quiz_06_solution

# quiz_06_solution - EML 5104 Classical Thermodynamics,...

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Unformatted text preview: EML 5104 Classical Thermodynamics, Spring 2010 Quiz 6, 19 Feb 2010 Name (Print): ___________________________________ UF ­ID: ___________________________________ Please sign the honor pledge: "On my honor, I have neither given nor received unauthorized aid in doing this assignment." _________________________________ Signature Only mark one of the multiple choice answers. Use additional paper if necessary. Q1: The Redlich ­Kwong equation is given by: Attention: the Redlich Kwong equation was given incorrectly in the Quiz. The difference in the numerical results is however marginal (v b) instead of (v+b) in the nominator of the second term. Both versions will be accepted. RT a R 2Tc5 / 2 RT ; a = a' ; b = b' c v − b v (v + b) T pc pc a ' = 0.42748, b ' = 0.08664, R = 8.314 J/(mol K) p= For hydrogen gas the critical temperature Tc = 33.2 K and the critical pressure pc = 13.0 bar. i) Calculate pressure for T=300 K, and v=0.001 m3/mol. ii) calculate the compressibility factor Z for the conditions given in i) (4 P) a = a' p= R 2Tc5 / 2 RT = 0.14436; b = b ' c = 1.8396 ⋅ 10 −5 pc pc RT a = 2.53 ⋅ 10 6 Pa v − b v (v + b) T a ' = 0.42748, b ' = 0.08664, R = 8.314 J/(mol K) Q2: The enthalpy difference between two states in single phase regions is given by : 2 2 ⎛ ∂v ⎞ . Assuming the equation of state is known and cp(p0,T) is given, draw the most h2 − h1 = ∫ c p dT + ∫ ⎜ v − T dp ∂T p ⎟ ⎠ 1 1⎝ appropriate integration path between state 1 and state 2. Hint: divide path 1 ­2 into 3 portions (4 P). Q3: Using the equation for the enthalpy difference given in Q2, give simplified expressions for the enthalpy differences of the 3 portions of path 1 ­2 (see Q2) (3 P). Turn page! EML 5104 Classical Thermodynamics, Spring 2010 p0 ha − h1 = p1 ∫ ⎜v −T ⎝ ⎛ 1 ∂v ⎞ dp ∂T p ⎟ ⎠ hb − ha = ∫ c p dT T1 p2 T2 hb − h2 = p0 ∫ ⎜v −T ⎝ ⎛ 2 ∂v ⎞ dp ∂T p ⎟ ⎠ Q4: The Clapeyron equation is valid in (2 P): a) Phase change regions b) Single phase regions c) Supercritical regions d) All of the above Q5: Estimate the enthalpy of evaporation of water at 52.5 °C using the Clapeyron equation: dp dT and steam table data: Tsat (°C) 50 55 (4 P) hg − h f dp = dT sat T vg − v f psat (bar) 0.1235 0.1576 vg (m3/kg) 12.032 9.568 = sat hg − h f T vg − v f ( ) vf (m3/kg) 1.0121⋅10 ­3 1.0146⋅10 ­3 hg − h f ( ≈ T (v g ) Δp −v ) ΔT f = 325.65 ⋅ (12.032 + 9.568 ) − 1.0121 ⋅ 10 −3 + 1.0146 ⋅ 10 −3 sat ( ( )) 2⋅ 15760 − 12350 5 = 2.3984 ⋅ 10 6 J/(kg ⋅ K) Q6: Based on the Clapeyron equation given in Q5 derive the Clausius ­Clapeyron equation by assuming that vf is negligible and vg can be expressed using the ideal gas equation (3 P) dp dT = sat hg − h f T vg − v f ( ) vg ≈ RT / p vf ≈ 0 dp dT ≈ sat hg − h f RT 2 / p hg − h f ≈ RT 2 d ln p dT sat Points: _______/ 20 P Turn page! ...
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