Question
Download Solution PDFFor a given value TH (source temperature) for a reversed Carnot cycle, the variation of TL (sink temperature) for different values of COP is represented by which one of the following graphs?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
A reversed Carnot cycle yield a refrigeration cycle, whose COP is given by:
\(COP = \frac{{Desired\;effect}}{{work\;input}}\;\;\; = \;\frac{{{T_L}}}{{{T_H} - {T_L}}}\)
where TL = temperature of the sink and TH = temperature of the source.
Explanation:
\(COP = \;\frac{{{{\rm{T}}_{\rm{L}}}}}{{{{\rm{T}}_{\rm{H}}} - {{\rm{T}}_{\rm{L}}}}}{\rm{\;\;}} = \frac{1}{{\frac{{{{\rm{T}}_{\rm{H}}}}}{{{{\rm{T}}_{\rm{L}}}}} - 1}}\)
As TL increases, the ratio \(\frac{T_H}{T_L}\) decreases, therefore COP increases.
Slope = \(\frac{d(COP)}{d(T_L)}\)
\(\frac{d(COP)}{d(T_L)}=\frac{(T_H\;-\;T_L)\times 1-(T_L)\times (-1)}{(T_H-T_L)^2}\)
\(\frac{d(COP)}{d(T_L)}= \;\frac{{{{\rm{T}}_{\rm{H}}}}}{{{{\rm{(T}}_{\rm{H}}} - {{\rm{T}}_{\rm{L}}}})^2}{\rm{\;\;}}\)
i.e. slope increase as TL increases.
\(COP = \frac{{Desired\;effect}}{{work\;input}}\;\;\; = \;\frac{{{T_L}}}{{{T_H} - {T_L}}}\)
Also, if we assume COP in y-axis and TL in x-axis then the above formula will look like
\(COP = \frac{{Desired\;effect}}{{work\;input}}\;\;\; = \;\frac{{{x}}}{{{y} - {x}}}\)
Thus when x = 0, y = 0 i.e. COP will be zero when TL is zero.
According to the above relation, the best-suited curve is option B, i.e. passing through origin and increasing.
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