Problem 13.14

# Problem 13.14 - c 1 = 352.9 m/s c 2 = 343.6 m/s V 1 = 109.8...

This preview shows page 1. Sign up to view the full content.

Problem 13.14 [Difficulty: 3] Given: Data on flow in a passage Find: Mach numbers at entrance and exit; area ratio of duct Solution: The given or available data is: R = 286.9 J/kg-K k = 1.4 T 1 = 310 K p 1 = 200 kPa T 2 = 294 K T 02 = 316 K p 2 = 125 kPa Equations and Computations: Since the flow is adiabatic, the stagnation temperature is constant: T 01 = 316 K Solving for the Mach numbers at 1 and 2 using Eq. 13.7b (using built-in function IsenMfromT ( Tratio , k )) Then M 1 = 0.311 M 2 = 0.612 Using the ideal gas equation of state, we can calculate the densities of the gas: ρ 1 = 2.249 kg/m 3 ρ 2 = 1.482 kg/m 3 From static temperatures and Eq. 12.18
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: c 1 = 352.9 m/s c 2 = 343.6 m/s V 1 = 109.8 m/s V 2 = 210.2 m/s Since flow is steady, the mass flow rate must be equal at 1 and 2. So the area ratio may be calculated from the densities and velocities: A 2 /A 1 = 0.792 Note that we can not assume isentropic flow in this problem. While the flow is adiabatic, it is not reversible. There is a drop in stagnation pressure from state 1 to 2 which would invalidate the assumption of isentropic flow....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online