Solution Manual for Fundamentals of Momentum, Heat and Mass Transfer, 6th Edition

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Solution Manual for Fundamentals of Momentum, Heat and Mass Transfer, 6th Edition - Page 1 preview image

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Chapter 1 End of Chapter Problem Solutions1.1n = 4 x 1020molecules/in3== 1.32x104in./sA=(10-3in)2NA =nA=1.04x108m/s

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1.2Flow Properties: Velocity, Pressure Gradient, StressFluid Properties: Pressure, Temperature, Density, Speed of Sound, Specific Heat

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1.3mass of solid=ρsvsmass of fluid= ρfvfmix===

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1.4Given=7For P1= 1 atm= 1.01P= 3001(1.01)7–3000 = 217 atm

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1.5At Constant Temperature= constant= constantFor 10% increase inPmust also increase by 10 %

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1.6Since density varies as&=nM(M=Molecular wt.)n250,000= nS.L.= 4 x 1020= 2.5 x 1016

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1.7=x+y=+=x+y=+Q.E.D.

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1.8=+===Q.E.D.

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1.9 Transformation from (x,y) to (r,)=+=+r2= x2+y2=so:==========+

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1.10=++= ()+ (+)+= (+)+(+)+Thus:=+

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1.11P=+P(a,b)=2+(sin1cos1)=2{[+ 2)+()

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1.12T(x,y)= Toe-1/4[(coscosh)+(sinsinh)T(a,b)= Toe-1/4[(coscosh1)+(sinsinh1)= Toe-1/4[+]=[(1+e-2)+

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1.13In problem 1.12 T(x,y) is dimensionally homogeneous (D.H.)P(x,y) in Prob 1.11 will be D.H. if~LBfs2/ ft4or using the conversion factor gc

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1.14= 3x2y+4y2A scalar field is given by the function:(a)Findat the point (3,5)For the value ofat the point (3,5)(b)Find the component ofthat makes a -60angle with the axis at the point (3,5)Let the unit vector be represented byAt the point (3,5) this becomes:

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