HP 50g Graphing manual Ideal Gas State Change 5, Isothermal Expansion 5, Polytropic Processes 5

Example:

Given: T=16.85_°C, P=1_atm, V=25_1, MW=36_g/gmol.

Solution: n=1.0506_gmol, m=3.7820E2_kg.

Ideal Gas State Change (5, 2)

Equation:

Pf ⋅ Vf Pi ⋅ Vi

---------------- = ---------------

Tf Ti

Example:

Given: Pi=1.5_kPa, Pf=1.5kPa, Vi=2_l, Ti=100_°C, Tf=373.15_K.

Solution: Vf=2_1.

Isothermal Expansion (5, 3)

These equations apply to an ideal gas.

Equations:

 Vf 	m = n ⋅ MW
W = n ⋅ R ⋅ T ⋅ LN ------ 
Vi

Example:

Given: Vi=2_l, Vf=125_l, T=300_°C, n=0.25_gmol, MW=64_g/gmol.

Solution: W=4926.4942_J, M=0.016_kg.

Polytropic Processes (5, 4)

These equations describe a reversible pressurevolu me change of an ideal gas such that P ∗Vn is constant. Special cases include isothermal processes (n=1), isentropic processes (n=k, the specific heat ratio), and constantpressure processes (n=0).

Equations:

					n – 1
		 Vf  –n			-----------
Pf	=		Tf	=	 Pf  n
-----		 ------ 	-----		 ----- 
Pi		Vi	Ti		Pi

Example:

Given: Pi=15_psi, Pf=35_psi, Vi=1_ft^3, Vf=0.50_ft^3, Ti=75_°F.

Solution: n=1.2224, Tf=164.1117_°F.

Isentropic Flow (5, 5)

The calculation differs at velocities below and above Mach 1. The Mach number is based on the speed of sound in the compressible fluid.

526 Equation Reference