Modern Compressible Flow Solutions Chapter 1

The following are solutions to the problems found in Chapter 1 of John D. Anderson's 2004 book Modern Compressible Flow.
==Problem 1==
At the nose of a missile in flight, the pressure and temperature are 5.6 atm and 850 °R, respectively. Calculate the density and specific volume.
===Given===
*P = 5.6 atm
*T = 850 °R
===Assuming===
*Perfect gas
*air
*P & T are statics, not totals
*Rair = 1716 (ft-lbf)/(sl-°R)
===Find===
#ρ
#v
===Solution===
\rho={\frac{P}{RT}}={\frac{(5.6 atm)(2116 {\frac{psf}{atm}})}{(1716{\frac{ft-lbf}{sl-^{\circ}R}})(850 ^{\circ}R)}}
v=\frac{1}{\rho}
==Problem 2==
In the reservoir of a supersonic wind tunnel, the pressure and temperature of air are 10 atm and 320 K, respectively. Calculate the density, the number density, and the mole-mass ratio.
===Given===
*P = 10 atm
*T = 320 K
===Assuming===
*Perfect gas (Z ≈ 0.998 ≈ 1.0)
*Rair = 287 J/(kg-K)
*k = \frac{R_u}{N_A}=1.38*10^{-23}\frac{J}{K}
===Find===
#ρ
#n
#η
===Solution===
p=\rho RT\rho =\frac{p}{RT}=\frac{\left(10atm\right)\left(101325\frac{Pa}{atm}\right)}
{\left(287\frac{J}{kg-K}\right)\left(320K\right)}
p=nkTn=\frac{p}{kT}=\frac{\left(10atm\right)\left(101325\frac{Pa}{atm}\right)}
{\left(1.38*10^{-23}\frac{J}{K}\right)\left(320K\right)}
p=\rho\eta R_uT\eta=\frac{p}{\rho R_uT}=\frac{\left(10atm\right)\left(101325\frac{Pa}{atm}\right)}{\left(11.03\frac{kg}{m^3}\right)\left(8314\frac{J}{kgmol-K}\right)\left(320K\right)}
==Problem 3==
For a calorically perfect gas, derive the relation cp - cv = R. Repeat the derivation for a thermally perfect gas.
===Derivation===
==Problem 4==
The pressure and temperature ratios across a given portion of a shock wave in air are p2/p1 = 4.5 and T2/T1 = 1.687, where 1 and 2 denote conditions ahead of and behind the shock wave, respectively. Calculate the change in entropy in units of (ft-lb)/(slug-°R) and J/(kg-K)
===Given===
*\frac{T_2}{T_1}=1.687
*\frac{P_2}{P_1}=4.5
===Assuming===
*Calorically perfect gas
*cp = 1006.1 J/(kg-K) = 6015 (ft-lbf)/(slug-°R)
*R = 287.1 J/(kg-K) = 1716.49 (ft-lbf)/(slug-°R)
===Find===
#ds in (ft-lb)/(slug-°R)
#ds in J/(kg-K)
===Solution===
:ds=c_p\ln\frac{T_2}{T_1}-R\ln\frac{P_2}{P_1}
ds = \left(6015\frac{ft-lbf}{slug-^{\circ}R}\right)\ln(1.687)-\left(1716.49\frac{ft-lbf}{slug-^{\circ}R}\right)\ln(4.5)
ds = \left(1006.1\frac{J}{kg-K}\right)\ln(1.687)-\left(287.1\frac{J}{kg-K}\right)\ln(4.5)
==Problem 5==
Assume that the flow of air through a given duct is isentropic. At one point in the duct, the pressure and temperature are p1 = 1800 lb/ft2 and T1 = 500°R, respectively. At a second point, the temperature is 400°R. Calculate the pressure and density at this second point.
===Given===
*p1 = 1800 lbf/ft2
*T1 = 500 °R
*T2 = 400 °R
===Assuming===
*Air
*Isentropic process from 1 to 2
*Calorically perfect gas
*γ = 1.4
*R = 1716 (ft-lbf)/(slug-°R)
===Find===
#p2
#ρ2
===Solution===
p_2=p_1\left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma-1}}=1800\frac{lbf}{ft^2}\left(\frac{400^{\circ}R}{500^{\circ}R}\right)^{\frac{1.4}{1.4-1}}
\rho_2=\rho_1\left(\frac{T_2}{T_1}\right)^{\frac{1}{\gamma-1}}=\frac{p_1}{RT_1}\left(\frac{T_2}{T_1}\right)^{\frac{1}{\gamma-1}}=\frac{1800\frac{lbf}{ft^2}}{\left(1716\frac{ft-lbf}{slug-^{\circ}R}\right)\left(500^{\circ}R\right)}\left(\frac{400^{\circ}R}{500^{\circ}R}\right)^{\frac{1}{1.4-1}}
==Problem 6==
Consider a room that is 20 ft long, 15 ft wide, and 8 ft high. For standard sea level conditions, calculate the mass of air in the room in slugs. Calculate the weight in pounds.
===Given===
*V = l*w*h = 2400 ft3
*p = 14.696 lbf/in2 = 2116.2 lbf/ft2
*T = 518.67 °R
===Assuming===
*Air
*R = 1716 (ft-lbf)/(slug-°R)
===Find===
#m
#W
===Solution===
pV=mRTm=\frac{pV}{RT}=\frac{\left(2116.2\frac{lbf}{ft^2}\right)\left(2400ft^3\right)}{\left(1716\frac{ft-lbf}{slug-^{\circ}R}\right)\left(518.67^{\circ}R\right)}
W=mg=(5.706slugs)\left(32.174\frac{ft}{s^2}\right)
==Problem 7==
In the infinitesimal neighborhood surrounding a point in a inviscid flow, the small change in pressure, dp, that corresponds to a small change in velocity, dV, is given by the differential equation dp = -\rhoVdV.
Using this relation, derive a differential relation for the fractional change in density, d\rho/\rho, as a function of the fractional change in velocity, dV/V, with the compressibility \tau as a coefficient.
The velocity at a point in an isentropic flow of air is 10 m/s (a low speed flow), and the density and pressure are 1.23 kg/m3 and 1.01 x 105 N/m2, respectively. The fractional change in velocity at the point is 0.01. Calculate the fractional change in density.
Repeat part 2, except for a local velocity at the point of 1000 m/s (a high speed flow). Compare this result with that from part 2, and comment on the differences.
==References==