Mixing steam at different pressures

  • Context: Archived 
  • Thread starter Thread starter Morey
  • Start date Start date
  • Tags Tags
    Mixing Steam
Click For Summary
SUMMARY

The discussion focuses on the calculation of the mass flow rate of throttled steam required to achieve a final mixture of dry saturated steam at 800 kPa. The initial steam flow is 3.5 kg/s at 650 kPa and 90% dryness. Using steam tables and the enthalpy equation, the enthalpy rate of the entering steam is calculated as 8278.2 kJ/s. By applying a heat balance equation, the mass of throttled steam needed is determined to be 48.7 kg/s.

PREREQUISITES
  • Understanding of steam tables and thermodynamic properties of steam
  • Knowledge of enthalpy calculations in thermodynamics
  • Familiarity with mass and energy balance equations
  • Basic principles of steam mixing and throttling processes
NEXT STEPS
  • Study the properties of steam at various pressures using steam tables
  • Learn about the application of the first law of thermodynamics in fluid systems
  • Explore the effects of throttling on steam properties and performance
  • Investigate advanced steam mixing techniques in industrial applications
USEFUL FOR

Mechanical engineers, thermodynamics students, and professionals involved in steam system design and optimization will benefit from this discussion.

Morey
Messages
1
Reaction score
0

Homework Statement


steam flow through a line is 3.5 kg/s at 650 kps and .9 dry. It enters a mixing line where it exits at 800kpa, dry and saturated. The mixing steam is throttle from 2000kps to 800kpa. What is the mass of throttling steam added to have a final mixture of 800 kpa dry steam.


Homework Equations


steam tables
q=mcΔT


The Attempt at a Solution



@650: hf=684 hfg=2076 hg=2552
@800: hf=721 hfg=2048 hg=2769
@2000:hf=908 hfg=1890 hg=2798
 
Physics news on Phys.org
To do this, we need to assume that the throttled steam was saturated. Otherwise the problem can't be solved.

Enthalpy rate of entering 650 kpa stream = 3.5((0.1)(684)+(0.9)(2552))=8278.2 kJ/s
Specific enthalpy of 2000 kPa steam = 2798 kJ/kg
Specific enthalpy of 800 kPa product steam =2769 kJ/kg

From heat balance on system, 8278.2 + 2798 w = (3.5+w) 2769

w = 48.7 kg/s