Question

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.

Answer 1

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

temperature = 368.92 K

cross-sectional area = 21.176 cm²

Step-by-step 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 =
`\rho` A V ..............1

and

pV = nRT ..........2

so we can say

`(n)/(V) = (p)/(RT)` ........3

so
`\rho = (p)/(RT)`

n =
`(p)/(RT) (AV)` ..........4

put her value we get

0.5 =
`(200* 10^3)/(287* 325) (6* 10^(-4)* V )`

solve we get

V = 388.64 m/s

and

we apply now here steady flow energy that is

`n (h1 + (v1^2)/(2000) +(gz1)/(1000) ) +Q = n (h2 + (v2^2)/(2000) +(gz2)/(1000) ) + W` ..........5

here no heat and work

so Q = 0 and W = 0

we get here h1 - h2 that is

h1 - h2 =
`(1)/(2000) (v2^2-v1^2)` ..............6

Cp(T1-T2) =
`(1)/(2000) (v2^2-v1^2)`

put here value and we get

0.008 ( 325 - T2) =
`(1)/(2000) (250^2-388.64^2)`

T2 = 368.92 K

and

for Area2 we put value in equation 4

0.5 =
`(100* 10^3)/(287* 368.92) (A2* 250 )`

solve it we get

A2 = 21.176 cm²