Air at 80 kPa and 127 °C enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the air stream is decreased from 250 m/s to 40 m/s as it passes through the diffuser. Determine

(a) the exit temperature of the air, and
(b) the inlet and exit areas of the diffuser.

Respuesta :

Answer:

a) The exit temperature is 430 K

b) The inlet and exit areas are 0.0096 m² and 0.051 m²

Explanation:

a) Given:

T₁ = 127°C = 400 K

At 400 K, h₁ = 400.98 kJ/kg (ideal gas properties table)

The energy equation is:

[tex]q-w=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2} }{2} +delta-p[/tex]

For a diffuser, w = Δp = 0

The diffuser is adiabatic, q = 0

Replacing:

[tex]0-0=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2} }{2} +0[/tex]

Where

V₁ = 250 m/s

V₂ = 40 m/s

Replacing:

[tex]0-0=h_{2} -400x10^{3} +\frac{40^{2}-250^{2} }{2} +0\\h_{2} =431430 J/kg=431.43kJ/kg[/tex]

Using tables, at 431.43 kJ/kg the temperature is 430 K

b) The inlet area is:

[tex]m=\frac{P_{1} }{RT_{1} } A_{1} V_{1} \\\frac{6000}{3600} =\frac{80}{0.287*400} A_{1} *250\\A_{1} =0.0096m^{2}[/tex]

The exit area is:

[tex]m=\frac{P_{2} }{RT_{2} } A_{2} V_{2} \\\frac{6000}{3600} =\frac{100}{0.287*430} A_{2} *40\\A_{2} =0.051m^{2}[/tex]

A) The exit temperature of air is; T₂ = 430 K

B) The inlet and exit areas of the diffuser are;

A₁  = 0.0096 m²

A₂ = 0.051 m²

A) We are given;

T₁ = 127°C = 400 K

V₁ = 250 m/s

V₂ = 40 m/s

From table A-17 attached which shows ideal gas properties of air, At a temperature of 400 K, h₁ = 400.98 kJ/kg = 400980 J/kg

Now, formula for the energy equation is;

q - w = h₂ - h₁ + ¹/₂(V₂² - V₁²) + ΔP

We are told that the diffuser is adiabatic and as such;

q = 0 and w = ΔP = 0

Thus;

0 - 0 = h₂ - 400.98 * 10 + ¹/₂(40² - 250²) + 0

0 = h₂ - 400980 - 30450

h₂ = 431430 J/kg = 431.43 KJ/kg

From the table attached, the temperature that corresponds to h₂ = 431.43 KJ/kg is T₂ = 430 K

B) To get the inlet area A₁, we will use the formula;

m = (P₁/RT₁) × A₁V₁

where;

m = 6000 kg/h = 1.67 Kg/s

P₁ = 80 kPa

R is a constant = 0.287 kJ/kg.k

T₁ = 400 K

V₁ = 250 m/s

Thus;

1.67 = (80/(0.287 × 400)) × (A₁ × 250)

Solving this gives;

A₁  = 0.0096 m²

For exit area;

m = (P₂/RT₂) × A₂V₂

where;

m = 6000 kg/h = 1.67 Kg/s

P₂= 100 kPa

R is a constant = 0.287 kJ/kg.k

T₂ = 430 K

V₂ = 40 m/s

1.67 = (100/(0.287 × 430)) × (A₂ × 400)

Solving this gives;

A₂ = 0.051 m²

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