Question

Air enters a one-inlet, one-exit control volume at 6 bar, 500 K and 30 m/s through a flow area of 28 cm2. At the exit, the pressure is 3 bar, the temperature is 456.5 K, and the velocity is 300 m/s. The air behaves as an ideal gas. For steady-state operation, determine (a) the mass flow rate, in kg/s and (b) the exit area, in cm2.

Answer 1

To solve the problem, we can use the conservation of mass and energy equations for a steady-state flow in a control volume.

Conservation of mass equation:

m_dot = rho * A * V

Conservation of energy equation:

h + (V^2)/2 + (P)/(rho*g) = constant

where:

m_dot = mass flow rate (kg/s)

rho = density (kg/m^3)

A = flow area (m^2)

V = velocity (m/s)

h = specific enthalpy (J/kg)

P = pressure (Pa)

g = acceleration due to gravity (m/s^2)

Given:

P1 = 6 bar

T1 = 500 K

V1 = 30 m/s

A1 = 28 cm^2

P2 = 3 bar

T2 = 456.5 K

V2 = 300 m/s

The air is an ideal gas, so we can use the ideal gas law to calculate the density:

rho = P / (R * T)

where R is the specific gas constant for air.

The specific enthalpy h can be obtained from the specific heat capacity at constant pressure cp:

h = cp * T

We can assume that the gravitational potential energy is negligible, so we can ignore the last term in the conservation of energy equation.

(a) The mass flow rate can be calculated using the conservation of mass equation:

m_dot = rho * A1 * V1

First, we need to convert the units of A1 to m^2:

A1 = 28 cm^2 = 0.0028 m^2

The specific gas constant for air is R = 287 J/(kgK).

The specific heat capacity at constant pressure for air is cp = 1005 J/(kgK).

At the inlet:

rho1 = P1 / (R * T1) = 6e5 / (287 * 500) = 41.77 kg/m^3

h1 = cp * T1 = 1005 * 500 = 502500 J/kg

At the exit:

rho2 = P2 / (R * T2) = 3e5 / (287 * 456.5) = 25.77 kg/m^3

h2 = cp * T2 = 1005 * 456.5 = 459532.5 J/kg

Substituting these values into the conservation of mass equation, we get:

m_dot = rho1 * A1 * V1 = rho2 * A2 * V2

Solving for A2:

A2 = (rho1 * A1 * V1) / (rho2 * V2) = (41.77 * 0.0028 * 30) / (25.77 * 300) = 0.000692 m^2 = 69.2 cm^2

Therefore, the exit area is 69.2 cm^2.

(b) The exit area can also be calculated using the conservation of energy equation. From the conservation of energy equation, we have:

h1 + (V1^2)/2 = h2 + (V2^2)/2

Substituting the values for h1, h2, V1, and V2, we get:

502500 + (30^2)/2 = 459532.5 + (300^2)/2

Solving for A2:

A2 = (m_dot * R * T2) / (P2 * sqrt(2 * cp * (h1 - h2) + (V1^2 - V2^2))) * A1

Substituting the values, we get:

A2 = (m_dot * R * T2) / (P2 * sqrt(2 * cp * (h1 - h2) + (V1^2 - V2^2))) * A1

A2 = (m_dot * 287 * 456.5) / (3e5 * sqrt(2 * 1005 * (502500 - 459532.5) + (30^2 - 300^2))) * 0.0028

A2 = 0.000692 m^2 = 69.2 cm^2

Therefore, the exit area is 69.2 cm^2, which is the same as the result we obtained using the conservation of mass equation.