In the absence of compressed liquid property tables it is common to approximate
ID: 1861305 • Letter: I
Question
In the absence of compressed liquid property tables it is common to approximate properties by using the saturated liquid values. The purpose of this exercise is to compare actual values to the approximate values. Compare the values of specific volume (v), internal energy (u), and enthalpy (h) for water at 10 MPa and 25C using the compressed liquid tables and the saturated liquid approximation. What are the percent errors associated with the approximation? What does the error and the percent error change to when the proper approximation for the enthalpy is used?Please show all work.
In the absence of compressed liquid property tables it is common to approximate properties by using the saturated liquid values. The purpose of this exercise is to compare actual values to the approximate values. Compare the values of specific volume (v), internal energy (u), and enthalpy (h) for water at 10 MPa and 25C using the compressed liquid tables and the saturated liquid approximation. What are the percent errors associated with the approximation? What does the error and the percent error change to when the proper approximation for the enthalpy is used?
Please show all work.
In the absence of compressed liquid property tables it is common to approximate properties by using the saturated liquid values. The purpose of this exercise is to compare actual values to the approximate values. Compare the values of specific volume (v), internal energy (u), and enthalpy (h) for water at 10 MPa and 25C using the compressed liquid tables and the saturated liquid approximation. What are the percent errors associated with the approximation? What does the error and the percent error change to when the proper approximation for the enthalpy is used?
Please show all work.
Explanation / Answer
From compressed liquid table, at 10 MPa, 25 deg C we get
v = 0.00099875 m^3/kg
u = 104.1075 kJ/kg
h = 114.0925 kJ/kg
from saturated liquid tables, at 25 deg C and quality = 0 (sat.liq.) we get
v = 0.0010029 m^3/kg
u = 104.88 kJ/kg
h = 104.89 kJ/kg
% error in v = (0.0010029 - 0.00099875)/0.00099875 *100 = 0.415 %
% error in u = (104.88 - 104.1075)/104.1075 *100 = 0.742 %
% error in h = (104.89 - 114.0925)/114.0925 *100 = -8.06 %
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