Fundamentals Of Thermal-fluid Sciences In Si Units
Fundamentals Of Thermal-fluid Sciences In Si Units
5th Edition
ISBN: 9789814720953
Author: Yunus Cengel, Robert Turner, John Cimbala
Publisher: McGraw-Hill Education
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Chapter 19, Problem 82P
To determine

The tube surface temperature necessary to heat water.

The tube surface temperature necessary to heat engine oil.

The tube surface temperature necessary to heat liquid mercury.

Expert Solution & Answer
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Explanation of Solution

Given:

The diameter of tube is (D) is 25mm.

The length of tube is (L) is 15m.

The inlet temperature (Ti) of fluid is 50oC.

The outlet temperature (Te) of fluid is 150oC.

The mass flow rate (m˙) is 0.01kg/s.

Calculation:

Calculate the average temperature.

    Tavg=Ti+Te2=50°C+150°C2=100°C

For water:

Refer Table A-15 “Properties of saturated water”.

Obtain the following properties of water corresponding to the temperature of 100°C as follows:

ρ=957.9kg/m3

k=0.679W/mK

Pr=1.75

μ=0.282×103kg/ms

Calculate the Reynolds number for water.

    Re=4m˙πDμ=4×0.01kg/sπ(25mm103m1mm)(0.282×103kg/ms)=1806.01

The value of Reynolds number is less than 2000. Therefore the flow is laminar flow.

Calculate the hydrodynamic entry length for water.

    LH=0.05ReD=0.05×1806.01×25mm103m1mm=0.05×1806.01×0.025m=2.2575m

Calculate the thermal entry length for water.

    LT=Pr×LH=1.75×2.2575m=3.9506m

The value of hydrodynamic entry length and thermal entry length is less than 15m. Therefore the flow is fully developed flow.

The value of Nusselt number for fully developed laminar flow is 3.66.

    Nu=3.66

Calculate the heat transfer coefficient.

    h=Nu(kD)=3.66(0.679W/mK25mm103m1mm)=99.41W/m2K

Calculate the surface area of tube.

    As=πDL=π(25mm103m1mm)(15m)=1.178m2

Calculate the exit temperature of fluid.

    Ts=TeTiexp[hAsm˙cp]1exp[hAsm˙cp]=150°C(50°C)exp[(99.41W/m2°C)×(1.178m2)(0.01kg/s)(4217J/kgK)]1exp[(99.41W/m2°C)×(1.178m2)(0.01kg/s)(4217J/kgK)]=150°C(50°C)exp[2.7769]1exp[2.7769]=156.6°C

Thus, the tube surface temperature is 156.6°C.

For engine oil:

Refer Table A-19 “Properties of liquids”.

Obtain the following properties of the engine oil corresponding to the temperature of 100°C as follows:

k=0.1367W/mK

cp=2220J/kgK

Pr=279.1

μ=0.01718kg/ms

Calculate the Reynolds number for engine oil.

    Re=4m˙πDμ=4×0.01kg/sπ(25mm103m1mm)(0.01718kg/ms)=29.64

The value of Reynolds number is less than 2000. Therefore the flow is laminar flow.

Calculate the hydrodynamic entry length for engine oil.

    LH=0.05ReD=0.05×29.64×25mm103m1mm=0.037m

Calculate the thermal entry length for engine oil.

    LT=Pr×LH=279.1×0.037m=10.34m

Calculate the heat transfer coefficient for engine oil.

    h=Nu(kD)=3.66(0.1367W/mk25mm103m1mm)=20.01W/m2°C

Calculate the exit temperature of fluid.

    Ts=TeTiexp[hAsm˙cp]1exp[hAsm˙cp]=150°C(50°C)exp[(20.01W/m2°C)×(1.178m2)(0.01kg/s)(2220J/kgK)]1exp[(20.01W/m2°C)×(1.178m2)(0.01kg/s)(2220J/kgK)]=150°C(50°C)exp[1.06126]1exp[1.06126]=202.9°C

Thus, the tube surface temperature is 202.9°C.

For liquid mercury:

Refer Table A-20 "Properties of liquid metals”.

Obtain the following properties of liquid mercury corresponding to the temperature of 100°C as follows:

k=9.467W/mK

cp=137.1J/kgK

Pr=0.018

μ=1.245×103kg/ms

Calculate the Reynolds number for liquid mercury.

    Re=4m˙πDμ=4×0.01kg/sπ(25mm103m1mm)(1.245×103kg/ms)=409.07

The value of Reynolds number is less than 2000. Therefore the flow is laminar flow.

Calculate the hydrodynamic entry length for liquid mercury.

    LH=0.05ReD=0.05×409.07×25mm103m1mm=0.05×1806.01×0.025m=0.511m

Calculate the thermal entry length for liquid mercury.

    LT=Pr×LH=0.018×0.511m=0.0092m

Calculate the heat transfer coefficient for liquid mercury.

    h=Nu(kD)=3.66(9.467W/mk25mm103m1mm)=1385.98W/m2°C

Calculate the exit temperature of fluid.

    Ts=TeTiexp[hAsm˙cp]1exp[hAsm˙cp]=150°C(50°C)exp[(1385.98W/m2°C)×(1.178m2)(0.01kg/s)(137.1J/kgK)]1exp[(1385.98W/m2°C)×(1.178m2)(0.01kg/s)(137.1J/kgK)]=150°C(50°C)exp[1190.87]1exp[1190.87]=150°C

Thus, the tube surface temperature is 150°C.

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Chapter 19 Solutions

Fundamentals Of Thermal-fluid Sciences In Si Units

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