Engineering Steam Turbine with Open Regenerative Feed Heater

[PaxFinnica96]

Homework Statement

A steam turbine operates with a condenser pressure of 0.04 bar and has a single open-type regenerative feed heater. How much steam, at a pressure of 2 bar and a dryness fraction of 0.98, must be bled from the turbine to produce 1 kg of saturated feedwater at 2 bar?

Relevant Equations

m h3 + (1 – m) h6 = h7

I would very much appreciate anyone to cast their eye over my attempt at solving this problem which I've been struggling with the past few days. Many thanks in advance.

Assuming that:

• energy input from open-type regenerative feed heater pump is negligible
• no energy is lost
• the heater pressure is the same as the bled pressure

h6 = hf at 0.04 bar = 121.41 kJ kg-1

h7 = hf at 2 bar = 505 kJ kg-1
Enthalpy at exit of turbine…

h3 = hf + x hfg at 2 bar (from steam tables)

h3 = 505 + 0.98 (1795) = 2264.1
Balancing the

m (2264.1) = (1 – m) (121.41) + 505

m = 0.26259

So, mass of steam bled from the turbine to produce 1kg of saturated feedwater as 2 bar would be 0.26259 kg

 

[Chestermiller]

Homework Statement:: A steam turbine operates with a condenser pressure of 0.04 bar and has a single open-type regenerative feed heater. How much steam, at a pressure of 2 bar and a dryness fraction of 0.98, must be bled from the turbine to produce 1 kg of saturated feedwater at 2 bar?
Relevant Equations:: m h3 + (1 – m) h6 = h7

I would very much appreciate anyone to cast their eye over my attempt at solving this problem which I've been struggling with the past few days. Many thanks in advance.

Assuming that:

• energy input from open-type regenerative feed heater pump is negligible
• no energy is lost
• the heater pressure is the same as the bled pressure

h6 = hf at 0.04 bar = 121.41 kJ kg-1

h7 = hf at 2 bar = 505 kJ kg-1
Enthalpy at exit of turbine…

h3 = hf + x hfg at 2 bar (from steam tables)

h3 = 505 + 0.98 (1795) = 2264.1
Balancing the

m (2264.1) = (1 – m) (121.41) + 505

This equation should read

m (2264.1) + (1 – m) (121.41) = 505

 

[PaxFinnica96]

Many thanks for your help, I get 0.179 kg