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Masja [62]
3 years ago
7

At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r

eceiving energy at the rate QC from a cold reservoir at temperature TC. a. If TH = 13°C and TC = 2°C, determine the coefficient of performance. b. If QH = = 10.5 kW, 8.75 kW QC , and TC = 0°C, determine TH, in °C. c. If the coefficien
Engineering
2 answers:
ludmilkaskok [199]3 years ago
8 0

Answer:

a) COP_{R} = 25.014, b) T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)

Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}

COP_{R} = 25.014

b) The respective coefficient of performance is determined:

COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}

COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}

COP_{R} = 5

But:

COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}

The temperature at hot reservoir is found with some algebraic help:

COP_{R} \cdot (T_{H}-T_{L})=T_{L}

T_{H}-T_{L} = \frac{T_{L}}{COP_{R}}

T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}}  \right)

T_{H} = 273.15\,K \cdot \left(1+\frac{1}{5}  \right)

T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)

gregori [183]3 years ago
3 0

Answer:

a) COP = 26

b) TH = 327.6 K

c) TC = 297.3 K

Explanation:

a) TH = temperature hot reservoir = 13°C = 286 K

Tc = temperature cold reservoir = 2°C = 275 K

The coefficient of performance is

COP=\frac{1}{1-\frac{T_{C} }{T_{H} } } =\frac{1}{1-275/286} =26

b) given:

QH = 10.5 kW

QC = 8.75 kW

TC = 0°C = 273 K

\frac{Q_{H} }{Q_{C} } =\frac{T_{H} }{T_{C} } \\T_{H} =\frac{Q_{H}T_{C}  }{Q_{C} } =\frac{10.5*273}{8.75} =327.6K

c) From the COP formula we clear TC:

10=\frac{1}{1-T_{C}/300 } \\T_{C} =297.3k

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