Ask question

Ask question

All categories

snow_tiger [21]

4 years ago

8

A heat engine absorbs 2500 J of heat from a hot reservoir and expels 1000 J to a cold reservoir. When it is run in reverse, with

the same reservoirs, the engine pumps 2500 J of heat to the hot reservoir, requiring 1500 J of work to do so. Find the ratio of the work done by the heat engine to the work done by the pump. Is the heat engine reversible?

1 answer:

Alexxandr [17]4 years ago

3 0

To solve this problem, we must simply use the concept of Total Energy transferred both in terms of work and heat. This is basically conjugated in the first law of thermodynamics.

If we take the heat absorbed as positive and the expelled as negative we have that the total work done in the heat engine is

$W_1 = 2500-1000$

$W_1 = 1500J$

For the case of the engine pumps the Energy absorbed is

$W_2 = 1500J$

In this way the ratio between the two would be

$Ratio = \frac{W_1}{W_2} = \frac{1500}{1500} = 1$

So it is reversible, because the state of efficiency of the body is totally efficient.

You might be interested in

Substance Specific Heat (cal/g°C) Specific Heat (J/g°C) water 1.00 4.18 steam 0.96 4.02 alcohol 0.59 2.47 ice 0.50 2.09 wood 0.4

Jobisdone [24]

Answer:

C) 43,2°C

Explanation:

<em>Sensible heat</em> is the amount of thermal energy that is required to change the temperature of an object, the equation for calculating the heat change is given by:

Q=msΔT

where:

Q, heat that has been absorbed or realeased by the substance [J]
m, mass of the substance [g]
s, specific heat capacity [J/g°C]
ΔT, changes in the substance temperature [°C]

To solve the problem, we clear ΔT of the equation and then replace our data:

Q=890 [J],

m=16,6 [g],

s=2,74 [J/g°C]

Q=msΔT.......................ΔT=Q/ms

Δ $T=\frac{890 J}{16,6g*2,47\frac{J}{gC}}=21,7$ °C

As:

ΔT=Tfinal-Tinitial

Tfinal=ΔT+Tinitial

Tfinal=21,7+21,5=43,2°C

The final temperature of the ethanol is 43,2°C.

4 0

4 years ago

The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels along the smooth guide. Determine its s

Gnoma [55]

Answer:

The speed at point B is 5.33 m/s

The normal force at point B is 694 N

Explanation:

The length of the spring when the collar is in point A is equal to:

$lA=\sqrt{0.2^{2}+0.2^{2} }=0.2\sqrt{2}m$

The length in point B is:

lB=0.2+0.2=0.4 m

The equation of conservation of energy is:

$(Tc+Ts+Vc+Vs)_{A}=(Tc+Ts+Vc+Vs)_{B}$ (eq. 1)

Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2

in point B: Ts=0, Vc=0, Tc = 1/2 mcVB^2, Vs=1/2k(lB-lul)^2

Replacing in eq. 1:

$\frac{1}{2}m_{c}v_{A}^{2}+0+m_{c}gh_{A}+ \frac{1}{2}k(l_{A}-l_{ul}) ^{2}=\frac{1}{2}m_{c}v_{B}^{2}+0+0+\frac{1}{2}k(l_{B}-l_{ul}) ^{2}$

Replacing values and clearing vB:

vB = 5.33 m/s

The balance forces acting in point B is:

Fc-NB-Fs=0

$\frac{m_{C}v_{B}^{2} }{R}-N_{B}-k(l_{B}-l_{ul})=0$

Replacing values and clearing NB:

NB = 694 N

6 0

3 years ago

Read 2 more answers

A unidirectional E-Glass fiber-epoxy composite material contains 61% by volume E-Glass fibers stressed under isostrain condition

zalisa [80]

Answer:

The total load carried by the fiber will be "98%".

Explanation:

The given values are:

$V_{f}=0.61$

$V_{m}=1-V_{f}$

$=1-0.61$

$=0.39$

$E_{f}=10 \ Mpa$

$\sigma_{f}=0.35 \ Mpa$

$E_{m}=0.45 \ Mpa$ , $\sigma_{m}=9\times 10^{-3} \ Mpa$

As we know,

⇒ $E_{e}=fE_{f}+mE_{m}$

On putting the estimated values, we get

⇒ $=0.61\times 10+0.39\times 0.95$

⇒ $=6.27 \ Mpa$

Now,

⇒ $\sigma_{c}=f\sigma_{f}+m\sigma_{m}$

On putting the estimated values, we get

⇒ $=0.61\times 0.35+0.39\times 0.009$

⇒ $=0.217 \ Mpa$

Therefore,

The load carried by fiber,

$=\frac{f\sigma_{f}}{\sigma_{c}}$

$=\frac{0.35\times 0.61}{0.217}$

$=0.98$ i.e., 98%

4 0

4 years ago

If the feedforward path of a control system contains at least one integrating element, then the output continues to change as lo

Thepotemich [5.8K]

Answer:

The attached system shows that there’s an integrator between the point where disturbance enters the system and error measuring element. A any time when R(s)=0 then

$\frac {C(s)}{D(s)}=\frac {G(s)}{1+G_c(s)G(s)}$ and considering that $E(s)=D(s)-G_c(s)C(s)$ then

$\frac {E(s)}{D(s)}=1-(\frac {C(s)}{D(s)})G_c(s)$

$\frac {E(s)}{D(s)}=1-(\frac {G(s)}{1+G_c(s)D(s)})G_c(s)$

$\frac {E(s)}{D(s)}=\frac {1}{1+G_c(s)G(s)}$

$E(s)=\frac {D(s)}{1+G_c(s)G(s)}$

For ramp disturbance d(t)=at

$D(s)=\frac {a}{s^{2}}$ therefore, the steady state error is given by

$e(\infty)= \lim_{s \to 0} s E(s)$

$e(\infty)= \lim_{s \to 0} s [\frac {D(s)}{1+G_c(s)G(s)}]$

$e(\infty)= \lim_{s \to 0} s [\frac {a}{s^{2}+s^{2}G_c(s)G(s)}]$

$e(\infty)= \lim_{s \to 0} s [\frac {a}{s+sG_c(s)G(s)}]$

$e(\infty)= \lim_{s \to 0} s [\frac {a}{sG_c(s)G(s)}]$

Whenever $G_c(s)$ has a double intergrator, the error $e(\infty)$ becomes zero

3 0

3 years ago

Rr4rrrjrjdjdjdjdndndndnndjdjdjndjdjdkdkd

vodomira [7]

Answer:

37 degrees Fahrenheit

4 0

3 years ago

Read 2 more answers