CHE222_solutions5

# CHE222_solutions5 - ChE 222 Problem Set#5 Due February...

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ChE 222: Problem Set #5 Due: February 29,2008 (8:OO am) ~7) 1 .) Koretsky, Problem 2.34 [ l~) 2.) Koretsky, Problem 2.37 C 1 a) 3.) Koretsky, Problem 2.49 ( (7) 4.) Koretsky, Problem 2.5 1

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2.34 First, a sketch of the process is useful: 30 bar 20 bar %. C, 100 Oc 150 To find the heat in we will apply the 1st law. Assuming steady state, the open system energy balance with one stream in and one stream out can be written: which upon rearranging is: Thus this problem reduces to finding the change in the thermodynamic property, enthalpy from the inlet to the outlet. We know 2 intensive properties at both the inlet and outlet so the values for the other properties (like enthalpy!) are already constrained. From Table A.2.1, we have an expression for the ideal gas heat capacity: with Tin (K). Since this expression is limited to ideal gases any change in temperature must be under ideal conditions. From the definition of heat capacity: By integrating and substituting the temperatures, we obtain:
2.37 First start with the energy balance around the nozzle. Assume that heat transfer and potential

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## This homework help was uploaded on 04/18/2008 for the course CHE 222 taught by Professor Benjamin during the Spring '08 term at SDSMT.

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CHE222_solutions5 - ChE 222 Problem Set#5 Due February...

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