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: McGraw-Hill Education
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Chapter 7, Problem 7.11P
Interpretation Introduction

Interpretation:

The temperature of steam at the nozzle entrance under the given conditions needs to be calculated

Concept Introduction:

  • The energy balance equation that relates the inlet and outlet enthalpies and velocities is given based on the first law of thermodynamics as:

  (H1+ u12 2)inlet(H2+ u22 2)outlet(or)ΔH + Δu22 = 0 ----(1)

  • For an adiabatic process, there is no loss or gain of heat i.e. q = 0
  • ΔS =qT ΔS=S2-S1 = 0-----(2)

  

Temperature at the nozzle inlet 182.5C0

Given Information:

Outlet Temperature of air T2= 15C0

Inlet velocity, u1 = negligible = 0 m/s

Outlet velocity, u2 = 580 m/s

Explanation:

Step 1:

Calculate the change in enthalpy

Based on equation (1) we have:

  ΔH =  u22u122 ΔH= (580)2(0)22 m2/s2 = 168200 m2/s2Since, 1 kJ/kg = 1000 m2/s2ΔH = 168.2kJ/kgMolar mass of air = 0.02897 kg/molThus, ΔH = 168.2kJ/kg×0.02897 kg/mol = -4.873 kJ/mol

Step 2:

Calculate the inlet temperature, T1

For an ideal gas the change in enthalpy is related to the change in temperature through the specific heat Cp;

  ΔH = Cp(T2-T1)where Cp = 7/2 R = 7/2×0.008314 kJ/mol-K = 0.0291 kJ/mol-kT1 = T2 - ΔHCp = (15 +273)K -(-4.873 kJ/mol)0.0291 kJ/mol.K = 455.46 KIn terms of degrees C we have:T1 = 455.46-273 = 182.5C0

Thus, the temperature at the nozzle inlet 182.5C0

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