At steady state, air at 200 kPa, 325 K, and mass flow rate of 0.5 kg/s enters an insulated duct having differing inlet and exit cross-sectional areas. The inlet cross-sectional area is 6 cm2. At the duct exit, the pressure of the air is 100 kPa and the velocity is 250 m/s. Neglecting potential energy effects and modeling air as an ideal gas with constant cp = 1.008 kJ/kg. K, determine

(a) the velocity of the air at the inlet, in m/s.
(b) the temperature of the air at the exit, in K.
(c) the exit cross-sectional area, in cm2.

velocity of the air at the inlet = 388.64 m/s

temperature = 368.92 K

cross-sectional area = 21.176 cm²

Explanation:

given data

pressure p1 = 200 kPa

temperature  t1 = 325 K

mass flow rate n = 0.5 kg/s

inlet cross-sectional area = 6 cm²

pressure of the air = 100 kPa

velocity = 250 m/s.

cp = 1.008 kJ/kg.K

solution

as we know that

mass = A V     ..............1

and

pV = nRT    ..........2

so we can say

     ........3

so  

n =    ..........4

put her value we get

0.5 =  

solve we get

V = 388.64 m/s

and

we apply now here steady flow energy that is

    ..........5

here no heat and work

so Q = 0 and W = 0

we get here h1 - h2 that is

h1 - h2 =     ..............6

Cp(T1-T2) =

put here value and we get

0.008 ( 325 - T2) =  

T2 = 368.92 K

and

for Area2 we put value in equation 4

0.5 =  

solve it we get

A2 = 21.176 cm²

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation: