Question

Air fl ows isentropically through a duct. At section 1, the pressure and temperature are 250 kPa and 1258C, and the velocity is 200 m/s. At section 2, the area is 0.25 m2 and the Mach number is 2.0. Determine (a) Ma1; (b) T2; (c) V2; and (d ) the mass fl ow.

Answer 1

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

Step-by-step 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