Question

Air flows isentropically through a duct. At a section, the area is 0.05 m² , velocity v = 180 m/s and temperature T = 470 K. Compute the stagnation temperature T_o
1. 486 K
2. 286 K
3. 556 K
4. 508 K

Answer 1

Correct Answer - Option 1 : 486 K

Concept:

Stagnation Temperature:

The stagnation temperature of the flowing fluid can be defined as the temperature attained when the fluid is isentropically decelerated to zero velocity.

\(C_pT_o=C_pT+\frac{V^2}{2} \)

\(\Rightarrow T_o=T+\frac{V^2}{2C_p}\)

Calculation:

Given:

T = 470 K,  v = 180 m/s , C_P = 1.005 KJ/ Kg-K

\(T_o=T+\frac{V^2}{2C_p}\)

\(T_o=470+\frac{180^2}{2\times1.005\times 10^3}\)

To = 486.12 K