SM_3B - Chapter 3 Section B Non-Numerical Solutions 3.2...

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Chapter 3 - Section B - Non-Numerical Solutions 3.2 Differentiate Eq. (3.2) with respect to P and Eq. (3.3) with respect to T : µ ∂β P T =− 1 V 2 µ V P T µ V T P + 1 V µ 2 V P T = βκ + µ 2 V P T µ ∂κ T P = 1 V 2 µ V T P µ V P T 1 V µ 2 V T P =− βκ µ 2 V P T Addition of these two equations leads immediately to the given equation. One could of course start with Eq. (3.4) and apply the condition for an exact differential, but this topic is not covered until Chapter 6. 3.3 The Tait equation is given as: V = V 0 µ 1 AP B + P where V 0 , A , and B are constants. Application of Eq. (3.3), the definition of κ , requires the derivative of this equation: µ V P T = V 0 · A B + P + AP ( B + P ) 2 ¸ = AV 0 B + P µ 1 + P B + P Multiplication by 1 / V in accord with Eq. (3.3), followed by substitution for V 0 / V by the Tait equa- tion leads to: κ = AB ( B + P ) [ B + ( 1 A ) P ] 3.7 ( a ) For constant T , Eq. (3.4) becomes: dV V =− κ dP Integration from the initial state ( P 1 , V 1 ) to an intermediate state ( P , V ) for constant κ gives: ln V V 1 =− κ( P P 1 ) Whence, V = V 1 exp[ κ( P P 1 ) ] = V 1 exp ( κ P ) exp P 1 ) If the given equation applies to the process, it must be valid for the initial state; then, A ( T ) = V 1 exp P 1 ) , and V = A ( T ) exp ( κ P ) ( b ) Differentiate the preceding equation: dV =− κ A ( T ) exp ( κ P ) dP Therefore, W =− Z V 2 V 1 PdV = κ A ( T ) Z P 2 P 1 P exp ( κ P ) dP = A ( T ) κ [ P 1 + 1 ) exp ( κ P 1 ) P 2 + 1 ) exp ( κ P 2 ) ] 545
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With V 1 = A ( T ) exp ( κ P 1 ) and V 2 = A ( T ) exp ( κ P 2 ) , this becomes: W = 1 κ [ P 1 + 1 ) V 1 P 2 + 1 ) V 2 ] or W = P 1 V 1 P 2 V 2 + V 1 V 2 κ 3.11 Differentiate Eq. (3.34c) with respect to T : T µ 1 δ δ P [ ( 1 δ)/δ ] 1 dP dz + P ( 1 δ)/δ dT dz = T µ 1 δ δ P ( 1 δ)/δ P dP dz + P ( 1 δ)/δ dT dz = 0 Algebraic reduction and substitution for dP / dz by the given equation yields: T P µ 1 δ δ ( M ρ g ) + dT dz = 0 For an ideal gas T ρ/ P = 1 / R . This substitution reduces the preceding equation to: dT dz =− M g R µ δ 1 δ 3.12 Example 2.12 shows that U 2 = H 0 . If the gas is ideal, H 0 = U 0 + P 0 V 0 = U 0 + RT 0 and U 2 U 0 = RT 0 For constant C V , U 2 U 0 = C V ( T 2 T 0 ) and C V ( T 2 T 0 ) = RT 0 Whence, T 2 T 0 T 0 = R C V = C P C V C V When C P / C V is set equal to γ , this reduces to: T 2 = γ T 0 This result indicates that the final temperature is independent of the amount of gas admitted to the tank, a result strongly conditioned by the assumption of no heat transfer between gas and tank. 3.13 Isobaric case
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SM_3B - Chapter 3 Section B Non-Numerical Solutions 3.2...

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