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A room is mechanically ventilated through four air-conditioning ducts. The opening area of each
duct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between the
ambient air and supply air is 10 °C. Volumetric specific heat of air is 1250 J/m3 °C. Assuming one
Ton of refrigeration (TR) equals 3.5 kW, the cooling load of the room in TR will be________
duct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between the
ambient air and supply air is 10 °C. Volumetric specific heat of air is 1250 J/m3 °C. Assuming one
Ton of refrigeration (TR) equals 3.5 kW, the cooling load of the room in TR will be________
Correct answer is between '2.42,2.55'. Can you explain this answer?
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A room is mechanically ventilated through four air-conditioning ducts....
Degrees Celsius. The density of air is 1.225 kg/m^3. The specific heat capacity of air is 1005 J/kg·K. Calculate the volume flow rate of air in the room in m^3/s.
To calculate the volume flow rate of air in the room, we need to find the mass flow rate of air first.
The mass flow rate can be calculated using the formula:
mass flow rate = density × velocity × area
mass flow rate = 1.225 kg/m^3 × 0.5 m/s × 0.35 sqm × 4 ducts
mass flow rate = 0.8575 kg/s
Next, we can calculate the volume flow rate using the mass flow rate and the specific heat capacity of air. The formula to calculate the volume flow rate is:
volume flow rate = mass flow rate / (density × specific heat capacity × temperature difference)
volume flow rate = 0.8575 kg/s / (1.225 kg/m^3 × 1005 J/kg·K × 10 K)
volume flow rate = 0.0691 m^3/s
Therefore, the volume flow rate of air in the room is 0.0691 m^3/s.
To calculate the volume flow rate of air in the room, we need to find the mass flow rate of air first.
The mass flow rate can be calculated using the formula:
mass flow rate = density × velocity × area
mass flow rate = 1.225 kg/m^3 × 0.5 m/s × 0.35 sqm × 4 ducts
mass flow rate = 0.8575 kg/s
Next, we can calculate the volume flow rate using the mass flow rate and the specific heat capacity of air. The formula to calculate the volume flow rate is:
volume flow rate = mass flow rate / (density × specific heat capacity × temperature difference)
volume flow rate = 0.8575 kg/s / (1.225 kg/m^3 × 1005 J/kg·K × 10 K)
volume flow rate = 0.0691 m^3/s
Therefore, the volume flow rate of air in the room is 0.0691 m^3/s.
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A room is mechanically ventilated through four air-conditioning ducts. The opening area of eachduct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between theambient air and supply air is 10 °C. Volumetric specific heat of air is 1250 J/m3 °C. Assuming oneTon of refrigeration (TR) equals 3.5 kW, the cooling load of the room in TR will be________Correct answer is between '2.42,2.55'. Can you explain this answer?
Question Description
A room is mechanically ventilated through four air-conditioning ducts. The opening area of eachduct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between theambient air and supply air is 10 °C. Volumetric specific heat of air is 1250 J/m3 °C. Assuming oneTon of refrigeration (TR) equals 3.5 kW, the cooling load of the room in TR will be________Correct answer is between '2.42,2.55'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A room is mechanically ventilated through four air-conditioning ducts. The opening area of eachduct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between theambient air and supply air is 10 °C. Volumetric specific heat of air is 1250 J/m3 °C. Assuming oneTon of refrigeration (TR) equals 3.5 kW, the cooling load of the room in TR will be________Correct answer is between '2.42,2.55'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A room is mechanically ventilated through four air-conditioning ducts. The opening area of eachduct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between theambient air and supply air is 10 °C. Volumetric specific heat of air is 1250 J/m3 °C. Assuming oneTon of refrigeration (TR) equals 3.5 kW, the cooling load of the room in TR will be________Correct answer is between '2.42,2.55'. Can you explain this answer?.
A room is mechanically ventilated through four air-conditioning ducts. The opening area of eachduct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between theambient air and supply air is 10 °C. Volumetric specific heat of air is 1250 J/m3 °C. Assuming oneTon of refrigeration (TR) equals 3.5 kW, the cooling load of the room in TR will be________Correct answer is between '2.42,2.55'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A room is mechanically ventilated through four air-conditioning ducts. The opening area of eachduct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between theambient air and supply air is 10 °C. Volumetric specific heat of air is 1250 J/m3 °C. Assuming oneTon of refrigeration (TR) equals 3.5 kW, the cooling load of the room in TR will be________Correct answer is between '2.42,2.55'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A room is mechanically ventilated through four air-conditioning ducts. The opening area of eachduct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between theambient air and supply air is 10 °C. Volumetric specific heat of air is 1250 J/m3 °C. Assuming oneTon of refrigeration (TR) equals 3.5 kW, the cooling load of the room in TR will be________Correct answer is between '2.42,2.55'. Can you explain this answer?.
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A room is mechanically ventilated through four air-conditioning ducts. The opening area of eachduct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between theambient air and supply air is 10 °C. Volumetric specific heat of air is 1250 J/m3 °C. Assuming oneTon of refrigeration (TR) equals 3.5 kW, the cooling load of the room in TR will be________Correct answer is between '2.42,2.55'. Can you explain this answer?, a detailed solution for A room is mechanically ventilated through four air-conditioning ducts. The opening area of eachduct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between theambient air and supply air is 10 °C. Volumetric specific heat of air is 1250 J/m3 °C. Assuming oneTon of refrigeration (TR) equals 3.5 kW, the cooling load of the room in TR will be________Correct answer is between '2.42,2.55'. Can you explain this answer? has been provided alongside types of A room is mechanically ventilated through four air-conditioning ducts. The opening area of eachduct is 0.35 sqm. The air velocity in the duct is 0.5 m/s. The temperature difference between theambient air and supply air is 10 °C. Volumetric specific heat of air is 1250 J/m3 °C. Assuming oneTon of refrigeration (TR) equals 3.5 kW, the cooling load of the room in TR will be________Correct answer is between '2.42,2.55'. Can you explain this answer? theory, EduRev gives you an
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