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Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of air
D = 0.256 * 10 -4 m2/s
µ = 1.86 * 10 -5 kg/m s
c p = 1.005 k J/kg degree Celsius
p r = 0.701
p =1.165 kg/m3
D = 0.256 * 10 -4 m2/s
µ = 1.86 * 10 -5 kg/m s
c p = 1.005 k J/kg degree Celsius
p r = 0.701
p =1.165 kg/m3
- a)0.1076 m/s
- b)0.2076 m/s
- c)0.3076 m/s
- d)0.4076 m/s
Correct answer is option 'A'. Can you explain this answer?
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Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat p...
Explanation: h m = 0.0296 (Re) -0.2 (V)/ (Sc) 0.667.
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Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat p...
To estimate the value of the mass transfer coefficient, we can use the Sherwood number correlation for forced convection over a flat plate. The Sherwood number (Sh) is defined as the dimensionless mass transfer coefficient, and it can be related to the Reynolds number (Re) and Schmidt number (Sc) as follows:
Sh = 0.023 * Re^0.8 * Sc^0.33
To calculate the Reynolds number, we need to determine the velocity and characteristic length. The velocity is given as 45 m/s, and the characteristic length is the length of the flat plate, which is 0.5 m. Therefore:
Re = (velocity * length) / kinematic viscosity
= (45 * 0.5) / (0.256 * 10^-4)
= 87,890
Next, we need to determine the Schmidt number, which is the ratio of momentum diffusivity to mass diffusivity:
Sc = kinematic viscosity / mass diffusivity
= (0.256 * 10^-4) / ? (unknown value)
Unfortunately, the water vapor diffusivity in air is not provided, so we cannot calculate the Schmidt number or the mass transfer coefficient without this information.
Sh = 0.023 * Re^0.8 * Sc^0.33
To calculate the Reynolds number, we need to determine the velocity and characteristic length. The velocity is given as 45 m/s, and the characteristic length is the length of the flat plate, which is 0.5 m. Therefore:
Re = (velocity * length) / kinematic viscosity
= (45 * 0.5) / (0.256 * 10^-4)
= 87,890
Next, we need to determine the Schmidt number, which is the ratio of momentum diffusivity to mass diffusivity:
Sc = kinematic viscosity / mass diffusivity
= (0.256 * 10^-4) / ? (unknown value)
Unfortunately, the water vapor diffusivity in air is not provided, so we cannot calculate the Schmidt number or the mass transfer coefficient without this information.
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Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of airD = 0.256 * 10-4m2/sµ = 1.86 * 10-5kg/m scp= 1.005 k J/kg degree Celsiuspr= 0.701p =1.165 kg/m3a)0.1076 m/sb)0.2076 m/sc)0.3076 m/sd)0.4076 m/sCorrect answer is option 'A'. Can you explain this answer?
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Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of airD = 0.256 * 10-4m2/sµ = 1.86 * 10-5kg/m scp= 1.005 k J/kg degree Celsiuspr= 0.701p =1.165 kg/m3a)0.1076 m/sb)0.2076 m/sc)0.3076 m/sd)0.4076 m/sCorrect answer is option 'A'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of airD = 0.256 * 10-4m2/sµ = 1.86 * 10-5kg/m scp= 1.005 k J/kg degree Celsiuspr= 0.701p =1.165 kg/m3a)0.1076 m/sb)0.2076 m/sc)0.3076 m/sd)0.4076 m/sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of airD = 0.256 * 10-4m2/sµ = 1.86 * 10-5kg/m scp= 1.005 k J/kg degree Celsiuspr= 0.701p =1.165 kg/m3a)0.1076 m/sb)0.2076 m/sc)0.3076 m/sd)0.4076 m/sCorrect answer is option 'A'. Can you explain this answer?.
Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of airD = 0.256 * 10-4m2/sµ = 1.86 * 10-5kg/m scp= 1.005 k J/kg degree Celsiuspr= 0.701p =1.165 kg/m3a)0.1076 m/sb)0.2076 m/sc)0.3076 m/sd)0.4076 m/sCorrect answer is option 'A'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of airD = 0.256 * 10-4m2/sµ = 1.86 * 10-5kg/m scp= 1.005 k J/kg degree Celsiuspr= 0.701p =1.165 kg/m3a)0.1076 m/sb)0.2076 m/sc)0.3076 m/sd)0.4076 m/sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of airD = 0.256 * 10-4m2/sµ = 1.86 * 10-5kg/m scp= 1.005 k J/kg degree Celsiuspr= 0.701p =1.165 kg/m3a)0.1076 m/sb)0.2076 m/sc)0.3076 m/sd)0.4076 m/sCorrect answer is option 'A'. Can you explain this answer?.
Solutions for Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of airD = 0.256 * 10-4m2/sµ = 1.86 * 10-5kg/m scp= 1.005 k J/kg degree Celsiuspr= 0.701p =1.165 kg/m3a)0.1076 m/sb)0.2076 m/sc)0.3076 m/sd)0.4076 m/sCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemical Engineering.
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Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of airD = 0.256 * 10-4m2/sµ = 1.86 * 10-5kg/m scp= 1.005 k J/kg degree Celsiuspr= 0.701p =1.165 kg/m3a)0.1076 m/sb)0.2076 m/sc)0.3076 m/sd)0.4076 m/sCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of airD = 0.256 * 10-4m2/sµ = 1.86 * 10-5kg/m scp= 1.005 k J/kg degree Celsiuspr= 0.701p =1.165 kg/m3a)0.1076 m/sb)0.2076 m/sc)0.3076 m/sd)0.4076 m/sCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of airD = 0.256 * 10-4m2/sµ = 1.86 * 10-5kg/m scp= 1.005 k J/kg degree Celsiuspr= 0.701p =1.165 kg/m3a)0.1076 m/sb)0.2076 m/sc)0.3076 m/sd)0.4076 m/sCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of airD = 0.256 * 10-4m2/sµ = 1.86 * 10-5kg/m scp= 1.005 k J/kg degree Celsiuspr= 0.701p =1.165 kg/m3a)0.1076 m/sb)0.2076 m/sc)0.3076 m/sd)0.4076 m/sCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Chemical Engineering tests.
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