Hot combustion gases enter the nozzle of a turbojet engine at 260 kpa, 747oc, and 80 m/s. the gases exit at a pressure of 85 kpa. assuming an isentropic efficiency of 92 percent and treating the combustion gases as air, determine
a. the exit velocity
b. the exit temperature

Hot combustion gases are accelerated in a 92% efficient adiabatic nozzle from low velocity to a specified velocity. The exit velocity and the exit temp are to be determined.

Given:

T1 = 1020 K à h1 = 1068.89 kJ/kg, Pr1 = 123.4

P1 = 260 kPa

T1 = 747 degrees Celsius

V1 = 80 m/s ->nN = 92% -> P2 = 85 kPa

Solution:

From the isentropic relation,

Pr2 = (P2 / P1)PR1 = (85 kPa / 260 kPa) (123.4) = 40.34 = h2s = 783.92 kJ/kg

There is only one inlet and one exit, and thus, m1 = m2 = m3. We take the nozzle as the system, which is a control volume since mass crosses the boundary.

h2a = 1068.89 kJ/kg – (((728.2 m/s)­2 – (80 m/s)2) / 2) (1 kJ/kg / 1000 m2/s2) = 806.95 kJ/kg\

From the air table, we read T2a  = 786.3 K

explanation:

as the height from which the ruler was dropped increased, the speed of the ruler at ryan's hand increased.

k2s for potassium sulphide