The passage of a wave through a medium is...?

Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the co

ladessa [460]

Answer:

$\dot W_{in} = 49.386\,kW$

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

$\dot W_{in} + \dot m \cdot c_{p}\cdot (T_{1}-T_{2}) = 0$

The power consumed by the compressor can be calculated by the following expression:

$\dot W_{in} = \dot m \cdot c_{v}\cdot (T_{2}-T_{1})$

Let consider that air behaves ideally. The density of air at inlet is:

$P\cdot V = n\cdot R_{u}\cdot T$

$P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

$\rho = \frac{P\cdot M}{R_{u}\cdot T}$

$\rho = \frac{(104\,kPa)\cdot (28.02\,\frac{kg}{kmol})}{(8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )\cdot (292\,K)}$

$\rho = 1.2\,\frac{kg}{m^{3}}$

The mass flow through compressor is:

$\dot m = \rho \cdot \dot V$

$\dot m = (1.2\,\frac{kg}{m^{3}})\cdot (0.15\,\frac{m^{3}}{s} )$

$\dot m = 0.18\,\frac{kg}{s}$

The work input is:

$\dot W_{in} = (0.18\,\frac{kg}{s} )\cdot (1.005\,\frac{kJ}{kg\cdot K})\cdot (565\,K-292\,K)$

$\dot W_{in} = 49.386\,kW$

5 0

3 years ago

1. How do you describe the path that the light passes through when it travels?

Ber [7]

A:

The white light is divided into its component hues - red, orange, yellow, green, blue, and violet - as it passes through the prism. Dispersion is the splitting of visible light into its many hues.

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B:

Dispersion is the splitting of visible light into its many hues. Each hue is associated with a certain wave frequency, as stated in the Light and Color unit, and different frequencies of light waves bend at different rates while passing through a prism.

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C:

The white light is divided into its component hues - red, orange, yellow, green, blue, and violet - as it passes through the prism. Dispersion is the splitting of visible light into its many hues.

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D:

The sun and the weather combine to create a rainbow. As it passes from air to denser water, light enters a water droplet, slowing and bending. The light bounces off the interior of the droplet, separating the wavelengths—or colors—that make up the droplet. A rainbow is created as light escapes the droplet.

5 0

2 years ago

Un bombero alejado d = 31.0 m de un edificio en llamas dirige un chorro de agua desde una manguera contra incendios a nivel del

Mila [183]

Answer:

El chorro golpea el edificio a una altura de 15.943 metros con respecto al suelo.

Explanation:

El chorro de agua exhibe un movimiento parabólico, dado que este tiene una inclinación inicial y la única aceleración es debida a la gravitación terrestre. Las ecuaciones cinemáticas que modelan el fenómeno son:

Distancia horizontal (en metros)

$x = x_{o} + v_{o}\cdot t \cdot \cos \theta$

Donde:

$x_{o}$ - Posición horizontal inicial, medida en metros.

$t$ - Tiempo, medido en segundos.

$v_{o}$ - Velocidad inicial, medida en metros por segundo.

$\theta$ - Angulo de inclinación del chorro de agua, medido en grados sexagesimales.

Distancia vertical (en metros)

$y = y_{o} + v_{o}\cdot t \cdot \sin \theta + \frac{1}{2}\cdot g \cdot t^{2}$

Donde:

$y_{o}$ - Posición vertical inicial, medida en metros.

$g$ - Constante gravitacional, medida en metros por segundo al cuadrado.

Partiendo de la primera ecuación, se despeja el tiempo:

$t = \frac{x - x_{o}}{v_{o}\cdot \cos \theta}$

Si $x = 31\,m$ , $x_{o} = 0\,m$ , $v_{o} = 40\,\frac{m}{s}$ and $\theta = 33^{\circ}$ , entonces:

$t = \frac{31\,m-0\,m}{\left(40\,\frac{m}{s} \right)\cdot \cos 33^{\circ}}$

$t = 0.924\,s$

La altura máxima se calcula por sustitución directa de términos en la segunda ecuación. Si $y_{o} = 0\,m$ , $v_{o} = 40\,\frac{m}{s}$ , $t = 0.924\,s$ y $g = -9.807\,\frac{m}{s^{2}}$ , entonces:

$y = 0\,m + \left(40\,\frac{m}{s} \right)\cdot (0.924\,s)\cdot \sin 33^{\circ} + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (0.924\,s)^{2}$