Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 19, Problem 86P
To determine

The outlet mean temperature of water.

The surface temperature of the parallel plates.

The total rate of heat transfer.

Expert Solution & Answer
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Explanation of Solution

Given:

The value of width of plate is (a) is 1m.

The value of length of plate is (b) is 10m.

The mass flow rate (m˙) is 0.58kg/s

The free stream velocity (V) is 5m/s.

Calculation:

Refer to table A-15, “Properties of saturated water”.

Obtain the following properties of water corresponding to temperature 30°C as follows:

ρ=997kg/m3k=0.6129W/mKcp=4178J/kgKPr=5.68ν=0.8315×106m2/s

Calculate the average velocity of flow (Vavg) using the relation.

    Vavg=m˙ρAc=0.58kg/s(997kg/m3)(1m×(12.5mm103m1mm))=0.0465m/s

Calculate the perimeter (P) using the relation.

    P=2(a+b)=2(1m+(12.5mm103m1mm))=2.025m

Calculate the hydraulic diameter of the tube (Dh) using the relation.

    Dh=4AcP=4(ab)2.025m=4(1m×(12.5mm103m1mm))2.025m=0.0246m

Calculate the Reynolds number of flow (Re) using the relation.

    Re=VavgDhν=(0.0465m/s)×(0.0246m)(0.8315×106m2/s)=1376

The value of Reynolds number is less than 2300. Therefore the flow is laminar flow.

Calculate the ratio of dimensions a and b using the relation.

    ab=1m0.0125m=80

The Nusselt number value corresponding to ab=80 for fully developed laminar flow is 8.24.

    Nu=8.24

Calculate the heat transfer coefficient (h) using the relation.

    h=Nu(kD)=8.24(0.6129W/mK0.0246m)=205.29W/m2K

Calculate the total surface area (As) using the relation.

    As=L×2(a+b)=10m×2(1m+(12.5mm103m1mm))=20.25m2

Calculate the value of hAsm˙cp using the relation.

    hAsm˙cp=[(205.29W/m2K)×(20.25m2)(0.58kg/s)(4178J/kgK)]=1.71

Calculate the outlet temperature using interpolation from Figure 19-27 of book corresponding to hAsm˙cp value.

    5hAsm˙cp51=99.5Te99.570.65[1.71]51=99.5Te99.570.6Te=75.766°C

Thus, the outlet mean temperature of water is 75.766°C.

Calculate the surface temperature of fluid (Ts) using the relation.

    Ts=TeTiexp[hAsm˙cp]1exp[hAsm˙cp]=75.766°C(20°C)exp[1.71]1exp[1.71]=88°C

Thus, the surface temperature of the parallel plates is 88°C.

Refer to table A-23, “Properties of gases at 1atm pressure”.

Obtain the following properties of H2 gas corresponding to temperature 100°C:

ρ=0.06584kg/m3k=0.2095W/mKcp=14473J/kgKPr=0.7196ν=1.582×104m2/s

Calculate the Reynolds number of flow (Re) using the relation.

    Re=VavgLν=(5m/s)×(10m)(1.582×104m2/s)=316055.62

Calculate the Nusselt number (Nu) using the relation.

    Nu=0.332Re0.5Pr0.33=0.332(316055.62)0.5(0.677)0.33=164.10

Calculate the heat transfer coefficient (h) using the relation.

    h=Nu(kD)=164.10(0.21981W/mK10m)=3.60W/m2K

Calculate the heat transfer rate (Q) using the relation.

    Q=hAsΔTm=hAsΔTeΔTilnΔTΔTi=(3.60W/m2K)(20.25m2)(88°C75.766°C)(88°C20°C)ln(88°C75.766°C)(88°C20°C)=72.9W/K×(12.234°C+273)K(68°C+273)K1.7153=2370.06W

Thus, the uniform heat flux on tube surface is 2370.06W.

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Chapter 19 Solutions

Fundamentals of Thermal-Fluid Sciences

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