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2 years ago

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Air enters a nozzle steadily at 200 kPa and 65C with a velocity of 35 m/s and exits at 95 kPa and 240 m/s. The heat loss from th

e nozzle to the surrounding medium at 17C is estimated to be 3 kJ/kg. Determine a) the exit temperature b) the exergy destroyed during this process.

1 answer:

Naddika [18.5K]2 years ago

5 0

Answer:

34°C

36.9 KJ/kg

Explanation:

solution:

the initial enthalpy is determined by the following equation:

$h_{1}$ =338.404 kJ/kg

next the final enthalpy is determined from the energy balance:

$h_{1} +\frac{v_{1} ^{2}}{y} =h_{2} +\frac{v_{2} ^{2}}{y}+q\\h_{2}=h_{1} +v_{1} ^{2}-{v_{2} ^{2}/2-q\\=338.404 kJ/kg+{35^{2}-{\frac{240 ^{2}}{2} } -3kJ/kg\\=307.6125 kJ/kg\\$

from the final enthalpy the final temperature is determined by the following equation:

$T_{2} =T_{1} +\frac{T_{2}-T_{1}}{h_{2}-h_{1}} (h_{2}-h_{1})\\=305 K+\frac{310-305}{310.24-305.22}(307.215-305.22)K\\ =307 K\\=34 C$

The energy destroyed calculated from the enthalpy generation

$x_{dest} =T_{0} s_{gen} \\=T_{0}(c_{p} ln\frac{T_{2}}{T_{1}} -Rln\frac{P_{2}}{P_{1}}+\frac{q}{T_{0}}\\ =36.9 KJ/kg$

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(c) T = T = 15.50°C

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Length = 25 m

Entry temperature = 200°C

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velocity = 2.5 m/s.

Step 1: Calculating the mean temperature;

(200 + 15)/2

Mean temperature = 107.5°C = 380.5 K

The properties of air at mean temperature 380.5 K are given as:

v = 24.2689*10⁻⁶m²/s

a = 35.024*10⁻⁶m²/s

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Step 2: Calculating the prantl number using the formula;

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Nu = hD/k

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v = 14.82 *10⁻⁶m²/s

k = 0.0253 W/m.k

a = 20.873 *10⁻⁶m²/s

Pr(outside) = v/a

= 14.82 *10⁻⁶/20.873 *10⁻⁶

= 0.71

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Nu = 0.26Re^0.6 * Pr^0.37 * (Pr(outside)/Pr)^1/4

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2) The muzzle velocity of the ball is approximately 5.272 meters per second.

3) The maximum height of the ball is 1.417 meters.

Explanation:

1) Which of the following statements are true?

a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

True, statement indicates that there is no air resistence and no friction between ball and the inside of the gun because the first never touches the latter one.

b) The forces of gravity and the spring have potential energies associated with them.

False, force of gravity do work on the ball and spring receives a potential energy at being deformated by the ball.

c) No conservative forces act in this problem after the ball is released from the spring gun.

False, the absence of no conservative forces is guaranteed for the entire system according to the statement of the problem.

2) According to the statement, we understand that spring is deformed and once released and just after reaching its equilibrium position, the muzzle velocity is reached. As spring deformation is too small in comparison with height, we can neglect changes in gravitational potential energy. By Principle of Energy Conservation, we describe the motion of the ball by the following expression:

$U_{k, 1}+K_{1}=U_{k,2}+K_{2}$ (Eq. 1)

Where:

$U_{k,1}$ , $U_{k,2}$ - Initial and final elastic potential energies of spring, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After using definitions of elastic potential and translational kinetic energies, we expand the equation above as:

$\frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2}) = \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})$

And the final velocity is cleared:

$m\cdot (v_{2}^{2}-v_{1}^{2}) = k\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2}-v_{1}^{2} =\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2} =v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }$ (Eq. 2)

Where:

$v_{1}$ , $v_{2}$ - Initial and final velocities of the ball, measured in meters per second.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the ball, measured in kilograms.

$x_{1}$ , $x_{2}$ - Initial and final position of spring, measured in meters.

If we know that $v_{1} = 0\,\frac{m}{s}$ , $k = 667\,\frac{N}{m}$ , $m = 1.50\,kg$ , $x_{1} = -0.25\,m$ and $x_{2} = 0\,cm$ , the muzzle velocity of the ball is:

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\left(\frac{667\,\frac{N}{m} }{1.50\,kg} \right)\cdot [(-0.25\,m)^{2}-(0\,m)^{2}]}$

$v_{2}\approx 5.272\,\frac{m}{s}$

The muzzle velocity of the ball is approximately 5.272 meters per second.

3) After leaving the toy gun, the ball is solely decelerated by gravity. We construct this model by Principle of Energy Conservation:

$U_{g,2}+K_{2} = U_{g,3}+K_{3}$ (Eq. 3)

Where:

$U_{g,2}$ , $U_{g,3}$ - Initial and gravitational potential energies of the ball, measured in joules.

$K_{2}$ , $K_{3}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After applying definitions of gravitational potential and translational kinetic energies, we expand the equation above and solve the resulting for the final height:

$m\cdot g \cdot (h_{3}-h_{2}) = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{3}^{2})$

$h_{3}-h_{2}=\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$

$h_{3} = h_{2} +\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$ (Eq. 4)

$h_{2}$ , $h_{3}$ - Initial and final heights of the ball, measured in meters.

$v_{2}$ , $v_{3}$ - Initial and final velocities of the ball, measured in meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

If we get that $v_{2} = 5.272\,\frac{m}{s}$ , $v_{3} = 0\,\frac{m}{s}$ , $h_{2} = 0\,m$ and $g = 9.807\,\frac{m}{s^{2}}$ , the maximum height of the ball is:

$h_{3} = 0\,m+\frac{\left(5.272\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h_{3} = 1.417\,m$

The maximum height of the ball is 1.417 meters.

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