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Find the number of odd and even vertices in each network. Imagine travelling each network to see if it can be traced without backtracking.

Short Answer

Expert verified

The number of odd and even vertices is 2 and 6.

Step by step solution

01

Step 1. Given information:

The given figure is as follows:

02

Step 2. Concept use.

A vertex with odd number of edges attached to it is called odd vertex and a vertex with even number of edges attached to it is called even vertex.

03

Step 3. Applying the concept. 

In the given network, by observing, the total number of vertices =8.

A vertex with odd number of edges attached to it is called odd vertex and a vertex with even number of edges attached to it is called even vertex.

Hence, the number of odd vertices =2.

The number of even vertices =6.

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Most popular questions from this chapter

The number of odd vertices will tell you whether or not a network can be traced without backtracking. Do you see how? If not, read on.

suppose that a given network can be traced without backtracking.

a. Consider a vertex that is neither the start nor end of a journey through this network. Is such a vertex odd or even?

b. Now consider the two vertices at the start and finish of a journey through this network. Can both of these vertices be odd? Even?

c. Can just one of the start and finish vertices be odd?

Find the measure of central โˆ 1.

In exercises 8-13 find the angle or the arc named.

โˆ GQF

To get more than an infinitesimal amount of work out of a Carnot engine, we would have to keep the temperature of its working substance below that of the hot reservoir and above that of the cold reservoir by non-infinitesimal amounts. Consider, then, a Carnot cycle in which the working substance is at temperature Thw as it absorbs heat from the hot reservoir, and at temperature Thwas it expels heat to the cold reservoir. Under most circumstances the rates of heat transfer will be directly proportional to the temperature differences:

Qhฮ”t=KThโˆ’ThwandQcฮ”t=KTcwโˆ’Tc

I've assumed here for simplicity that the constants of proportionality (K)are the same for both of these processes. Let us also assume that both processes take the

same amount of time, so the ฮ”tโ€ฒsare the same in both of these equations.

(a) Assuming that no new entropy is created during the cycle except during the two heat transfer processes, derive an equation that relates the four temperatures Th,Tc,Thw,andTcw

(b) Assuming that the time required for the two adiabatic steps is negligible, write down an expression for the power (work per unit time) output of this engine. Use the first and second laws to write the power entirely in terms of the four temperatures (and the constant (K)), then eliminate Tcwusing the result of part (a).

(c) When the cost of building an engine is much greater than the cost of fuel (as is often the case), it is desirable to optimize the engine for maximum power output, not maximum efficiency. Show that, for fixed ThandTc, the expression you found in part (b) has a maximum value at Thw=12Th+ThTc. (Hint: You'll have to solve a quadratic equation.) Find the corresponding expression for TCW

(d) Show that the efficiency of this engine is 1โˆ’Tc/Th. Evaluate this efficiency numerically for a typical coal-fired steam turbine with Th=600โˆ˜CTc=25โˆ˜C, and compare to the ideal Carnot efficiency for this temperature range. Which value is closer to the actual efficiency, about 40%, of a real coal-burning power plant?

Given PAยฏand PBยฏ are tangents to โŠ™o.

Use the diagram at the right to explain how the corollary on page 333

follows from Theorem9-1.

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