Question

Steam at 400C has a specific volume of 0.02m3/kg. Determine the pressure of the steam based on a) the ideal gas equation b) the generalized compressibility chart and c) the steam tables.

Answer 1

Answer:

by ideal gas pressure = 15529.475 kPa

by compressibility chart pressure = 12576 kPa

by steam tables Pressure = 12517 kPa

Explanation:

given data

temperature T = 400°C = 673 K

volume v = 0.02 m³/kg

to find out

pressure by ideal gas, compressibility chart and steam tables

solution

we know here by table

gas constant R is 0.4615 kJ/ kg-K

and critical temp Tc = 647.1 K

and critical pressure Pc = 22064 kPa

so by ideal gas pressure is

pressure = R×T / v

pressure = 0.4615 × 673  / 0.02

pressure = 15529.475 kPa

and

by compressibility chart

temperature reduce is = T/ Tc

temperature reduce Tr = 673 / 647.1

Tr = 1.040 K

so pseudo reduce volume is here

reduce volume Vr = v / ( RTc/Pc)

reduce volume Vr =$\frac{0.02}{\frac{461.5(647.1)}{22064*10^{3} } }$

0.02 / ( 461.5(647.1) / 22064×10³)

reduce volume = 1.48

and we know by compressibility chart

reduce pressure Pr is 0.57

so

pressure = Pr × Pc

pressure = 0.57 × 22064 ×  10³

pressure = 12576 kPa

and

from steam table

pressure is 12.5 MPa at 673 K and 0.020030 m³/kg

pressure is 15 MPa at 673 K and 0.015671 m³/kg

so

pressure P is

$\frac{0.02 - 0.020030}{0.015671 - 0.020030}$ = $\frac{ P - 12.5}{15 - 12.5}$

so

Pressure = 12517 kPa