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The cooling capacity of cooling and dehumidifying coil is 128 kW. Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters to the cooling coil with an enthalpy of 85 kJ/kg of dry air and humidity ratio is 20 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio is 10 grams/kg of dry air. What will be the enthalpy (in kJ/kg d.a) of condensate water leaving the coil?
- a)60.24
- b)50.12
- c)46.15
- d)66.67
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
The cooling capacity of cooling and dehumidifying coil is 128 kW. Atm...
Cooling capacity = 128 kW
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ma = 3 kg/s
h1 = 85 kJ/kg
ω1 = 20 g/kg of d.a.
h2 = 43 kJ/kg
ω2 = 10g/kg of d.a.
h3 = enthalpy of condensate leaving the coil?
Inlet mass of water vapour
mv1 = ma × ω1
= 3 × 30 × 10-3
= 60 × 10-3 kg
Outlet mass of water vapour
mv2 = ma × ω2
= 3 × 10 × 10-3
= 30 × 10-3 kg
Required cooling capacity = Enthalpy of inlet air - Enthalpy of outlet air + Enthalpy of condensate water
Qc = mah1 - mah2 +mvh3
⇒ 128 = 3 85-43 + 30 × 10-3 × h3
⇒ h3 = 66.67 kJ/kg
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The cooling capacity of cooling and dehumidifying coil is 128 kW. Atm...
Problem Analysis:
We are given the cooling capacity of the cooling and dehumidifying coil as 128 kW. We are also given the inlet and outlet conditions of the air flowing through the coil, including the enthalpy and humidity ratio. We need to determine the enthalpy of the condensate water leaving the coil.
Solution:
Step 1: Calculate the mass flow rate of the air
Given:
Flow rate of air = 3 kg/s (on dry basis)
Humidity ratio = 20 grams/kg of dry air
To find the mass flow rate of dry air, we need to convert the humidity ratio from grams/kg of dry air to kg/kg of dry air:
Humidity ratio = 20 grams/kg of dry air = 0.02 kg/kg of dry air
Mass flow rate of dry air = Flow rate of air * Humidity ratio
Mass flow rate of dry air = 3 kg/s * 0.02 kg/kg of dry air = 0.06 kg/s
Step 2: Calculate the change in enthalpy of the air
Given:
Inlet enthalpy = 85 kJ/kg of dry air
Outlet enthalpy = 43 kJ/kg of dry air
Change in enthalpy = Outlet enthalpy - Inlet enthalpy
Change in enthalpy = 43 kJ/kg of dry air - 85 kJ/kg of dry air = -42 kJ/kg of dry air
Step 3: Calculate the cooling load
The cooling load is the product of the cooling capacity and the mass flow rate of dry air:
Cooling load = Cooling capacity * Mass flow rate of dry air
Cooling load = 128 kW * 0.06 kg/s = 7.68 kW
Step 4: Calculate the enthalpy of condensate water
The enthalpy of condensate water can be calculated using the equation:
Enthalpy of condensate water = Cooling load / Mass flow rate of condensate water
Since the specific heat capacity of water is 4.186 kJ/kg·°C, we can use the equation:
Enthalpy of condensate water = Cooling load / (Mass flow rate of condensate water * Specific heat capacity of water)
Given:
Specific heat capacity of water = 4.186 kJ/kg·°C
Cooling load = 7.68 kW
We need to convert the cooling load from kW to kJ/s:
Cooling load = 7.68 kW * 1000 W/kW = 7680 W = 7.68 kJ/s
We also need to calculate the mass flow rate of condensate water. Since the mass flow rate of dry air is equal to the sum of the mass flow rate of dry air and the mass flow rate of condensate water, we can use the equation:
Mass flow rate of condensate water = Mass flow rate of dry air - Mass flow rate of dry air
Mass flow rate of condensate water = 0.06 kg/s - 3 kg/s = -2.94 kg/s (negative sign indicates condensation)
Enthalpy of condensate water = Cooling load / (Mass flow rate of condensate water *
We are given the cooling capacity of the cooling and dehumidifying coil as 128 kW. We are also given the inlet and outlet conditions of the air flowing through the coil, including the enthalpy and humidity ratio. We need to determine the enthalpy of the condensate water leaving the coil.
Solution:
Step 1: Calculate the mass flow rate of the air
Given:
Flow rate of air = 3 kg/s (on dry basis)
Humidity ratio = 20 grams/kg of dry air
To find the mass flow rate of dry air, we need to convert the humidity ratio from grams/kg of dry air to kg/kg of dry air:
Humidity ratio = 20 grams/kg of dry air = 0.02 kg/kg of dry air
Mass flow rate of dry air = Flow rate of air * Humidity ratio
Mass flow rate of dry air = 3 kg/s * 0.02 kg/kg of dry air = 0.06 kg/s
Step 2: Calculate the change in enthalpy of the air
Given:
Inlet enthalpy = 85 kJ/kg of dry air
Outlet enthalpy = 43 kJ/kg of dry air
Change in enthalpy = Outlet enthalpy - Inlet enthalpy
Change in enthalpy = 43 kJ/kg of dry air - 85 kJ/kg of dry air = -42 kJ/kg of dry air
Step 3: Calculate the cooling load
The cooling load is the product of the cooling capacity and the mass flow rate of dry air:
Cooling load = Cooling capacity * Mass flow rate of dry air
Cooling load = 128 kW * 0.06 kg/s = 7.68 kW
Step 4: Calculate the enthalpy of condensate water
The enthalpy of condensate water can be calculated using the equation:
Enthalpy of condensate water = Cooling load / Mass flow rate of condensate water
Since the specific heat capacity of water is 4.186 kJ/kg·°C, we can use the equation:
Enthalpy of condensate water = Cooling load / (Mass flow rate of condensate water * Specific heat capacity of water)
Given:
Specific heat capacity of water = 4.186 kJ/kg·°C
Cooling load = 7.68 kW
We need to convert the cooling load from kW to kJ/s:
Cooling load = 7.68 kW * 1000 W/kW = 7680 W = 7.68 kJ/s
We also need to calculate the mass flow rate of condensate water. Since the mass flow rate of dry air is equal to the sum of the mass flow rate of dry air and the mass flow rate of condensate water, we can use the equation:
Mass flow rate of condensate water = Mass flow rate of dry air - Mass flow rate of dry air
Mass flow rate of condensate water = 0.06 kg/s - 3 kg/s = -2.94 kg/s (negative sign indicates condensation)
Enthalpy of condensate water = Cooling load / (Mass flow rate of condensate water *
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The cooling capacity of cooling and dehumidifying coil is 128 kW. Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters to the cooling coil with an enthalpy of 85 kJ/kg of dry air and humidity ratio is 20 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio is 10 grams/kg of dry air. What will be the enthalpy (in kJ/kg d.a) of condensate water leaving the coil?a)60.24b)50.12c)46.15d)66.67Correct answer is option 'D'. Can you explain this answer?
Question Description
The cooling capacity of cooling and dehumidifying coil is 128 kW. Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters to the cooling coil with an enthalpy of 85 kJ/kg of dry air and humidity ratio is 20 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio is 10 grams/kg of dry air. What will be the enthalpy (in kJ/kg d.a) of condensate water leaving the coil?a)60.24b)50.12c)46.15d)66.67Correct answer is option 'D'. 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 The cooling capacity of cooling and dehumidifying coil is 128 kW. Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters to the cooling coil with an enthalpy of 85 kJ/kg of dry air and humidity ratio is 20 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio is 10 grams/kg of dry air. What will be the enthalpy (in kJ/kg d.a) of condensate water leaving the coil?a)60.24b)50.12c)46.15d)66.67Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The cooling capacity of cooling and dehumidifying coil is 128 kW. Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters to the cooling coil with an enthalpy of 85 kJ/kg of dry air and humidity ratio is 20 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio is 10 grams/kg of dry air. What will be the enthalpy (in kJ/kg d.a) of condensate water leaving the coil?a)60.24b)50.12c)46.15d)66.67Correct answer is option 'D'. Can you explain this answer?.
The cooling capacity of cooling and dehumidifying coil is 128 kW. Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters to the cooling coil with an enthalpy of 85 kJ/kg of dry air and humidity ratio is 20 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio is 10 grams/kg of dry air. What will be the enthalpy (in kJ/kg d.a) of condensate water leaving the coil?a)60.24b)50.12c)46.15d)66.67Correct answer is option 'D'. 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 The cooling capacity of cooling and dehumidifying coil is 128 kW. Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters to the cooling coil with an enthalpy of 85 kJ/kg of dry air and humidity ratio is 20 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio is 10 grams/kg of dry air. What will be the enthalpy (in kJ/kg d.a) of condensate water leaving the coil?a)60.24b)50.12c)46.15d)66.67Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The cooling capacity of cooling and dehumidifying coil is 128 kW. Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters to the cooling coil with an enthalpy of 85 kJ/kg of dry air and humidity ratio is 20 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio is 10 grams/kg of dry air. What will be the enthalpy (in kJ/kg d.a) of condensate water leaving the coil?a)60.24b)50.12c)46.15d)66.67Correct answer is option 'D'. Can you explain this answer?.
Solutions for The cooling capacity of cooling and dehumidifying coil is 128 kW. Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters to the cooling coil with an enthalpy of 85 kJ/kg of dry air and humidity ratio is 20 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio is 10 grams/kg of dry air. What will be the enthalpy (in kJ/kg d.a) of condensate water leaving the coil?a)60.24b)50.12c)46.15d)66.67Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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ample number of questions to practice The cooling capacity of cooling and dehumidifying coil is 128 kW. Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters to the cooling coil with an enthalpy of 85 kJ/kg of dry air and humidity ratio is 20 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio is 10 grams/kg of dry air. What will be the enthalpy (in kJ/kg d.a) of condensate water leaving the coil?a)60.24b)50.12c)46.15d)66.67Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice GATE tests.
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