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4 years ago

A refrigeration cycle has Qout = 1000 Btu and Wcycle = 300 Btu. Determine the coefficient of performance for the cycle.

Physics

1 answer:

DENIUS [597]4 years ago

5 0

Answer:

The coefficient of performance for the cycle is 2.33.

Explanation:

Given that,

Output energy $Q_{out}=1000\ Btu$

Work done $W_{cycle}=300\ Btu$

We need to calculate the coefficient of performance

Using formula of the coefficient of performance

$COP=\dftrac{Q_{in}}{W_{cycle}}$

We need to calculate the $Q_{in}$

$W_{cycle}=Q_{out}-Q_{in}$

Put the value into the formula

$300=1000-Q_{in}$

$Q_{in}=300-1000$

$Q_{in}=700\ Btu$

Now put the value of $Q_{in}$ into the formula of COP

$COP=\dfrac{700}{300}$

$COP=\dfrac{7}{3}=2.33$

Hence, The coefficient of performance for the cycle is 2.33.

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3 years ago

Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci

ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

$A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})]$ (1)

Donde:

$w$ - Ancho de la placa, en centímetros.

$l$ - Longitud de la placa, en centímetros.

$\alpha$ - Coeficiente de dilatación, en $\frac{1}{^{\circ}C}$ .

$T_{o}$ - Temperatura inicial, en grados Celsius.

$T_{f}$ - Temperatura final, en grados Celsius.

Si sabemos que $w = 65\,cm$ , $l = 78\,cm$ , $\alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $T_{o} = 20\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , entonces el área de la placa a la temperatura final: