Respuesta :
Answer:
The answer is "[tex]3.74 \ cm\ \ and \ \ 0.186 \frac{kg}{s}[/tex]"
Explanation:
Given data:
Initial temperature of tank [tex]T_1 = 300^{\circ}\ C= 573 K[/tex]
Initial pressure of tank [tex]P_1= 400 \ kPa[/tex]
Diameter of throat [tex]d* = 2 \ cm[/tex]
Mach number at exit [tex]M = 2.8[/tex]
In point a:
calculating the throat area:
[tex]A*=\frac{\pi}{4} \times d^2[/tex]
[tex]=\frac{\pi}{4} \times 2^2\\\\=\frac{\pi}{4} \times 4\\\\=3.14 \ cm^2[/tex]
Since, the Mach number at throat is approximately half the Mach number at exit.
Calculate the Mach number at throat.
[tex]M*=\frac{M}{2}\\\\=\frac{2.8}{2}\\\\=1.4[/tex]
Calculate the exit area using isentropic flow equation.
[tex]\frac{A}{A*}= (\frac{\gamma -1}{2})^{\frac{\gamma +1}{2(\gamma -1)}} (\frac{1+\frac{\gamma -1}{2} M*^2}{M*})^{\frac{\gamma +1}{2(\gamma -1)}}[/tex]
Here: [tex]\gamma[/tex] is the specific heat ratio. Substitute the values in above equation.
[tex]\frac{A}{3.14}= (\frac{1.4-1}{2})^{-\frac{1.4+1}{2(1.4 -1)}} (\frac{1+\frac{1.4-1}{2} (1.4)^2}{1.4})^{\frac{1.4+1}{2(1.4-1)}} \\\\A=\frac{\pi}{4}d^2 \\\\10.99=\frac{\pi}{4}d^2 \\\\d = 3.74 \ cm[/tex]
exit diameter is 3.74 cm
In point b:
Calculate the temperature at throat.
[tex]\frac{T*}{T}=(1+\frac{\Gamma-1}{2} M*^2)^{-1}\\\\\frac{T*}{573}=(1+\frac{1.4-1}{2} (1.4)^2)^{-1}\\\\T*=411.41 \ K[/tex]
Calculate the velocity at exit.
[tex]V*=M*\sqrt{ \gamma R T*}[/tex]
Here: R is the gas constant.
[tex]V*=1.4 \times \sqrt{1.4 \times 287 \times 411.41}\\\\=569.21 \ \frac{m}{s}[/tex]
Calculate the density of air at inlet
[tex]\rho_1 =\frac{P_1}{RT_1}\\\\=\frac{400}{ 0.287 \times 573}\\\\=2.43\ \frac{kg}{m^3}[/tex]
Calculate the density of air at throat using isentropic flow equation.
[tex]\frac{\rho}{\rho_1}=(1+\frac{\Gamma -1}{2} M*^2)^{-\frac{1}{\Gamma -1}} \\\\\frac{\rho *}{2.43}=(1+\frac{1.4-1}{2} (1.4)*^2)^{-\frac{1}{1.4-1}} \\\\\rho*= 1.045 \ \frac{kg}{m^3}[/tex]
Calculate the mass flow rate.
[tex]m= \rho* \times A* \times V*\\\\= 1.045 \times 3.14 times 10^{-4} \times 569.21\\\\= 0.186 \frac{kg}{s}[/tex]
