The correct temperature is 125°C
Answer:
A) M_a1 = 0.5
B) T2 = 232.17 K
C) V2 = 611 m/s
D) m' = 187 kg/s
Explanation:
We are given;
Pressure; P1 = 250 kPa
Temperature; T1 = 125°C = 398 K
Speed; v1 = 200 m/s
Area; A2 = 0.25 m²
M_a2 = 2
A) Formula for M_a1 is given by;
M_a1 = v/a1
Where;
v is speed
a1 = √kRT
k is specific heat capacity ratio of air = 1.4
R is a gas constant with a value of R = 287 J/kg·K
T is temperature
Thus;
M_a1 = 200/√(1.4 × 287 × 398)
M_a1 = 200/399.895
M_a1 = 0.5
B) To find T2, let's first find the Stagnation pressure T0
Thus;
T0/T1 = 1 + ((k - 1)/2) × (M_a1)²
T0 = T1(1 + ((k - 1)/2) × (M_a1)²)
T0 = 398(1 + ((1.4 - 1)/2) × (0.5)²)
T0 = 398(1 + (0.2 × 0.5²))
T0 = 398 × 1.05
T0 = 417.9 K
Now,similarly;
T0/T2 = 1 + ((k - 1)/2) × (M_a2)²
T2 = T0/[(1 + ((k - 1)/2) × (M_a2)²)]
T2 = 417.9/(1 + (0.2 × 2²))
T2 = 417.9/1.8
T2 = 232.17 K
C) V2 is gotten from the formula;
T0 = T2 + (V2)²/(2C_p)
Cp of air = 1005 J/Kg.K
Thus;
V2 = √(2C_p)[T0 - T2]
V2 = √((2 × 1005) × (417.9 - 232.17))
V2 =√373317.3
V2 = 611 m/s
D) mass flow is given by the formula;
m' = ρA2•V2
Where;
ρ is Density of air with an average value of 1.225 kg/m³
m' = 1.225 × 0.25 × 611
m' = 187 kg/s
