorithm:

The client nodes are totally dependent on the central node or server so if there is a failure in the central server, the client node is then totally shut down.

Advantages of decentralized routing algorithm:

There is a random distribution of data on all the processing node or server which automatically creates a form of balance within the system. This leads to minimal or no down time processing client request.

Disadvantages of decentralized routing algorithm:

Due to the nature or fact that there are multiple processing system for different client node, it is difficult to detect which client node or request processing server is faulty. This can lead to delay in the fixing of fault in the system should it arise.

Why do we prefer a ""decentralized"" algorithm for routing messages through the internet?

The major reason why decentralized algorithm routing is preferred is because of the level of security attached to it. Each processing servers are secured independently and there is privilege of utilizing independent networking system.

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What can your employer do to protect you from overhead power lines?
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Answer:

Have the power company install insulated sleeves (also known as “eels”) over power lines.​

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Explanation:

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5 0
3 years ago
A Rankine steam power plant is considered. Saturated water vapor enters a turbine at 8 MPa and exits at condenser at 10 kPa. The
Ray Of Light [21]

Answer:

0.31

126.23 kg/s

Explanation:

Given:-

- Fluid: Water

- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%

- Pump: Isentropic

- Net cycle-work output, Wnet = 100 MW

Find:-

- The thermal efficiency of the cycle

- The mass flow rate of steam

Solution:-

- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.

First process: Isentropic compression by pump

       P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )

      h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )

       s1 = s-P1 = 0.6492 KJ/kg.K

       v1 = v-P1 = 0.001010 m^3 / kg

       

       P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )

       s2 = s1 = 0.6492 KJ/kg.K   .... ( compressed liquid )

- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:

   

                           w_p = v_1*( P_2 - P_1 )\\\\w_p = 0.001010*( 8000 - 10 )\\\\w_p = 8.0699 \frac{KJ}{kg}

- From the following relation we can determine ( h2 ) as follows:

                          h2 = h1 + wp

                          h2 = 191.81 + 8.0699

                          h2 = 199.88 KJ/kg

                           

Second Process: Boiler supplies heat to the fluid and vaporize

- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).

- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).

              P3 = 8 MPa

              T3 = ?  ( assume fluid exist in the saturated vapor phase )

              h3 = hg-P3 = 2758.7 KJ/kg

              s3 = sg-P3 = 5.7450 KJ/kg.K

- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:

                          q_s = h_3 - h_2\\\\q_s = 2758.7 -199.88\\\\q_s = 2558.82 \frac{KJ}{kg}

Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).

- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.

- Under the isentropic conditions the steam exits the turbine at the following conditions:

             P4 = 10 KPa

             s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )

             

- Compute the quality of the mixture at condenser inlet by the following relation:

                           x = \frac{s_4 - s_f}{s_f_g} \\\\x = \frac{5.745- 0.6492}{7.4996} \\\\x = 0.67947

- Determine the isentropic ( h4s ) at this state as follows:

                          h_4_s = h_f + x*h_f_g\\\\h_4_s = 191.81 + 0.67947*2392.1\\\\h_4_s = 1817.170187 \frac{KJ}{kg}        

- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:

                         h4 = h_3 - n_t*(h_3 - h_4_s ) \\\\h4 = 2758.7 - 0.85*(2758.7 - 181.170187 ) \\\\h4 = 1958.39965 \frac{KJ}{kg} \\

- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.

                        w_t = h_3 - h_4\\\\w_t = 2758.7-1958.39965\\\\w_t = 800.30034 \frac{KJ}{kg}

- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.

                       W_n_e_t = flow(m) * ( w_t - w_p )\\\\flow ( m ) = \frac{W_n_e_t}{w_t - w_p} \\\\flow ( m ) = \frac{100000}{800.30034-8.0699} \\\\flow ( m ) = 126.23 \frac{kg}{s}

Answer: The mass flow rate of the steam would be 126.23 kg/s

- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):

                        n_t_h = \frac{W_n_e_t}{flow(m)*q_s} \\\\n_t_h = \frac{100000}{126.23*2558.82} \\\\n_t_h = 0.31

Answer: The thermal efficiency of the cycle is 0.31

       

   

7 0
4 years ago
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MAXImum [283]
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Answer:

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