Compressible Flow Example Problems

==Example 5==
:Air is contained in a piston-cylinder arrangement (see sketch). Heat is added very slowly, and the piston is allowed to move out for each increment of heat added. The piston is essentially frictionless.
:P_{initial} = 1atm
:T_{initial} = 294.4K
#Is this an isentropic process since the heat transfer is reversible and the piston (work taken out) motion is frictionless?
#Can all of the heat put into the system be transformed into work?
#Does this violate the second law?
===Solution===
No, heat was transfered; therefore, it is not an isentropic process. Isentropic requires adiabatic and reversible process.
Yes, if the process keeps the temperature of the air constant, then the first law says that:\delta Q=\Delta E+\delta WIf the air is ideal E = E(T), and if T is kept constant by the allowed motion, then:\Delta E = 0\delta Q = \delta WSo, a reversible process has converted 100% of the heat into work.
No, the second law only refers to heat engines that operate in a cycle. (ie the process would include the piston being returned to its initial position, and to accomplish this, heat would have to be removed from the system.
==Example 6==
:For the closed system of Example 5, a process is carried out whereby 100 J of heat is put into the system. The energy of the system increases by 50 J.
#How much work was done?
#How much did the system temperature change, if the initial mass of air was 1 kg, and what is the temperature at the second state?
===Solution===
 \Delta E = mc_v\Delta T = mc_v(T_2-T_1) \Delta T = \frac{\Delta E}{mc_v} = \frac{\Delta E}{m\left(\frac{c_p}{\gamma}\right)}\Delta T = \frac{50 J}{(1kg)\left(\frac{1000\frac{J}{kg-K}}{1.4}\right)}=Solve hereT_2 = 294.4K + \Delta T
==Example 7==
:A stream of hot air expands isentropically through a turbine. THe inlet pressure and temperature are 500 psia and 1000F. The air leaves the turbine with a velocity of 100m/s at atmospheric pressure.
#What is the temperature of the air leaving the turbine?
#How much work was done by the turbine per unit mass flow?
===Assuming===
\gamma = 1.4
c_p = 0.24 \frac{Btu}{lbm-^{\circ}R}
===Solution===
==Example 8==
:Air at 1 atm and 70F is raised in pressure and temperature to 1000 psia and 500F.
#Determine the entropy change in the unit of mass of the air for the process.
===Assuming===
\gamma = 1.4
c_p = 0.24 \frac{Btu}{lbm-^{\circ}R}
W_m = 28.97 \frac{lbm}{lbmol}
===Solution===
s_2-s_1=c_p\ln\left(\frac{T_2}{T_1}\right)-R\ln\left(\frac{P_2}{P_1}\right)R = \frac{R_u}{W_m} = \frac{1545.4}{(28.97\frac{lbm}{lbmol})(778\frac{ft-lbm}{Btu})}=0.0686\frac{Btu}{lbm-^{\circ}R}s_2-s_1=0.24\frac{Btu}{lbm-^{\circ}R}\ln\left(\frac{960}{530}\right)-\left(0.0686\frac{Btu}{lbm-^{\circ}R}\right)\ln\left(\frac{1000psia}{14.696psia}\right)=-0.1468\frac{Btu}{lbm-^{\circ}R}
==Example 9==
:If the air in Example 8 had its pressure and temperature changed such that s2 = s1, could we say that the process was isentropic?
===Solution===
:No, not unless we knew that the process was both adiabatic and reversible. The fact that the entropy change was zero does not by itself tell us that the process was isentropic.
==Example 10==
:A rocket motor has a combustion process that produces a gas with the properties:
*\gamma = 1.2
*W_m = 20 \frac{lbm}{lbmol}
:The chamber stagnation pressure and temperature are:
*P_tc = 1000 psia
*T_tc = 5000^{\circ}F
*M_e = 10
#Determine the pressure, temperature, density, and velocity of the gas flow in the nozzle throat.
#Determine the pressure, temperature, density, and velocity of the gas flow in the nozzle exit.
===Solution===
P^* = P_{tc}(1+\frac{\gamma -1}{2}M^{*2})^{-\frac{\gamma}{\gamma - 1}} = 1000psia\left(1+\frac{1.2 -1}{2}(1)^2\right)^{-\frac{1.2}{1.2 - 1}}P^* = 564.47psiaT^* = T_{tc}(1+\frac{\gamma -1}{2}M^{*2})^{-1} = 5459.67^{\circ}R\left(1+\frac{1.2 -1}{2}(1)^2\right)^{-1}T^* = 4963.6^{\circ}R\rho ^* = \frac{P^*}{RT^*} = \frac{(564.47psia)\left(144\frac{psf}{psia}\right)}{\left(\frac{1545.4\frac{ft-lbf}{lbm-^{\circ}R}}{20\frac{lbm}{lbmol}}\right)(4963.6^{\circ}R)}=0.2119lbm/ft^3V^* = M^*a^* = M^*\sqrt{\gamma RT^*} = (1)\sqrt{(1.2)\left(2486.08\frac{ft^2}{s^2-^{\circ}R}\right)(4963.6^{\circ}R)} = 3848.1\frac{ft}{s}
Repeat calculations for nozzle exit.
==Example 11==
What is the energy, in Joules, of a 1 kg mass traveling at a velocity of 12169 ft/s? What is the energy in Btu?
===Solution===
:KE = \frac{1}{2}mV^2
:V = \frac{12169\frac{ft}{s}}{3.2808 \frac{ft}{m}}=3709.16 \frac{m}{s}
:KE = \frac{1}{2}(1kg)\left(3709.16 \frac{m}{s}\right)^2