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2 years ago

12

: The maximum theoretical efficiency of a Carnot engine operating between reservoirs at the steam point and at room temperature

is about A :

1 answer:

Hunter-Best [27]2 years ago

5 0

Answer:

The value is $\eta = 0.2145$ or 21.45%

Explanation:

From the question we are told that

The first reservoir is at steam point $T_s = 100^o C = 100 + 273 = 373 \ K$

The second reservoir is at room temperature $T_r = 20^o C = 293 \ K$

Generally the maximum theoretical efficiency of a Carnot engine is mathematically evaluated as

$\eta = 1- \frac{T_r}{T_s}$

=> $1 - \frac{ 293}{373}$

=> $\eta = 0.2145$

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Which statement about moons is true?

kotegsom [21]

Answer:

A

Explanation:

All of the other answers don't make much sense

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3 years ago

Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm

Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: <em>Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.</em>

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

$\eta\cdot \dot W = \dot K$

$\dot W = \frac{\dot K}{\eta}$ (Eq. 1)

Where:

$\eta$ - Efficiency of fan, dimensionless.

$\dot W$ - Electric power supplied fan, measured in watts.

$\dot K$ - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

$\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2}$ (Eq. 2)

Where:

$\rho_{a}$ - Density of air, measured in kilograms per cubic meter.

$\dot V$ - Volume flow, measured in cubic meters per second.

$A_{s}$ - Cross-sectional area of fan, measured in square meters.

If we know that $\rho_{a} = 1.20\,\frac{kg}{m^{3}}$ , $\dot V = 0.05\,\frac{m^{3}}{s}$ , $\eta = 0.3$ and $A_{s} = 20\times 10^{-4}\,m^{2}$ , the power needed to be supplied by the fan is:

$\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}$

$\dot K = 62.5\,W$

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

5 0

3 years ago

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and the volume V satisfy th

jarptica [38.1K]

Answer: The volume is decreasing at a rate of 80 cm3/min

Explanation: Please see the attachments below

8 0

2 years ago

3. A model rocket takes 0.05 seconds to speed up from rest to its maximum velocity of 80 m/s.

nikklg [1K]

Answer:

$1600 \frac{m}{s^2}$

Explanation:

Acceleration is defined as the change in velocity divided by the time it took to produce such change. The formula then reads:

$a = \frac{change-in-velocity}{time} = \frac{Vf-Vi}{t}$

Where Vf is the final velocity of the object, (in our case 80 m/s)

Vi is the initial velocity of the object (in our case 0 m/s because the object was at rest)

and t is the time it took to change from the Vi to the Vf (in our case 0.05 seconds.

Therefore we have:

$a = \frac{80 m/s - 0 m/s}{0.05 sec} = 1600 \frac{m}{s^2}$

Notice that the units of acceleration in the SI system are $\frac{m}{s^2}$ (meters divided square seconds)

7 0

2 years ago

Read 2 more answers

What is the average velocity of atoms in 2.00 mol of neon (a monatomic gas)

Dimas [21]

Answer:

v = 876 m/s

Explanation:

It is given that,

Number of mol of Neon is 2 mol

Temperature, T = 308 K

Mass, m = 0.02 kg

Value of R - 8.31 J/mol-K

We need to find the average velocity of atoms in 2.00 mol of neon. Neon is a monoatomic gas. Let v is the velocity. So,

$\dfrac{1}{2}mv^2=\dfrac{3}{2}nRT\\\\v=\sqrt{\dfrac{3nRT}{m}} \\\\v=\sqrt{\dfrac{3\times 2\times 8.314\times 308}{0.02}} \\\\v=876.47\ m/s$

So, the correct option is (B).

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2 years ago

Read 2 more answers