Consider adiabatic flow of air through a duct. At a given point in the duct, velocity of air is 300 m/s, temperature is 330 K and pressure is 180 kPa. Assume that the air behaves as a perfect gas with constant 𝒄_𝒑 =1.005 kJ/kg.K. The stagnation temperature at this point is _________ K (round off to two decimal places).

The correct answer is 374.77.

Problem Breakdown:
We are given the following properties for the adiabatic flow of air through a duct at a given point:
- Velocity of air, V = 300 m/s
- Static temperature, T = 330 K
- Static pressure, P = 180 kPa
- Specific heat at constant pressure, c_p = 1.005 kJ/kg·K = 1005 J/kg·K

Method 1: Direct Definition of Stagnation Temperature
For a gas with constant specific heats, the stagnation temperature (T₀) is related to the static temperature (T) and velocity (V) by the energy equation:

$T_0=T+\frac{V^2}{2c_p}$

Substituting the given values into this equation:

$T_0=330+\frac{{300}^2}{2\times 1005}$

$T_0=330+\frac{90000}{2010}$

$T_0=330+44.776=374.776\text{K}$

Rounding off to two decimal places, we get:
T₀ = 374.77 K (or 374.78 K depending on intermediate rounding).

Method 2: Using Mach Number
Alternatively, we can compute the stagnation temperature using the Mach number (M) and the ratio of specific heats (γ = 1.4, with gas constant R = 287 J/kg·K):
1. Calculate the speed of sound (a):

$a=\sqrt{\gamma RT}=\sqrt{1.4\times 287\times 330}\approx 364.13\text{m/s}$

2. Calculate the Mach number (M):

$M=\frac{V}{a}=\frac{300}{364.13}\approx 0.8239$

3. Use the stagnation temperature relation:

$\frac{T_0}{T}=1+\frac{\gamma -1}{2}{M}^2$

$T_0=330\times (1+0.2\times {0.8239}^2)\approx 374.77\text{K}$

Thus, both methods yield a stagnation temperature at this point of 374.77 K.