A refrigerator using Carnot cycle requires 1.25 kW per tonne of refrigeration to maintain a temperature of \u221230\u00b0C. Find: (a) COP of the Carnot refrigerator, (b) temperature at which heat is rejected; and (c) Heat rejected per tonne of refrigeration.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Given data:<\/strong><\/p>\n\n\n\n (a) COP of the Carnot Refrigerator:<\/strong><\/p>\n\n\n\n The coefficient of performance (COP) for a Carnot refrigerator is given by: The power input ( W ) relates to the refrigeration effect (( Q_L )) as: (b) Temperature at which heat is rejected (( T_H )):<\/strong><\/p>\n\n\n\n From the COP formula: (c) Heat rejected per tonne of refrigeration (( Q_H )):<\/strong><\/p>\n\n\n\n Heat rejected is the sum of refrigeration effect and work input: (a) COP = 2.8<\/strong> The Carnot refrigerator represents an idealized refrigeration cycle operating between two temperature reservoirs. It provides the theoretical maximum performance. The COP, a measure of efficiency, depends on the temperature difference between the heat source (condenser) and the sink (evaporator).<\/p>\n\n\n\n Given that the refrigeration effect is 1 tonne, equivalent to 3.5 kW, the power input is 1.25 kW. From the relation ( \\text{COP} = \\frac{Q_L}{W} ), the Carnot refrigerator’s COP is calculated as 2.8, indicating that for every 1 kW of input, 2.8 kW of heat is extracted.<\/p>\n\n\n\n To find the rejection temperature (( T_H )), the relationship between the evaporator temperature (( T_L )), COP, and ( T_H ) is used. A higher rejection temperature reduces the system’s efficiency as it increases the temperature gradient. For this case, ( T_H ) is calculated as approximately ( 329.8 \\, \\text{K} ) (or ( 56.8^\\circ \\text{C} )).<\/p>\n\n\n\n Finally, the heat rejected per tonne of refrigeration (( Q_H )) is the sum of the refrigeration effect (( Q_L )) and the input work (( W )). This represents the energy expelled to the surroundings. For the given system, ( Q_H = 4.75 \\, \\text{kW} ) (or ( 285 \\, \\text{kJ\/min} )).<\/p>\n\n\n\n This example underscores the Carnot cycle’s theoretical potential to optimize refrigeration while highlighting practical constraints like operational losses that prevent real systems from achieving Carnot efficiency.<\/p>\n","protected":false},"excerpt":{"rendered":" A refrigerator using Carnot cycle requires 1.25 kW per tonne of refrigeration to maintain a temperature of \u221230\u00b0C. Find: (a) COP of the Carnot refrigerator, (b) temperature at which heat is rejected; and (c) Heat rejected per tonne of refrigeration. The Correct Answer and Explanation is : Solution: Given data: (a) COP of the Carnot […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-185397","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185397","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=185397"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185397\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185397"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185397"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185397"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Solution:<\/h3>\n\n\n\n
\n
[
\\text{COP} = \\frac{T_L}{T_H – T_L}
]
where ( T_H ) is the temperature at which heat is rejected.<\/p>\n\n\n\n
[
\\text{COP} = \\frac{Q_L}{W}
]
Given ( W = 1.25 \\, \\text{kW} ) and ( Q_L = 210 \\, \\text{kJ\/min} = \\frac{210}{60} = 3.5 \\, \\text{kW} ),
[
\\text{COP} = \\frac{3.5}{1.25} = 2.8
]<\/p>\n\n\n\n
[
T_H = T_L \\left(1 + \\frac{1}{\\text{COP}}\\right)
]
Substituting:
[
T_H = 243 \\left(1 + \\frac{1}{2.8}\\right) = 243 \\times 1.357 = 329.8 \\, \\text{K}
]
[
T_H = 329.8 – 273 = 56.8^\\circ C
]<\/p>\n\n\n\n
[
Q_H = Q_L + W
]
Substituting:
[
Q_H = 3.5 + 1.25 = 4.75 \\, \\text{kW}
]
[
Q_H = 4.75 \\, \\text{kW} \\times 60 = 285 \\, \\text{kJ\/min}
]<\/p>\n\n\n\nFinal Answers:<\/h3>\n\n\n\n
(b) Heat rejection temperature = 56.8\u00b0C<\/strong>
(c) Heat rejected = 4.75 kW (285 kJ\/min)<\/strong><\/p>\n\n\n\n
\n\n\n\nExplanation (300 words):<\/h3>\n\n\n\n