a) 1 kg of steam contained in a horizontal frictionless piston and cylinder is h
ID: 786148 • Letter: A
Question
a) 1 kg of steam contained in a horizontal frictionless piston and cylinder is heated at a constant pressure of 1.013 bar from 125 C to such a temperature that the volume doubles. Calculate 1) the amount of heat that must be added to accomplish the change, 2) the final temperature of the steam 3) the work the steam does against its surroundings 4) the internal energy change and 5) the enthalpy change of the steam for this process>
b) Repeat part a) if the heating occurs at a constant volume to a pressure that is twice the initial pressure.>
c) Repeat the calculation of part a) assuming an ideal gas with Cp = 34.4 J/mol- K>
d) Repeat part b) assuming an ideal gas as in part c).>
Explanation / Answer
This problem is an application of the 1st Law to an isobaric expansion (constant P) using the steam tables for the physical properties (specific V, specific U, and specific H).
Given any 2 intensive properties, you can look up all the others in the steam tables. Begin by looking up properties of steam at 101.3 kPa and 125 degC. I found:
P 101300 Pascals
V 1.79334116 m^3/kg
T 125 deg C
U 2544.334961 kJ/kg
H 2726.000488 kJ/kg
S 7.484985352 kJ/kg-K
STATE Superheated vapor
Calculate the initial volume: V1 = (1.7933 m3/kg) * 1 kg = 1.7933 m3
Then the final volume is: V2 = 2 * V1 = 3.5867 m3
And the final specific volume is: V2hat = 3.5867 m3 / 1 kg = 3.5867 m3/kg.
Now, we know both the specific volume and the pressure at the final state (because the process is isobaric). We can use these two intensive properties to lookup any othee intensive properties in the steam tables. Interpolation may be required. OR you can use the NIST Webbook to avoid interpolation. Here is the web address:
http://webbook.nist.gov/chemistry/fluid/
I got the following properties for the final state:
P 101300 Pascals
V 3.58668232 m^3/kg
T 514.6740723 deg C
U 3155.648438 kJ/kg
H 3518.979492 kJ/kg
S 8.86747551 kJ/kg-K
STATE Superheated vapor
Now, we can apply the 1st Law to this process. If we assume changes in kinetic and potential energies are negligible, then the 1st Law is: Q = deltaH = m*deltaHhat = 1 kg * (3518.98 kJ/kg - 2726.00 kJ/kg ) = 792.98 kJ.
The 1st Law can also be expressed as: Q - Wtot = deltaU, where Wtot is the total amount of work that crosses the system boundary. This is what the problem wants you to determine. Solving for Wtot yields: Wtot = Q - deltaU.
Use the data from the steam tables to calculate deltaU = m * deltaUhat = 1 kg * (3155.65 kJ/kg - 2544.334961 kJ/kg ) = 611.31 kJ.
Finally, calculate Wtot = 792.98 kJ - 611.31 kJ = 181.67 kJ.
Fine !
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