A stream of oxygen enters a compressor at 298 K and 1.00 atm at a rate of 127m3/h and is compressed to 358 K and 1000 atm. Estimate the volumetric flow rate of compressed O2 using the compressibility-factor equation of state.

Question

contestada

Respuesta :

okpalawalter8 okpalawalter8 · Answer 1 · 2020-10-15T23:02:19+02:00

Answer:

The value is [tex]V_2 = 0.246 \ m^3/h[/tex]

Explanation:

From the question we are told that

The temperature at which the gas enters the compressor is

[tex]T_i = 298 \ K[/tex]

The pressure at which the gas enters the compressor is

[tex]P_I = 1.0 \ atm[/tex]

The volumetric rate at which the gas enters the compressor is

[tex]V = 127 m^3/h[/tex]

The temperature to which the gas is compressed to is

[tex]T_f = 358 \ K[/tex]

The pressure to which the gas is compressed to is

[tex]P_f= 1000 \ atm[/tex]

Generally the volumetric flow rate of compressed oxygen is evaluated from the compressibility-factor equation of state as

[tex]V_2 = V_1 *\frac{z_2}{z_1} * \frac{T_2}{T_1} * \frac{P_1}{P_2}[/tex]

Here [tex]z_1[/tex] is the inflow compressibility factor with value [tex]z_1 = 1[/tex]

Here [tex]z_1[/tex] is the outflow compressibility factor with value [tex]z_2 = 1.61[/tex]

So

[tex]V_2 = 127*\frac{1.61}{1} * \frac{358}{298} * \frac{1}{1000}[/tex]

[tex]V_2 = 0.246 \ m^3/h[/tex]