Respuesta :
Answer:
The answers to the question are as follows
a. Find the throat Mach number Mt. Mt* = 1.536
b. Find the inlet Mach number Ma. Ma = 1.91
c. Find the ratio of exit stagnation pressure to inlet stagnation pressure pob/poa [tex]\frac{P_{ob} }{P_{oa} } =[/tex] 0.19515
Explanation:
To solve this we select the appropriate relations linking the known variables and the required variables
a. We look for the relation between the Mach number at the normal shock and the throat thus
The [tex]M^{*} _{t} = M_{t}\sqrt{\frac{k+1}{2+(k-1)M_{t} ^{2} } }[/tex]
Where
Mt* = mach number at the throat
k = specific heat ratio
Mn =mach number at location at the normal shock
Thus when Mn = 1.8, and k = 1.4 for air, Mt* = 1.536
b.
Here, we look for the relationship between the Mack number at the inlet and the throat given the ratio of their areas thus
[tex]\frac{A}{A^{*} } = \frac{1}{M_{a} } {((\frac{2}{k+1})(1+\frac{k-1}{2}(M_{a}) ^{2} ) ^{\frac{0.5k+1}{k-1} }[/tex]
Here the ratio A/A* = 2
Solving for Ma where k = 1.4 gives the Mack number at inlet as
Ma = 1.91
c) Here, we first look for the Mack number at exit, then we look for the ratio of the exit to inlet pressures using the appropriate relations as follows
Mack number at exit = [tex]M_{a2} = \sqrt{\frac{(k-1)M^{2} _{a}+2 }{2k(M_{a} ^{2} - k + 1} }[/tex]
Solving the above we get Where Ma = 1.8, and k 1.4
Ma2 = Mack number at exit = 0.6165
Thus stagnation pressure the ratio = [tex]\frac{P_{ob} }{P_{oa} } = \frac{1+YM^{2} _{inlet} }{1+M_{exit} ^{2} }(\frac{1+\frac{y-1}{2} M_{exit} ^{2} }{1+\frac{Y-1}{2} M^{2} _{inlet} } )^{\frac{Y}{Y-1} }[/tex]
Where Minlet = 1.91, Mexit = 0.6165 and Y, the specific heat of the gas, = 1.005 kJ/kg·K
From where the stagnation ratio can be found by plugging in the values of Minlet and Moutlet thus
[tex]\frac{P_{ob} }{P_{oa} } =[/tex] 0.19515
