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An unsaturated air enters in an insulated chamber at 320C where it flows over a long sheet of water and becomes cooled to 260C, which is adiabatic saturation temperature. The total pressure of the mixture remains constant at 100 kPa. The corresponding pressure to saturation temperature is 3.363 kPa. The humidity ratio at the entry of chamber is (in kg vap/kg dry air)
Given data: Cp of air = 1.005 kJ/kgK
At 320C, hv1 = 2559.9 kJ/kg,
At 260C, hfg2 = 2439.9 kJ/kg, hf2 = 109.1 kJ/kg
Given data: Cp of air = 1.005 kJ/kgK
At 320C, hv1 = 2559.9 kJ/kg,
At 260C, hfg2 = 2439.9 kJ/kg, hf2 = 109.1 kJ/kg
Correct answer is between '0.017,0.021'. Can you explain this answer?
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Verified Answer
An unsaturated air enters in an insulated chamber at 320Cwhere it flow...
Given, p =100 kPa, ps = 3.363 kPa
Since the system is insulted and no work interaction, so using conservation of energy
ma1ha1 + mv1hv1 + (mv2−mv1) hf2 = ma2ha2 + m2hv2 (i)
ma = mass of dry air (kg)
mv = mass of water vapour (kg)
ha = specific enthalpy of dry air (kJ/kg)
hf = specific enthalpy of liquid water (kJ/kg)
hv = specific enthalpy of water vapour in air (kJ/kg)
Divide the eqn (i) by ma1, we get
ha1 + W1 hw1 + (W2− W1) hf2 = ha2 + W2hv2 (∵ ma1 = ma2) (ii)
Where, W = specific humidity
Humidity ratio at exit,
(ha1 − ha2) + W2 (hf2 − hv2) = W1 (hf2−hv1)
Cp (T2 – T1) + W2 hfg2 = W1 (hv1−hf2) (∵ hv2 = hg2 and hfg2 = hg2 − hf2)
1.005 × (26−32) + 0.0216×2439.9 = W1 × (2559.9−109.1)
W1 = 0.019 kg vap/kg dry air
Since the system is insulted and no work interaction, so using conservation of energy
ma1ha1 + mv1hv1 + (mv2−mv1) hf2 = ma2ha2 + m2hv2 (i)
ma = mass of dry air (kg)
mv = mass of water vapour (kg)
ha = specific enthalpy of dry air (kJ/kg)
hf = specific enthalpy of liquid water (kJ/kg)
hv = specific enthalpy of water vapour in air (kJ/kg)
Divide the eqn (i) by ma1, we get
ha1 + W1 hw1 + (W2− W1) hf2 = ha2 + W2hv2 (∵ ma1 = ma2) (ii)
Where, W = specific humidity
Humidity ratio at exit,
(ha1 − ha2) + W2 (hf2 − hv2) = W1 (hf2−hv1)
Cp (T2 – T1) + W2 hfg2 = W1 (hv1−hf2) (∵ hv2 = hg2 and hfg2 = hg2 − hf2)
1.005 × (26−32) + 0.0216×2439.9 = W1 × (2559.9−109.1)
W1 = 0.019 kg vap/kg dry air
Most Upvoted Answer
An unsaturated air enters in an insulated chamber at 320Cwhere it flow...
< b="" />Given:< />
- Initial temperature of unsaturated air (T1) = 32°C
- Final temperature of air after cooling (T2) = 26°C
- Adiabatic saturation temperature (T_ad) = 26°C
- Total pressure (P_total) = 100 kPa
- Saturation pressure at adiabatic saturation temperature (P_sat) = 3.363 kPa
- Specific heat capacity of air (Cp) = 1.005 kJ/kgK
- Enthalpy at 32°C (hv1) = 2559.9 kJ/kg
- Enthalpy of vaporization at 26°C (hfg2) = 2439.9 kJ/kg
- Enthalpy of liquid water at 26°C (hf2) = 109.1 kJ/kg
< b="" />To find:< />
Humidity ratio at the entry of the chamber
< b="" />Solution:< />
< b="" />Step 1: Determine the enthalpy change due to cooling:< />
The enthalpy change (Δh) can be calculated using the specific heat capacity (Cp) and the temperature change (ΔT) as follows:
Δh = Cp * ΔT
ΔT = T1 - T2 = 32°C - 26°C = 6°C
Δh = 1.005 kJ/kgK * 6°C = 6.03 kJ/kg
< b="" />Step 2: Calculate the enthalpy at adiabatic saturation temperature:< />
The enthalpy at adiabatic saturation temperature (h_ad) can be calculated using the enthalpy change (Δh) and the enthalpy at the initial temperature (hv1) as follows:
h_ad = hv1 - Δh
h_ad = 2559.9 kJ/kg - 6.03 kJ/kg = 2553.87 kJ/kg
< b="" />Step 3: Determine the enthalpy of vaporization at the adiabatic saturation temperature:< />
The enthalpy of vaporization at the adiabatic saturation temperature (hfg_ad) can be calculated using the enthalpy of vaporization at the final temperature (hfg2) and the enthalpy at the adiabatic saturation temperature (h_ad) as follows:
hfg_ad = hfg2 + h_ad
hfg_ad = 2439.9 kJ/kg + 2553.87 kJ/kg = 4993.77 kJ/kg
< b="" />Step 4: Determine the humidity ratio:< />
The humidity ratio (W) can be calculated using the enthalpy of vaporization at the adiabatic saturation temperature (hfg_ad) and the enthalpy of liquid water at the adiabatic saturation temperature (hf_ad) as follows:
W = (hfg_ad / (hfg_ad - hf_ad)) * (P_total / (P_total - P_sat))
hf_ad = hf2 = 109.1 kJ/kg
W = (4993.77 kJ/kg / (4993.77 kJ/kg - 109.1 kJ/kg)) * (100 kPa / (
- Initial temperature of unsaturated air (T1) = 32°C
- Final temperature of air after cooling (T2) = 26°C
- Adiabatic saturation temperature (T_ad) = 26°C
- Total pressure (P_total) = 100 kPa
- Saturation pressure at adiabatic saturation temperature (P_sat) = 3.363 kPa
- Specific heat capacity of air (Cp) = 1.005 kJ/kgK
- Enthalpy at 32°C (hv1) = 2559.9 kJ/kg
- Enthalpy of vaporization at 26°C (hfg2) = 2439.9 kJ/kg
- Enthalpy of liquid water at 26°C (hf2) = 109.1 kJ/kg
< b="" />To find:< />
Humidity ratio at the entry of the chamber
< b="" />Solution:< />
< b="" />Step 1: Determine the enthalpy change due to cooling:< />
The enthalpy change (Δh) can be calculated using the specific heat capacity (Cp) and the temperature change (ΔT) as follows:
Δh = Cp * ΔT
ΔT = T1 - T2 = 32°C - 26°C = 6°C
Δh = 1.005 kJ/kgK * 6°C = 6.03 kJ/kg
< b="" />Step 2: Calculate the enthalpy at adiabatic saturation temperature:< />
The enthalpy at adiabatic saturation temperature (h_ad) can be calculated using the enthalpy change (Δh) and the enthalpy at the initial temperature (hv1) as follows:
h_ad = hv1 - Δh
h_ad = 2559.9 kJ/kg - 6.03 kJ/kg = 2553.87 kJ/kg
< b="" />Step 3: Determine the enthalpy of vaporization at the adiabatic saturation temperature:< />
The enthalpy of vaporization at the adiabatic saturation temperature (hfg_ad) can be calculated using the enthalpy of vaporization at the final temperature (hfg2) and the enthalpy at the adiabatic saturation temperature (h_ad) as follows:
hfg_ad = hfg2 + h_ad
hfg_ad = 2439.9 kJ/kg + 2553.87 kJ/kg = 4993.77 kJ/kg
< b="" />Step 4: Determine the humidity ratio:< />
The humidity ratio (W) can be calculated using the enthalpy of vaporization at the adiabatic saturation temperature (hfg_ad) and the enthalpy of liquid water at the adiabatic saturation temperature (hf_ad) as follows:
W = (hfg_ad / (hfg_ad - hf_ad)) * (P_total / (P_total - P_sat))
hf_ad = hf2 = 109.1 kJ/kg
W = (4993.77 kJ/kg / (4993.77 kJ/kg - 109.1 kJ/kg)) * (100 kPa / (
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An unsaturated air enters in an insulated chamber at 320Cwhere it flows over a long sheet of water and becomes cooled to 260C, which is adiabatic saturation temperature. The total pressure of the mixture remains constant at 100 kPa. The corresponding pressure to saturation temperature is 3.363 kPa. The humidity ratio at the entry of chamber is (in kg vap/kg dry air)Given data: Cpof air = 1.005 kJ/kgKAt 320C, hv1= 2559.9 kJ/kg,At 260C, hfg2= 2439.9 kJ/kg, hf2= 109.1 kJ/kgCorrect answer is between '0.017,0.021'. Can you explain this answer?
Question Description
An unsaturated air enters in an insulated chamber at 320Cwhere it flows over a long sheet of water and becomes cooled to 260C, which is adiabatic saturation temperature. The total pressure of the mixture remains constant at 100 kPa. The corresponding pressure to saturation temperature is 3.363 kPa. The humidity ratio at the entry of chamber is (in kg vap/kg dry air)Given data: Cpof air = 1.005 kJ/kgKAt 320C, hv1= 2559.9 kJ/kg,At 260C, hfg2= 2439.9 kJ/kg, hf2= 109.1 kJ/kgCorrect answer is between '0.017,0.021'. 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 An unsaturated air enters in an insulated chamber at 320Cwhere it flows over a long sheet of water and becomes cooled to 260C, which is adiabatic saturation temperature. The total pressure of the mixture remains constant at 100 kPa. The corresponding pressure to saturation temperature is 3.363 kPa. The humidity ratio at the entry of chamber is (in kg vap/kg dry air)Given data: Cpof air = 1.005 kJ/kgKAt 320C, hv1= 2559.9 kJ/kg,At 260C, hfg2= 2439.9 kJ/kg, hf2= 109.1 kJ/kgCorrect answer is between '0.017,0.021'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An unsaturated air enters in an insulated chamber at 320Cwhere it flows over a long sheet of water and becomes cooled to 260C, which is adiabatic saturation temperature. The total pressure of the mixture remains constant at 100 kPa. The corresponding pressure to saturation temperature is 3.363 kPa. The humidity ratio at the entry of chamber is (in kg vap/kg dry air)Given data: Cpof air = 1.005 kJ/kgKAt 320C, hv1= 2559.9 kJ/kg,At 260C, hfg2= 2439.9 kJ/kg, hf2= 109.1 kJ/kgCorrect answer is between '0.017,0.021'. Can you explain this answer?.
An unsaturated air enters in an insulated chamber at 320Cwhere it flows over a long sheet of water and becomes cooled to 260C, which is adiabatic saturation temperature. The total pressure of the mixture remains constant at 100 kPa. The corresponding pressure to saturation temperature is 3.363 kPa. The humidity ratio at the entry of chamber is (in kg vap/kg dry air)Given data: Cpof air = 1.005 kJ/kgKAt 320C, hv1= 2559.9 kJ/kg,At 260C, hfg2= 2439.9 kJ/kg, hf2= 109.1 kJ/kgCorrect answer is between '0.017,0.021'. 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 An unsaturated air enters in an insulated chamber at 320Cwhere it flows over a long sheet of water and becomes cooled to 260C, which is adiabatic saturation temperature. The total pressure of the mixture remains constant at 100 kPa. The corresponding pressure to saturation temperature is 3.363 kPa. The humidity ratio at the entry of chamber is (in kg vap/kg dry air)Given data: Cpof air = 1.005 kJ/kgKAt 320C, hv1= 2559.9 kJ/kg,At 260C, hfg2= 2439.9 kJ/kg, hf2= 109.1 kJ/kgCorrect answer is between '0.017,0.021'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An unsaturated air enters in an insulated chamber at 320Cwhere it flows over a long sheet of water and becomes cooled to 260C, which is adiabatic saturation temperature. The total pressure of the mixture remains constant at 100 kPa. The corresponding pressure to saturation temperature is 3.363 kPa. The humidity ratio at the entry of chamber is (in kg vap/kg dry air)Given data: Cpof air = 1.005 kJ/kgKAt 320C, hv1= 2559.9 kJ/kg,At 260C, hfg2= 2439.9 kJ/kg, hf2= 109.1 kJ/kgCorrect answer is between '0.017,0.021'. Can you explain this answer?.
Solutions for An unsaturated air enters in an insulated chamber at 320Cwhere it flows over a long sheet of water and becomes cooled to 260C, which is adiabatic saturation temperature. The total pressure of the mixture remains constant at 100 kPa. The corresponding pressure to saturation temperature is 3.363 kPa. The humidity ratio at the entry of chamber is (in kg vap/kg dry air)Given data: Cpof air = 1.005 kJ/kgKAt 320C, hv1= 2559.9 kJ/kg,At 260C, hfg2= 2439.9 kJ/kg, hf2= 109.1 kJ/kgCorrect answer is between '0.017,0.021'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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An unsaturated air enters in an insulated chamber at 320Cwhere it flows over a long sheet of water and becomes cooled to 260C, which is adiabatic saturation temperature. The total pressure of the mixture remains constant at 100 kPa. The corresponding pressure to saturation temperature is 3.363 kPa. The humidity ratio at the entry of chamber is (in kg vap/kg dry air)Given data: Cpof air = 1.005 kJ/kgKAt 320C, hv1= 2559.9 kJ/kg,At 260C, hfg2= 2439.9 kJ/kg, hf2= 109.1 kJ/kgCorrect answer is between '0.017,0.021'. Can you explain this answer?, a detailed solution for An unsaturated air enters in an insulated chamber at 320Cwhere it flows over a long sheet of water and becomes cooled to 260C, which is adiabatic saturation temperature. The total pressure of the mixture remains constant at 100 kPa. The corresponding pressure to saturation temperature is 3.363 kPa. The humidity ratio at the entry of chamber is (in kg vap/kg dry air)Given data: Cpof air = 1.005 kJ/kgKAt 320C, hv1= 2559.9 kJ/kg,At 260C, hfg2= 2439.9 kJ/kg, hf2= 109.1 kJ/kgCorrect answer is between '0.017,0.021'. Can you explain this answer? has been provided alongside types of An unsaturated air enters in an insulated chamber at 320Cwhere it flows over a long sheet of water and becomes cooled to 260C, which is adiabatic saturation temperature. The total pressure of the mixture remains constant at 100 kPa. The corresponding pressure to saturation temperature is 3.363 kPa. The humidity ratio at the entry of chamber is (in kg vap/kg dry air)Given data: Cpof air = 1.005 kJ/kgKAt 320C, hv1= 2559.9 kJ/kg,At 260C, hfg2= 2439.9 kJ/kg, hf2= 109.1 kJ/kgCorrect answer is between '0.017,0.021'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice An unsaturated air enters in an insulated chamber at 320Cwhere it flows over a long sheet of water and becomes cooled to 260C, which is adiabatic saturation temperature. The total pressure of the mixture remains constant at 100 kPa. The corresponding pressure to saturation temperature is 3.363 kPa. The humidity ratio at the entry of chamber is (in kg vap/kg dry air)Given data: Cpof air = 1.005 kJ/kgKAt 320C, hv1= 2559.9 kJ/kg,At 260C, hfg2= 2439.9 kJ/kg, hf2= 109.1 kJ/kgCorrect answer is between '0.017,0.021'. Can you explain this answer? tests, examples and also practice GATE tests.
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