Carnot Cycle
Consider a Carnot refrigerator operates between -40°F and 100°F. If the energy absorbed from the low temperature space is 250 Btu/hr, what is the net work done on the working substance?
Expand Hint
For a Carnot refrigeration cycle, the Coefficient of Performance (COP) is:
$$$COP_c=\frac{T_L}{(T_H-T_L)}$$$
where
$$T_H$$
and
$$T_L$$
are the absolute high and low temperatures respectively in either Kelvin or Rankine.
Hint 2
Refrigeration cycles are the reverse of heat engine cycles. Heat is moved from low to high temperatures, which requires work,
$$W$$
. For refrigerators and air conditioners, the Coefficient of Performance (COP) is:
$$$COP=\frac{Q_L}{W}$$$
where
$$Q_L$$
is the system’s rejected heat.
For a Carnot refrigeration cycle, the Coefficient of Performance (COP) is:
$$$COP_c=\frac{T_L}{(T_H-T_L)}$$$
where
$$T_H$$
and
$$T_L$$
are the absolute high and low temperatures respectively in either Kelvin or Rankine. To convert from Fahrenheit to Rankine:
$$$^{\circ}R\:=\:^{\circ}F+459.69$$$
Thus, the high and low temperatures are:
$$$^{\circ}R_H=100^{\circ}F+459.69=559.69^{\circ}R$$$
$$$^{\circ}R_L=-40^{\circ}F+459.69=419.69^{\circ}R$$$
Therefore,
$$$COP_c=\frac{419.69^{\circ}R}{(559.69^{\circ}R-419.69^{\circ}R)}=\frac{419.69}{140}=2.998$$$
Refrigeration cycles are the reverse of heat engine cycles. Heat is moved from low to high temperatures, which requires work,
$$W$$
. For refrigerators and air conditioners, the Coefficient of Performance (COP) is:
$$$COP=\frac{Q_L}{W}$$$
where
$$Q_L$$
is the system’s rejected heat. Solving for work:
$$$W=\frac{Q_L}{COP}=\frac{250Btu/hr}{2.998}=83.4\:Btu/hr$$$
83.4 Btu/hr
Time Analysis
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