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[3] Redo Example 6.9 (textbook pages 245-246) part (a) only except with t₁-200 °C. Determine down- stream state and ΔΗ.

[3] Redo Example 6.9 (textbook pages 245-246) part (a) only except with t₁-200 °C. Determine down- stream state and ΔΗ.

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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value from steam table for T1 = 200 C

H = 2826.8

S = 6.6922

[3] Redo Example 6.9 (textbook pages 245-246) part (a) only except with t₁=200 °C. Determine down- stream state and ΔΗ.
Transcribed Image Text:[3] Redo Example 6.9 (textbook pages 245-246) part (a) only except with t₁=200 °C. Determine down- stream state and ΔΗ.
Example 6.9 Superheated steam originally at P₁ and T₁ expands through a nozzle to an exhaust pres- sure P₂. Assuming the process is reversible and adiabatic, determine the downstream state of the steam and AH for P₁ = 1000 kPa, t₁ 250°C, and P₂ = 200 kPa. = Solution 6.9 Because the process is both reversible and adiabatic, there is no change in the entropy of the steam. For the initial temperature of 250°C at 1000 kPa, no entries appear in the tables for superheated steam. Interpolation between values for 240°C and 260°C yields, at 1000 kPa, H₁ = 2942.9 kJ-kg-1 11- For the final state at 200 kPa, Numerically, S₁ = 6.9252 kJ.kg-¹.K-1 Because the entropy of saturated vapor at 200 Pa is greater than S2, the final state must lie in the two-phase liquid/vapor region. Thus t2 is the saturation tempera- ture at 200 kPa, given in the superheat tables as t₂ = 120.23°C. Equation (6.96a) t2 applied to the entropy becomes: S₂ = (1-x₂)S₂ + x₂ S ₂ S₂ S₁ = 6.9252 kJ.kg-¹.K-1 = 6.9252 = 1.5301(1-x2)+7.1268x2 where the values 1.5301 and 7.1268 are entropies of saturated liquid and saturated vapor at 200 kPa. Solving, Finally, x₂ = 0.9640 The mixture is 96.40% vapor and 3.60% liquid. Its enthalpy is obtained by further application of Eq. (6.96a): H₂ = (0.0360)(504.7) + (0.9640)(2706.3) = 2627.0 kJ·kg−¹ AH = H₂ - H₁ = 2627.0 2942.9 = -315.9 kJ.kg-1 For a nozzle, under the stated assumptions the steady-flow energy balance, Eq. (2.31), becomes 1 AH+=Au² = 0 24u² Thus the decrease in enthalpy is exactly compensated by an increase in kinetic energy of the fluid. In other words, the velocity of a fluid increases as it flows through a nozzle, which is its usual purpose. Nozzles are treated further in Sec. 7.1.
Transcribed Image Text:Example 6.9 Superheated steam originally at P₁ and T₁ expands through a nozzle to an exhaust pres- sure P₂. Assuming the process is reversible and adiabatic, determine the downstream state of the steam and AH for P₁ = 1000 kPa, t₁ 250°C, and P₂ = 200 kPa. = Solution 6.9 Because the process is both reversible and adiabatic, there is no change in the entropy of the steam. For the initial temperature of 250°C at 1000 kPa, no entries appear in the tables for superheated steam. Interpolation between values for 240°C and 260°C yields, at 1000 kPa, H₁ = 2942.9 kJ-kg-1 11- For the final state at 200 kPa, Numerically, S₁ = 6.9252 kJ.kg-¹.K-1 Because the entropy of saturated vapor at 200 Pa is greater than S2, the final state must lie in the two-phase liquid/vapor region. Thus t2 is the saturation tempera- ture at 200 kPa, given in the superheat tables as t₂ = 120.23°C. Equation (6.96a) t2 applied to the entropy becomes: S₂ = (1-x₂)S₂ + x₂ S ₂ S₂ S₁ = 6.9252 kJ.kg-¹.K-1 = 6.9252 = 1.5301(1-x2)+7.1268x2 where the values 1.5301 and 7.1268 are entropies of saturated liquid and saturated vapor at 200 kPa. Solving, Finally, x₂ = 0.9640 The mixture is 96.40% vapor and 3.60% liquid. Its enthalpy is obtained by further application of Eq. (6.96a): H₂ = (0.0360)(504.7) + (0.9640)(2706.3) = 2627.0 kJ·kg−¹ AH = H₂ - H₁ = 2627.0 2942.9 = -315.9 kJ.kg-1 For a nozzle, under the stated assumptions the steady-flow energy balance, Eq. (2.31), becomes 1 AH+=Au² = 0 24u² Thus the decrease in enthalpy is exactly compensated by an increase in kinetic energy of the fluid. In other words, the velocity of a fluid increases as it flows through a nozzle, which is its usual purpose. Nozzles are treated further in Sec. 7.1.
Expert Solution
Step 1

given = P1 = 1000 KPa T1 = 200^c and P2 = 200 KPa

since the process is both reversible and adiabatic it means the change in entropy is zero

hence S1 = S2

at 1000 KPa and 200^c

given = 

S1 = 6.6922 Kg/KJ

H1 = 2826.8 KJ/Kg*K

At P2= 200kPa,

S2=S1= 6.6922 kJ/Kg.K

hf = Specific enthalpy of saturated liquid at 200 KPa pressure = 504.7 KJ/Kg

hg = Specific enthalpy of saturated vapor at 200 KPa pressure = 2706.3 KJ/Kg

sf = Specific entropy of saturated liquid at 200 KPa MPa pressure = 1.5300 KJ/KgK

sg = Specific entropy of saturated vapor at 200 KPa pressure = 7.1268 KJ/Kg

since the entropy of saturated vapor at 200 KPa is greater than S2 hence the final state must lie in two phase liquid/vapor region

 

 

 

 

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