Fundamentals of Heat and Mass Transfer
Fundamentals of Heat and Mass Transfer
7th Edition
ISBN: 9780470501979
Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine
Publisher: Wiley, John & Sons, Incorporated
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Textbook Question
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Chapter 7, Problem 7.1P

Consider the following fluids at a film temperature of 300 K in parallel flow over a flat plate with velocity of 1 m/s: atmospheric air, water, engine oil, and mercury.

  1. For each fluid, determine the velocity and thermal boundary layer thicknesses at a distance of 40 mm from the leading edge.
  2. For each of the prescribed fluids and on the same coordinates, plot the boundary layer thicknesses as a function of distance from the leading edge to a plate length of 40 mm.

(a)

Expert Solution
Check Mark
To determine

The velocity and thermal boundary layer thickness at a distance of 40mm from the leading edge.

Answer to Problem 7.1P

The thermal boundary layer thickness for air, water, oil and mercury is 4.48, 0.52, 1.27, and 1.17 respectively.

Explanation of Solution

Given Information:

Properties of air at 300K, 1atm .

  v=15.89×106m2/s , Pr=0.707 .

Properties of water at 300K, 1atm .

  v=0.858×106m2/s , Pr=5.83 .

Properties of engine oil at 300K, 1atm .

  v=550×106m2/s , Pr=6400 .

Properties of mercury at 300K, 1atm .

  v=0.113×106m2/s , Pr=0.0248 .

Concept used:

The equation of Reynolds number is Re=uL1v .

The formula for nusselt number is Nu=hLk .

The formula for average nusselt number is Nu¯=0.664Rex12Pr13 .

The formula of laminar δ and δt is δ=5xRex12 , δt=δPr13 .

Nusselt number:

It is the ratio of convective heat transfer to convective heat transfer across a boundary.

  NuL=ConvectiveheattransferConductiveheattransfer=hk/L=hLk

Where h is the convective heat transfer coefficient of the flow, L is the characteristic length, k is the thermal conductivity of the fluid.

Calculation:

Know that the equation of Reynolds number.

  Re=uL1v

From the figure:

  Fundamentals of Heat and Mass Transfer, Chapter 7, Problem 7.1P , additional homework tip  1

Put 0.04m from leading edge, 1m/s for velocity in Reynolds equation.

  Re=1m/s( 0.04)v=0.04m2/sv

Put 15.89×106m2/s for velocity and obtain Reynolds number for air.

  Reair=0.0415.89× 10 6=2517.30

Know that the formula of laminar δ and δt .

  δ=5xRex12 , δt=δPr13

Put 0.04 for length, 2517.30 for Reynolds number and obtain laminar δ .

  δ=5×0.04 ( 2517 ) 1 2 =0.003986m

Convert the unit of laminar δ from meter into millimeter.

  δ=3.99mm

Know that the formula of laminar δt .

  δt=δPr13

Put 3.99mm for laminar δ , 0.707 for Prandtl number in equation of laminar δt .

  δt=3.99 ( 0.707 ) 1 3 =4.48mm

Put 0.858×106m2/s for velocity and obtain Reynolds number for water.

  Rewater=0.040.858× 10 6=4.66×104

Know that the formula of laminar δ .

  δ=5xRex12

Put 0.04 for length, 4.66×104 for Reynolds number and obtain laminar δ .

  δ=5×0.04 ( 4.66× 10 4 ) 1 2 =0.93mm

Know that the formula of laminar δt .

  δt=δPr13

Put 0.93mm for laminar δ , 5.83 for Prandtl number in equation of laminar δt .

  δt=0.93 ( 5.83 ) 1 3 =0.52mm

Put 550×106m2/s for velocity and obtain Reynolds number for engine oil.

  Reengineoil=0.04550× 10 6=72.7

Know that the formula of laminar δ .

  δ=5xRex12

Put 0.04 for length, 72.7 for Reynolds number and obtain laminar δ .

  δ=5×0.04 ( 72.7 ) 1 2 =23.5mm

Know that the formula of laminar δt .

  δt=δPr13

Put 23.5mm for laminar δ , 6400 for Prandtl number in equation of laminar δt .

  δt=23.5 ( 6400 ) 1 3 =1.27mm

Put 0.013×106m2/s for velocity and obtain Reynolds number for mercury.

  Remercury=0.040.113× 10 6=3.54×105

Know that the formula of laminar δ .

  δ=5xRex12

Put 0.04 for length, 3.54×105 for Reynolds number and obtain laminar δ .

  δ=5×0.04 ( 3.54× 10 5 ) 1 2 =0.34mm

Know that the formula of laminar δt .

  δt=δPr13

Put 0.34mm for laminar δ , 0.0248 for Prandtl number in equation of laminar δt .

  δt=23.5 ( 0.0248 ) 1 3 =1.17mm

Conclusion:

Hence, the thermal boundary layer thickness for air, water, oil and mercury is 4.48, 0.52, 1.27, and 1.17 respectively.

b)

Expert Solution
Check Mark
To determine

To draw: the boundary layer thickness from the leading edge to a plate length of 40mm .

Answer to Problem 7.1P

Created the graph for determined values of thermal boundary layer thickness for air, water, oil and mercury is 4.48, 0.52, 1.27, and 1.17 respectively with the values oflaminar thermal boundary layer air=3.99mm , water=0.93mm , mercury=0.34mm .

Explanation of Solution

Given Information:

The thermal boundary layer thickness for air, water, oil and mercury is 4.48, 0.52, 1.27, and 1.17 respectively.

Entire length =40mm .

Boundary layer air=3.99mm

  water=0.93mm

  mercury=0.34mm

Concept used:

Plot the graph by Matlab software, obtained values by formulas correlation between Reynolds equation, Nusselt number, average nusselt number and boundary layer thickness.

Calculation:

Know that the determined values of thermal boundary layer thickness for air, water, oil, mercury are 4.48, 0.52, 1.27, and 1.17 respectively and determined values of the laminar thermal boundary layer are air=3.99mm , water=0.93mm , mercury=0.34mm .

Substitute the values of thermal boundary layer thickness and laminar thermal boundary on the graph.

  Fundamentals of Heat and Mass Transfer, Chapter 7, Problem 7.1P , additional homework tip  2

Conclusion:

Created the graph for determined values of thermal boundary layer thickness for air, water, oil and mercury is 4.48, 0.52, 1.27, and 1.17 respectively with the values oflaminar thermal boundary layer air=3.99mm , water=0.93mm , mercury=0.34mm .

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