Chapter04

Mechanics of Fluids

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50 CHAPTER 4 The Integral Forms of the Fundamental Laws 4.1 a) No net force may act on the system: Σ v F = 0. b) The energy transferred to or from the system must be zero: Q - W = 0 . c) If 3 32 ˆˆ ˆ 1 0() 0 n V V n ij =⋅= -= v is the same for all volume elements then Σ v v F D Dt V dm = , or Σ v v F D mV = ( ). Since mass is constant for a system Σ v v F m DV = . Since a F ma v v v v = = , . Σ 4.2 Extensive properties: Mass, m ; Momentum, mV v ; kinetic energy, 1 2 2 ; potential energy, mgh ; enthalpy, H . Associated intensive properties (divide by the mass): unity, 1; velocity, v V ; V 2 /2; gh ; H/m = h (specific enthalpy). Intensive properties: Temperature, T ; time, t ; pressure, p ; density, r ; viscosity, m . 4.3 ( B ) 4.4 System () tV = 1 c.v. = 1 System ttV + ∆= 1 V + 2 c.v. + 1 4.5 System = 1 V + 2 c.v. = 1 V + 2 System + 2 V + 3 c.v. + 1 V + 2 1 2 1 2 3 pump
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51 4.6 a) The energy equation (the 1 st law of Thermo). b) The conservation of mass. c) Newton’s 2 nd law. d) The energy equation. e) The energy equation. 4.7 4.8 4.9 4.10 $ $ $ . ( $ $ ) n i j i j 1 1 2 1 2 0 707 = - - = - + . $ . $ .5 $ n i j 2 0 866 0 = - . $ $ n j 3 = - . 1 11 ˆ ˆˆ ˆ 1 0 [ 0.707 ( ) ] 7.07 fps n V V n i ij =⋅= - + =- v V V n i i j n 2 2 2 10 0866 05 866 = = - = v $ $ ( . $ . $ ) . fps 3 32 ˆ 1 0() 0 n V V n ij -= v 4.11 flux = hr $ n VA v flux 1 = hr [ . ( $ $ )] $ / . - + = - 0 707 10 0 707 10 i j iA A flux 2 = ( . $ .5 $ ) $ / . 0 866 0 10 0 866 10 i j A - = flux 3 = hr ( $ ) $ - = j 10 0 3 ˆ n v ˆ n v ϖ ˆ n v ˆ n v ˆ n v ˆ n v ˆ n ˆ n ˆ n v v v v ˆ n ˆ n ˆ n ˆ n v v v v ˆ n ˆ n v v
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52 4.12 ( $ ) ( .5 $ . $ ) $ ( ) v B n A i j j = + × 15 0 0 866 10 12 = × × = 15 0 866 120 1559 . cm 3 Volume = 15 60 10 12 1559 sin cm 3 o × × = 4.13 The control volume must be independent of time. Since all space coordinates are integrated out on the left, only time remains; thus, we use an ordinary derivative to differentiate a function of time. But, on the right, we note that r and h may be functions of ( x, y, z, t ); hence, the partial derivative is used. 4.14 4.15 4.16 4.17 If fluid crosses the control surface only on areas A 1 and A 2 , r r r $ $ $ . . n VdA n VdA n VdA A A c s = + = v v v 0 2 1 For uniform flow all quantities are constant over each area: r r 1 1 1 2 2 2 0 2 1 $ $ n V dA n V dA A A + = v v Let A 1 be the inlet so $ n V V A 1 1 1 2 = - v and be the outlet so $ . n V V 2 2 2 = v Then - + = r r 1 1 1 2 2 2 0 V A V A or r r 2 2 2 1 1 1 A V A V = 1 2 1 system ( t ) is in volumes 1 and 2 c.v. (0) = c.v. ( t ) = volume 1 1 2 3 system ( t ) = V 1 + V 2 + V 3 c.v. ( t ) = V 1 + V 2 system boundary at ( t + t )
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53 4.18 Use Eq. 4.4.2 with m V representing the mass in the volume: 0 = + dm dt n VdA V c s r $ . . v = + - dm dt A V A V V r r 2 2 1 1 = + - dm dt Q m V r & . Finally, dm dt m Q V = - & . r 4.19 Use Eq. 4.4.2 with m S representing the mass in the sponge: 0 = + dm dt S r $ v = + + - dm dt A V A V A V S r r r 2 2 3 3 1 1 = + + - dm dt m A V Q S & . 2 3 3 1 r r Finally, dm dt Q m A V S = - - r r 1 2 3 3 & . 4.20 ( D ) 2 200 0.0 4 7 0 0.837 kg/s 0.28 7 293 p m A V AV RT rp = = = × ×= × & . 4.21 A 1 V 1 = A 2 V 2 . p × 125 144 2 . × 60 = p × 2 5 144 2 . V 2 . V 2 = 15 ft/sec . & . . m AV = = × r p 194 1 25 144 60 2 = 3.968 slug/sec . Q = AV = p 144 2 .
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Chapter04 - CHAPTER 4 The Integral Forms of the Fundamental...

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