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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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This note was uploaded on 09/09/2010 for the course EML 5104 taught by Professor Staff during the Spring '08 term at University of Florida.
 Spring '08
 Staff

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