Physics, 03.01.2020 04:31 haley5306

At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. the operating temperatures of the first are 750◦c and 440◦c respectively, and of the second 415◦c and 270◦c.

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Answer from: emmaraeschool

The rate at which coal burned is 111.12 kg/s.

Explanation:

Given that,

First initial temperature =750°C

First final temperature =440°C

Second initial temperature =415°C

Second final temperature =270°C

Suppose If the heat of combustion of coal is $2.8×10^{7}\ J/kg$ , at what rate must coal be burned if the plant is to put out 950 MW of power? Assume the efficiency of the engines is 65% of the Carnot efficiency.

The work done by first engine is

W=eQ

The work done by second engine is

W'=e'Q'

W'=e'Q(1-e)

Total out put of the plant is given by

$W+W'=950\ MW$

Put the value into the formula

$eQ+e'Q(1-e)=950\times10^{6}$ ....(I)

We need to calculate the efficiency of first engine

Using formula of efficiency

$e=65%(1-\dfrac{T_{L}}{T_{H}})$

$e=0.65(1-\dfrac{440+273}{750+273})$

e=0.196

We need to calculate the efficiency of second engine

Using formula of efficiency

$e'=65%(1-\dfrac{T_{L}}{T_{H}})$

$e'=0.65(1-\dfrac{270+273}{415+273})$

e'=0.136

Put the value of efficiency for first and second engine in the equation (I)

$Q(0.196+0.136(1-0.196))=950\times10^{6}$

$Q=\dfrac{950\times10^{6}}{(0.196+0.136(1-0.196))}$

$Q=3111.24\times10^{6}\ W$

We need to calculate the rate at which coal burned

Using formula of rate

$R=\dfrac{Q}{H_{coal}}$

$R=\dfrac{3111.24\times10^{6}}{2.8×10^{7}}$

$R=111.12\ kg/s$

Hence, The rate at which coal burned is 111.12 kg/s.

Answer from: Quest

Show him another sport or something

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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