Find BTU Required for Water Cooling ?

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SUMMARY

The discussion focuses on calculating the BTU/hr required to cool 30 liters of water from 50°C to 25°C in a closed, static vessel. The specific heat of water is established as 4.184 J/g°C, and the mass of the water is approximately 30,000 grams. The formula used is Heat added = specific heat x mass x (Tfinal - Tinitial), which is essential for determining the compressor capacity needed to achieve the cooling within a maximum time of 15 minutes. The conversion factors for BTU and temperature are also clarified, ensuring accurate calculations.

PREREQUISITES
  • Understanding of specific heat capacity and its application in thermal calculations.
  • Familiarity with the British Thermal Unit (BTU) as a measurement of heat energy.
  • Basic knowledge of temperature conversion between Celsius and Fahrenheit.
  • Ability to perform unit conversions, particularly between liters, kilograms, and pounds.
NEXT STEPS
  • Research the calculation of BTU/hr for various fluids, focusing on water cooling applications.
  • Learn about compressor specifications and how to select one based on cooling requirements.
  • Explore the principles of heat transfer in closed systems to enhance understanding of thermal dynamics.
  • Investigate the impact of time constraints on cooling efficiency and compressor performance.
USEFUL FOR

Engineers, HVAC professionals, and anyone involved in thermal management or cooling system design will benefit from this discussion, particularly those working with water cooling systems.

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Homework Statement


[/B]
Calculate the BTU/hr required for water to lower its temperature to find the compressor capacity required

- Cool 30 Liters of water from 50 degree Celsius(T1) to 25 Degree Celsius(T2) – ( 122 to 77 Fahrenheit )
-Water in a closed vessel, Static, not flowing.
-Max Time Available to cool the water - 10 to 15 mins (not sure if time is relevant for finding the BTU, But it will be ideal to find the compressor which can do the job in the above time)

Homework Equations



Specific heat equation
Heat added = specific heat x mass x (tfinal - tinitial)

The Attempt at a Solution


[/B]
Specific heat of water = 4.184
Mass = 30 litres = 30,000 grams(approx)
Tfinal - Tinitial = 25 Degree C

bit confused with the units used so I guess I am getting wrong answer
Kindly help to fix thisThanks
 
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Google "British thermal unit."
 
1 liter = 1kg = 2.205 lbm

1 C = 1.8 F

Heat capacity of water = ##1 \frac{BTU}{lbm\ F}##
 

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