Nitrogen traveling at 12 ms with a pressure of 14 bar abs at

Nitrogen, traveling at 12 m/s with a pressure of 14 bar abs. at a temperature of 800 K, enters a device with an area of 0.05 m². No work or heat transfer takes place. The temperature at the exit, where the area is 0.15 m², has dropped to 590 K. What are the velocity and pressure at the outlet section?

Solution

initial conditions

velocity at inlet of device, U₁=12m/s

pressure at inlet of device, P₁=14 Bar

temperature at inlet of device=800 K

outlet conditions

velocity at the outlet of device=?

pressure at the outlet of device=?

tremperature at the outlet of device=590 K

as the nitrogen is flowing from inlet to outlet and assming the density of nitrogen to be fixed we can apply continuity equation:A₁V₁=A₂V_{2 ,} where A₁ and A₂ are areas at the inlet and outlet of device which are given in question and V₁ and V₂ are velocities at inlet and outlet of device ,only inlet velocity is given andd outlet velocity can b determined by applying continuity equation, as-

A₁=0.05m²

A₂=0.15m²

V₁=12 m/s

by applying continuity equation:

0.05*12=0.15*V₂

V₂=(0.05*12)/0.15

V₂=4m/s

therefore velocity of nitrogen at exit of device is 4 m/s

as in problem it is assumed that the there is no heat transfer taking place so therefore we can apply reversible adiabatic equations

T₂/T₁=(P₂/P₁) ^k-1/k

where T₁ T₂ P₁ P₂ are temperatures and pressures at inlet and otlet of the device and k is the specific heat ratio of nitrogen which is, k=1.4

hence

T₂/T₁=(P₂/P₁)^1.4-1/1.4

590/800=(P₂/14)^0.4/1.4

14(59/80)^1.4/0.4=P₂

P₂=4.283 Bar

thus pressure of nitrogen at exit of device is 4.283 Bar.