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Air flows through a 10 cm internal diameter tube at the rate of 75 kg/hr. Measurements indicate that at a particular section in the tube, the pressure and mixing cup temperature of air are 1.5 bar and 325 K respectively while the tube wall temperature is 375 K. The heat transfer rate (in W) for one meter length in the region is ________. The general nondimensional correlations for turbulent flow in the tube is
Nu = 0.023 Re0.8 Pr0.4
where the fluid properties are evaluated at the bulk temperature.
For air at tf = 325 K and p 1.5 bar, the thermo-physical properties of air are 
μ = 1.967 x 10-5kg/ms
k = 0.02792 W/mK 
and Pr = 0.713 
    Correct answer is between '177.0,178.0'. Can you explain this answer?
    Verified Answer
    Air flows through a 10 cm internal diameter tube at the rate of 75 kg/...

    This is well in excess of the critical Reynolds number for flow in tubes (2500); thus the flow is turbulent and the given correlation applies
    i .e. Nu = 0.023 (13485)0.8(0.713)0.4 = 40.4409
    This gives the convective coefficient as
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    Most Upvoted Answer
    Air flows through a 10 cm internal diameter tube at the rate of 75 kg/...
    :

    - Density (ρ) = 6.21 kg/m³
    - Specific heat capacity at constant pressure (cp) = 1.005 kJ/kg·K
    - Thermal conductivity (k) = 0.034 W/m·K
    - Dynamic viscosity (μ) = 2.07 × 10⁻⁵ Pa·s
    - Prandtl number (Pr) = 0.7 (estimated)

    To calculate the heat transfer rate, we need to first determine the Reynolds number (Re) and Nusselt number (Nu) for the flow. The Reynolds number is given by:

    Re = ρvd/μ

    where v is the velocity of the flow and d is the diameter of the tube. Substituting the given values, we get:

    v = ṁ/ρA = (75 kg/hr) / (3600 s/hr) / (π/4) (0.1 m)² ρ = 9.45 m/s
    Re = (6.21 kg/m³) (9.45 m/s) (0.1 m) / (2.07 × 10⁻⁵ Pa·s) = 2.85 × 10⁵

    Next, the Nusselt number can be calculated using the given correlation:

    Nu = 0.023 Re⁰·⁸ Pr⁰·⁴

    Substituting the values, we get:

    Nu = 0.023 (2.85 × 10⁵)⁰·⁸ (0.7)⁰·⁴ = 163.7

    Finally, the heat transfer rate per unit length (q) can be calculated using the following formula:

    q = h A ΔT

    where h is the heat transfer coefficient, A is the cross-sectional area of the tube, and ΔT is the temperature difference between the fluid and the tube wall. The heat transfer coefficient can be calculated from the Nusselt number using the formula:

    h = k Nu / d

    Substituting the values, we get:

    h = (0.034 W/m·K) (163.7) / (0.1 m) = 55.7 W/m²·K

    The temperature difference between the fluid and the tube wall is:

    ΔT = Tfluid - Twall = 325 K - 375 K = -50 K

    (Note that the temperature gradient is negative because the wall temperature is higher than the fluid temperature.)

    Substituting the values, we get:

    q = (55.7 W/m²·K) (π/4) (0.1 m)² (-50 K) = -10.9 W/m

    (Note that the heat transfer rate is negative because heat is being transferred from the fluid to the tube wall, which results in a decrease in the fluid temperature.) Therefore, the heat transfer rate for one meter length in the region is approximately 10.9 W/m.
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    Air flows through a 10 cm internal diameter tube at the rate of 75 kg/hr. Measurements indicate that at a particular section in the tube, the pressure and mixing cup temperature of air are 1.5 bar and 325 K respectively while the tube wall temperature is 375 K. The heat transfer rate (in W) for one meter length in the region is ________. The general nondimensional correlations for turbulent flow in the tube isNu = 0.023 Re0.8 Pr0.4where the fluid properties are evaluated at the bulk temperature.For air at tf = 325 K and p 1.5 bar, the thermo-physical properties of air areμ = 1.967 x 10-5kg/msk = 0.02792 W/mKand Pr = 0.713Correct answer is between '177.0,178.0'. Can you explain this answer?
    Question Description
    Air flows through a 10 cm internal diameter tube at the rate of 75 kg/hr. Measurements indicate that at a particular section in the tube, the pressure and mixing cup temperature of air are 1.5 bar and 325 K respectively while the tube wall temperature is 375 K. The heat transfer rate (in W) for one meter length in the region is ________. The general nondimensional correlations for turbulent flow in the tube isNu = 0.023 Re0.8 Pr0.4where the fluid properties are evaluated at the bulk temperature.For air at tf = 325 K and p 1.5 bar, the thermo-physical properties of air areμ = 1.967 x 10-5kg/msk = 0.02792 W/mKand Pr = 0.713Correct answer is between '177.0,178.0'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Air flows through a 10 cm internal diameter tube at the rate of 75 kg/hr. Measurements indicate that at a particular section in the tube, the pressure and mixing cup temperature of air are 1.5 bar and 325 K respectively while the tube wall temperature is 375 K. The heat transfer rate (in W) for one meter length in the region is ________. The general nondimensional correlations for turbulent flow in the tube isNu = 0.023 Re0.8 Pr0.4where the fluid properties are evaluated at the bulk temperature.For air at tf = 325 K and p 1.5 bar, the thermo-physical properties of air areμ = 1.967 x 10-5kg/msk = 0.02792 W/mKand Pr = 0.713Correct answer is between '177.0,178.0'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Air flows through a 10 cm internal diameter tube at the rate of 75 kg/hr. Measurements indicate that at a particular section in the tube, the pressure and mixing cup temperature of air are 1.5 bar and 325 K respectively while the tube wall temperature is 375 K. The heat transfer rate (in W) for one meter length in the region is ________. The general nondimensional correlations for turbulent flow in the tube isNu = 0.023 Re0.8 Pr0.4where the fluid properties are evaluated at the bulk temperature.For air at tf = 325 K and p 1.5 bar, the thermo-physical properties of air areμ = 1.967 x 10-5kg/msk = 0.02792 W/mKand Pr = 0.713Correct answer is between '177.0,178.0'. Can you explain this answer?.
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