Solution Manual of Fundamentals of Momentum Heat and Mass Transfer 4th – 5th – 7

1. cl i ck h ere t o d ow nl oad Chapter 1 End of Chapter Problem Solutions 1.1 n = 4 x 1020 molecules/in3 == 1.32x104 in./s A= NA = (10-3 in)2 nA=1.04x108 m/s @solutionmanual1

2. cl i ck h ere t o d ow nl oad 1.2 Flow Properties: Velocity, Pressure Gradient, Stress Fluid Properties: Pressure, Temperature, Density, Speed of Sound, Specific Heat @solutionmanual1

3. cl i ck h ere t o d ow nl oad 1.3 mass of solid= ρsvs mass of fluid= ρfvf mix= = = @solutionmanual1

4. cl i ck h ere t o d ow nl oad 1.4 Given For P1= 1 atm P= 3001(1.01)7– 3000 = 217 atm = 7 = 1.01 @solutionmanual1

5. cl i ck h ere t o d ow nl oad 1.5 At Constant Temperature For 10% increase in P must also increase by 10 % = constant = constant @solutionmanual1

6. cl i ck h ere t o d ow nl oad 1.6 Since density varies as & =nM (M=Molecular wt.) n250,000= nS.L. = 4 x 1020 = 2.5 x 1016 @solutionmanual1

7. cl i ck h ere t o d ow nl oad 1.7 = x + y = + = x + y = + Q.E.D. @solutionmanual1

8. cl i ck h ere t o d ow nl oad 1.8 = = Q.E.D. = + = @solutionmanual1

9. cl i ck h ere t o d ow nl oad 1.9 Transformation from (x,y) to (r,) = r2 = x2+y2 = = + + so: = = = = = = = = = = + @solutionmanual1

10. cl i ck h ere t o d ow nl oad 1.10 = = ( = ( + ) Thus: = + + ) + ( + )+ + (+) + + @solutionmanual1

11. cl i ck h ere t o d ow nl oad 1.11 P= P(a,b)= = + 2 + (sin1cos1) 2{[ + 2)+ ( ) @solutionmanual1

12. cl i ck h ere t o d ow nl oad 1.12 T(x,y)= Toe-1/4 [ T(a,b)= Toe-1/4 [ = Toe-1/4[ = (cos cosh )+ (sin sinh ) (coscosh1)+ (sinsinh1) + ] [ (1+e-2) + @solutionmanual1

13. cl i ck h ere t o d ow nl oad 1.13 In problem 1.12 T(x,y) is dimensionally homogeneous (D.H.) P(x,y) in Prob 1.11 will be D.H. if ~ or using the conversion factor gc LBf s2/ ft4 @solutionmanual1

14. cl i ck h ere t o d ow nl oad 1.14 = 3x2y+4y2 A scalar field is given by the function: (a)Find at the point (3,5) For the value of at the point (3,5) (b)Find the component of that makes a -60 angle with the axis at the point (3,5) Let the unit vector be represented by At the point (3,5) this becomes: @solutionmanual1

15. cl i ck h ere t o d ow nl oad 1.15 For an ideal gas P= From Prob 1.3: @solutionmanual1

16. cl i ck h ere t o d ow nl oad 1.16 = Arsin(1- a) = b)= A max is given by d= 0 or Requiring For And for From Eq. 2: 4a2/r2=0 If a0, r then for which = 0, Subst. into Eq. 1 =0, 1-a2/r2=0 Giving a = r For = 1+ a2/r2 =0 impossible Thus conditions for max are r = a ) = Asin(1- + ) 1/2 dr+ d= 0 = =0 = 0: -(1+ ) + (1- ) = 0 (1) =0: (2) (3) @solutionmanual1

17. cl i ck h ere t o d ow nl oad 1.17 P=Po+ 2 = = + = + P = 2 2 2 2 @solutionmanual1

18. cl i ck h ere t o d ow nl oad 1.18 Vertical cylinder d=10m, h=6m V= @ 20w=998.2 kg/m3 m= wV= (998.2)(471.2)= 470350 kg @ 80w=971.8 kg/m3 m=(971.8)(471.2)= 457910 kg 12440 kg (10m)2(6m)= 471.2 m3 @solutionmanual1

19. cl i ck h ere t o d ow nl oad 1.19 Liquid = -V V= 1194 cm3= 1.194 x 10-3 m3 ∆V= -12 cm3= -1.2 x 10-7 m3 = -1.194 x 10-3 = +12440 MPa V= 1200 cm3 @ 1.25 MPa V=1188 cm3 @ 2.5 MPa T -V @solutionmanual1

20. cl i ck h ere t o d ow nl oad 1.20 = -V V=0.25 m3 V= -0.005 m3 P= 10 mPa = -0.25 T-V = 500 MPa @solutionmanual1

21. cl i ck h ere t o d ow nl oad 1.21 For H20: = -0.0075 -V ∆P= (2.205 GPa)(0.0075) = 0.0165 GPa= 16.5 MPa =2.205 GPa or ∆P = @solutionmanual1

22. cl i ck h ere t o d ow nl oad 1.22 For H20: = -V = P2=120 MPa 2.205 P1=100kPa or = = = 0.999 x 10-3 = 0.0999 percent @solutionmanual1

23. cl i ck h ere t o d ow nl oad 1.23 H20@ 68 (341 K) 0.123 [1-0.00139(341)] = 0.0647 N/m In a clean tube- =0 h = = 9.37 x 10-3m = 9.37 mm = @solutionmanual1

24. cl i ck h ere t o d ow nl oad 1.24 Parallel Glass Plates Gap=1.625 mm = 0.0735 N/m For a unit depth: Surface Tension Force = 2(1)cos Weight of H20 = (1)(1.625x10-3) For clean glass cos=1 Equating Forces: 2(1)= (1)(1.625x10-3) h = = 0.00922m= 9.22 mm @solutionmanual1

25. cl i ck h ere t o d ow nl oad 1.25 1.26 H20-Air-Glass Interface @40 Tube Radius= 1 mm h= = 0.123[1-0.0139(313)]=0.0695 N/m h = cos = 0.0143m (1.43 cm) @solutionmanual1

26. cl i ck h ere t o d ow nl oad 1.26 1.27 Soap Bubble- T=20 d=4mm = 0.025 N/m (Table 1.2) Force Balance for Bubble: r2r so = = 25 N/m2 = 25 Pa @solutionmanual1

27. cl i ck h ere t o d ow nl oad 1.29 @60C Tube Diameter= 0.55 mm h= For H20: 1.27 = 0.0662 N/m = 0.44 N/m = 0.0499m (4.99 cm Rise) For Hg: = -0.0157 m (1.57 cm Depression) @solutionmanual1

28. cl i ck h ere t o d ow nl oad 1.28 1.30 H20/ Glass Interface T=30C = 0.123[1-0.0139(303)] = 0.0712 N/m = 996 kg/m3 h 1 mm h = r = d = 2r = 2.915 cm = 0.0146 m (1.457 cm) = @solutionmanual1

29. cl i ck h ere t o d ow nl oad 1.29 1.31 Bubble Diameter = 0.25 cm = 0.0025 m, and so Radius = 0.00125 m Capillary Tube: Diameter = 0.2cm = 0.002 m, and so Radius = 0.001 m Beginning with: Rearrange and remember the unit conversion Pa=kg/ms2, Next, we can calculate the height of the fluid in the tube: @solutionmanual1

30. cl i ck h ere t o d ow nl oad 1.30 1.32 First, calculate the surface tension of water using the temperature of the water: Next, using the equation for the height of a fluid in a capillary, Rearranging and solving for the radius: Diameter is 2r so D=1.50 mm @solutionmanual1