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ArbitrLikvidat [17]

3 years ago

14

A heat engine that receives heat from a furnace at 1200°C and rejects waste heat to a river at 20°C has a thermal efficiency of

50 percent.
Determine the second-law efficiency of this power plant.

1 answer:

viktelen [127]3 years ago

3 0

Answer:

second-law efficiency = 62.42 %

Explanation:

given data

temperature T1 = 1200°C = 1473 K

temperature T2 = 20°C = 293 K

thermal efficiency η = 50 percent

solution

as we know that thermal efficiency of reversible heat engine between same temp reservoir

so here

efficiency ( reversible ) η1 = 1 - $\frac{T2}{T1}$ ............1

efficiency ( reversible ) η1 = 1 - $\frac{293}{1473}$

so efficiency ( reversible ) η1 = 0.801

so here second-law efficiency of this power plant is

second-law efficiency = $\frac{thernal\ efficiency}{0.801}$

second-law efficiency = $\frac{50}{0.801}$

second-law efficiency = 62.42 %

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The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet

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Answer:Counter,

0.799,

1.921

Explanation:

Given data

$T_{h_i}=200^{\circ}C$

$T_{h_o}=120^{\circ}C$

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$T_{c_o}=125^{\circ}C$

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

$m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]$

$\frac{m_hc_{ph}}{m_cc_{pc}}$ = $\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )$

we can see that heat capacity of hot fluid is minimum

Also from energy balance

$Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )$

$NTU=\frac{UA}{\left ( mc_p\right )_{h}}$ = $\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}$

$T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}$

$T_m=41.63^{\circ}C$

NTU=1.921

$And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}$

$\epsilon =\frac{1-0.2669}{1-0.0834}$

$\epsilon =0.799$

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In a black box experiment, when the amount of material exiting a closed system is less than the amount of material entering the

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When the material that exits is lesser in amount than that of the entering material in a black box experiment, the parts of the system need to be changed.

<h3>What happens in a black box experiment?</h3>

In a black box experiment, the experimenters need to make assumptions regarding the drawing of conclusions. One such conclusion is the amount of material that exits.

If such amount is lesser than the one that enters the system, such experiment concludes that it is the time to change the parts of the system.

Hence, option D holds true regarding the black box experiment.

Learn more about black box experiment here:

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Answer:

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

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Answer:

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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:

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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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3 years ago