Respuesta :
Answer:
The answers to the question are as follows
(1) velocity = 164.108 m/s
(2) Mach number, Ma = 3.05696
(3) temperature = 7172.51 K
(4) Area per unit mass flow = 0.1254 m²
Explanation:
(a) To solve the question, we list out the variables as follows
P₀ = 4 MPa,
T₀ = 2500 K
P₂ = 0.1 MPa
γ = 1.4
Molecular weight = 30
R = 0.287 kJ/kg
[tex]c_{p} = \frac{\gamma R}{\gamma -1}[/tex] = 1.005 kJ/kg
Also [tex]\frac{P_0}{P} =(1+\frac{(\gamma -1)C^{2.} }{2\gamma RT} )^{\frac{\gamma}{\gamma-1} }[/tex]
or 40 = [tex](1+{\frac{0.4C^{2} }{5763.821} )^{3.5 }[/tex] or c = 164.108 m/s
(b) The mach number can be calculated from
[tex]\frac{P_0}{P} = (1+\frac{(\gamma-1)(Ma)^{2}}{2} )^{\frac{\gamma}{\gamma-1} } }[/tex] That is[tex]40 = (1+\frac{0.4(Ma)^2}{2} )^{3.5}[/tex] or
Ma = 3.05696
(c) Since the process is isentropic, we have
[tex]\frac{P_0}{P} =(\frac{T_0}{T} )^{\frac{\gamma}{\gamma-1} }[/tex]
or T = 2500×
= 7172.51 K
(d) From A = mv₂/C₂
The area for unit mass flow rate =
v = RT/P = 0.287×7172.5×1000/100000 = 20.59 m³/kg
(Note: The 1000 product equation above is because the units were changed from kJ/kg to J/kg and so also for the 100000 in the denominator, that represents the pressure unit change from MPa to Pa)
and A = mv/C which for unit mass flow rate =
v/C Therefore A = 1kg/s ×20.59 m³/kg /164.108 m/s = 0.1254 m²
