A rigid container with 6 m × 9 m × 3 m can be considered a closed system in which the container's initial air pressure and temperature are 110 kPa and 32°C, respectively. A cooling coil is installed outside one of the container walls (Figure Q1) to reduce the container's air temperature at a specified value. The cooling coil removes 4500 J/s of heat from the container, and a 1000 W fan is supplied with electrical power to distribute the cold air in the container. The container absorbs 1500 W heat from warmer outdoor due to heat transfer through the walls. For air, take c, = 0.718 kJ/kg.K and R= 0.287 kJ/kg.K. a) Find the mass of the air in the container (kg). b) Determine the time required to reduce the container's air temperature to 24°C (minutes) and the corresponding electrical power to operate the fan (kWh). c) If the container's air is reduced to 26°C instead, determine the time required to accomplish this (minutes) and the corresponding electrical power to operate the fan (kWh). d) Calculate the percentage of saving in electricity power between part (a) and part (b) (%). Qou= 4,500 J/s Ws 1000-W Qm = 1,500 w Cooling coil

A rigid container with 6 m × 9 m × 3 m can be considered a closed system in which the container's initial air pressure and temperature are 110 kPa and 32°C, respectively. A cooling coil is installed outside one of the container walls (Figure Q1) to reduce the container's air temperature at a specified value. The cooling coil removes 4500 J/s of heat from the container, and a 1000 W fan is supplied with electrical power to distribute the cold air in the container. The container absorbs 1500 W heat from warmer outdoor due to heat transfer through the walls. For air, take c, = 0.718 kJ/kg.K and R= 0.287 kJ/kg.K. a) Find the mass of the air in the container (kg). b) Determine the time required to reduce the container's air temperature to 24°C (minutes) and the corresponding electrical power to operate the fan (kWh). c) If the container's air is reduced to 26°C instead, determine the time required to accomplish this (minutes) and the corresponding electrical power to operate the fan (kWh). d) Calculate the percentage of saving in electricity power between part (a) and part (b) (%). Qou= 4,500 J/s Ws 1000-W Qm = 1,500 w Cooling coil

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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