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2 years ago

Is an isothermal process necessarily internally reversible? Explain your answer with an example

Engineering

1 answer:

torisob [31]2 years ago

5 0

Answer:

please give me brainlist and follow

Explanation:

Example of an irreverseble isothermal process is mixing of two fluids on the same temperature - it requires a lot of energy to unmix Jack and coke. ... Example of an reversible process with changing temperature is isentropic expansion.

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Compared to 15 mph on a dry road, about how much longer will it take for

Marysya12 [62]

Answer:

8 to 10 times

Explanation:

For dry road

u= 15 mph ( 1 mph = 0.44 m/s)

u= 6.7 m/s

Let take coefficient of friction( μ) of dry road is 0.7

So the de acceleration a = μ g

a= 0.7 x 10 m/s ² ( g=10 m/s ²)

a= 7 m/s ²

We know that

v= u - a t

Final speed ,v=0

0 = 6.7 - 7 x t

t= 0.95 s

For snow road

μ = 0.4

de acceleration a = μ g

a = 0.4 x 10 = 4 m/s ²

u= 30 mph= 13.41 m/s

v= u - a t

Final speed ,v=0

0 = 30 - 4 x t'

t'=7.5 s

t'=7.8 t

We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.

8 to 10 times

7 0

3 years ago

Read 2 more answers

A motorist enters a freeway at 25 mi/h and accelerates uniformly to 65 mi/h. From the odometer in the car, the motorist knows th

Helga [31]

Answer:

a) 2.2 m/s² b) 8 s

Explanation:

a) Assuming that the acceleration is constant, we can use any of the kinematic equations to solve the question.

As we don´t know the time needed to accelerate, we can use the following equation:

vf2 – vo2 = 2*a*∆x

At first, we can convert the values of vf, vo and ∆x, to SI units, as follows:

vf = 65 mi/h* (1,605 m / 1mi) * (1h/3,600 sec) = 29 m/s

vo = 25 mi/h *(1,605 m / 1mi) * (1h/3,600 sec) = 11.2 m/s

∆x = 0.1 mi*(1,605 m / 1mi) = 160.5 m

Replacing these values in (1), and solving for a, we have:

a = (29 m/s – 11.2 m/s) / 321 m = 2.2 m/s2

b) In order to obtain the time needed to reach to 65 mi/h, we can rearrange the equation for the definition of acceleration, as follows:

vf = vo + at

Replacing by the values already known for vo, vf and a, and solving for t, we get:

t = vf-vo /a = (29 m/s – 11.2 m/s) / 2.2 m/s = 8 sec

5 0

3 years ago

Primary Creep: slope (creep rate) decreases with time

Igoryamba

Answer:

true

Explanation:

Creep is known as the time dependent deformation of structure due to constant load acting on the body.

Creep is generally seen at high temperature.

Due to creep the length of the structure increases which is not fit for serviceability purpose.

When time passes structure gain strength as the structure strength increases with time so creep tends to decrease.

When we talk about Creep rate for new structure the creep will be more than the old structure i.e. the creep rate decreases with time.

5 0

3 years ago

1.8 A water flow of 4.5 slug/s at 60 F enters the condenser of steam turbine and leaves at 140 F. Determine the heat transfer ra

Ann [662]

Answer:

$Hr=4.2*10^7\ btu/hr$

Explanation:

From the question we are told that:

Water flow Rate $R=4.5slug/s=144.78ib/sec$

Initial Temperature $T_1=60 \textdegree F$

Final Temperature $T_2=140 \textdegree F$

Let

Specific heat of water $\gamma= 1$

And

$\triangle T= 140-60$

$\triangle T= 80\ Deg.F$

Generally the equation for Heat transfer rate of water $H_r$ is mathematically given by

Heat transfer rate to water= mass flow rate* specific heat* change in temperature

$H_r=R* \gamma*\triangle T$

$H_r=144.78*80*1$

$H_r=11582.4\ btu/sec$

Therefore

$H_r=11582.4\ btu/sec*3600$

$Hr=4.2*10^7\ btu/hr$

3 0

3 years ago

HfrrghhJbfrfefefft nbgyjjiutrwdgju

Tasya [4]

Answer:

hshdhriwjajaldh skshdjdywuusg

Explanation:

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4 0

3 years ago