Q1 - 1D isentropic flow

Question 1

List the assumptions for the approximation of a real flow in a duct (pipe, nozzle) by a one-
dimensional isentropic flow.

Answer 1

1.Velocity, pressure and other variables are constant in a cross section.
2. Velocity is directed along the axis.
3. The flow is steady.
4. Neglect heat exchange between different parts of gas and between gas and the walls.
5.Neglect friction between the gas and the walls.
6. Disregard gravity force and other body forces.
7. Neglect the network.

Question 2

Derive Mass and energy conservation relation for 1D isentropic flow.

Answer 2

Mass: Double integral(DI) of rhovndA = 0, therefore v1rho1A1 = v2rho2A2.
Mass flow rate = vrhoA = const.

Energy: Double integral of horhovn*dA = 0. Perfect gas h0 = const.
T01 = T1 + (v1^2/2cp) = T2 + (v2^2/2cp) = T02

Question 3

Air enters a duct at 300 kPa, 100 Degrees C, and 10 m/s and leaves at 200 kPa and 200 m/s. The
flow is adiabatic. Show that the flow is not isentropic.

Answer 3

Using formula T1 + (v1^2/2cp) = T2 + (v2^2/2cp) , rearrange to find T2. If flow is isentropic, the relationship (p2/p1) = (T2/T1)^(k/(k-1)), should apply. If p2 is found to be different than the p2 given, then flow is not isentropic.

Question 4

Air enters a duct steadily at 300 kPa, 100oC, and 10 m/s and leaves 200 m/s. The inlet area of
the duct is 100 cm2. Considering the flow as isentropic
(e) compute the mass flow rate through the duct;

Answer 4

Use state equation p1= rho1T1Rg, and rearrange to find rho1(Pressure In PASCAL!!!).
Mass flow rate is A1rho1v1.

Question 5

Air enters a duct steadily at 300 kPa, 100oC, and 10 m/s and leaves 200 m/s. The inlet area of
the duct is 100 cm2. Considering the flow as isentropic.
(f) compute exit temperature and pressure of the air;

Answer 5

Using earlier Equations from the proof for non-isentropic flow, we can find T2 and P2 which are our exit values for air, assume that flow is isentropic so that our formula for (p2/p1) applies.

Question 6

Air enters a duct steadily at 300 kPa, 100oC, and 10 m/s and leaves 200 m/s. The inlet area of
the duct is 100 cm2. Considering the flow as isentropic.
(g) compute the exit area of the duct.

Answer 6

As we know m_dot1 = m_dot2, set rho1v1A1 = rho2v2A2. Find rho2 by using state equation P2 = rho2T2Rg, then rearrange to find A2.