Answer:
Exit temperature = 285 K
Explanation:
We are given that;
At inlet ;
T1 = 270K
V1 = 180 m/s
At exit;
V2 = 48.4 m/s
Now, from the table i attached, specific enthalpy for air at temperature of 270K is;
h1 = 270.11 Kj/Kg
Mow, Energy equation is given by;
Q' - W' = m'[(h2 - h1) + [(V2² - V1²)/2] + g(z2 - z1)]
The question tells us that potential energy effects are negligible. Thus, it means no work is exerted from or to the system and thus we solve as an adiabatic system.
Thus, Q', W', z1 and z2 are zero. Also, mass flow rate does not exist since there is no work.
So, we have;
0 = [(h2 - h1) + [(V2² - V1²)/2]
Since we are looking for h2 which is the specific enthalpy at exit, let's make it the subject;
h2 = h1 + [(V1² - V2²)/2]
h2 = 270.11 + [(180² - 48.4²)/2]
h2 = 270.11 Kj/Kg + 15.028 Kj/Kg = 285.138 Kj/Kg
From the table attached, at h2 = 285.138Kj/Kg, we have a temperature of approximately 285K
