Question

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

Answer 1

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:

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

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

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

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

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

Answer 2

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:

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