1.4 Units

The Importance of Unit Analysis

Unit analysis is perhaps one of the most important tasks in the solution of an engineering problem. Unit analysis is very easy to do, but it is far too often ignored. Significant errors can occur if units are incorrect. The chances of solving a system of equations where the units are incorrect is virtually nil. The loss of the Mars Climate Orbiter in 1999 is a classic example of unit analysis gone terribly wrong. The cost of this error was $125 million.

Example: Using CoolProp to Compute Fluid Properties

CoolProp is explained in Section 1.2 of the text. As shown in 1.2 Software on this website, the spreadsheet cell call using CoolProp is,

=PropsSI(code, input code 1, property1, input code 2, property2, fluid)

When using CoolProp, the input properties are in SI units as shown in 1.2 Software. Suppose you are working in the IP unit system and you wish to compute the enthalpy of Refrigerant-22 at -10°F and a pressure of 30 psia. Furthermore, you would like to have the enthalpy expressed in Btu/lbm. Therefore, to use CoolProp, the input pressure and temperature need to be converted to the SI system and the calculated enthalpy must be converted to the IP system. Units cannot be specified in a spreadsheet cell. Therefore, some preliminary "pencil and paper" work needs to be done to determine the proper conversions.

The temperature input to CoolProp must be in kelvin degrees (K) and the pressure must be in pascals (Pa). From Appendix A,

T[K] = (T[°F] + 459.67)/1.8

P[Pa] = P[psia]*[(1 kPa)/(0.145038 psia)]*[(1000 Pa)/(1 kPa)]

Using Appendix A again, the conversion of the enthalpy from J/kg to Btu/lbm is,

h{Btu/lbm] = h[J/kg]*[(0.429923 Btu/lbm)/(1 kJ/kg)]*[(1 kJ)/(1000 J)]

Figure 1.4.1 shows the spreadsheet result. Notice how the property call is written.