Question

Steam at a pressure of 20 bar and 250°C enters a turbine and leave finally at a pressure of 0.05 bar. Three feed water heated are used which are operating at 5 bar, 1.5 bar and 0.3 bar. Assume following data for steam.

At 20 bar and 250°C, h= 2905 kJ/kg, Enthalpy of steam entering first, second and third feed water heaters are 2600 kJ/kg, 2430 kJ/kg and 2210 kJ/kg respectively. Enthalpy of steam finally leaving the turbine is 2000 kJ/kg.

At 5 bar; $h_f$ = 640.1 kJ/kg.
At 1.5 bar, $h_f$ = 467.1 kJ/kg.
At 0.3 bar; $h_f$ = 289.3 kJ/kg.
At 0.05 bar, $h_f$ = 137.8 kJ/kg.

The mass of bled steam per kg of steam flow at third feed water heater operating at 0.3 bar is ___ kg. (Neglect all pump works)

• A. 0.084

Answer 1

The correct option is A 0.084

$h_1$ = 2905 kJ/kg,
$h_2$ = 2600 kj/kg,
$h_3$ = 2430 kJ/kg
$h_4$ = 2210 kJ/kg,
$h_5$ = 2000 kJ/kg
$h_6$ = $h_7$ = 137.8 kJ/kg
$h_8$ = $h_9$ = 289.3 kJ/kg
$h_{10}$ = $h_{11}$ = 467.8 kJ/kg
$h_{12}$ = $h_{13}$ = 640.1 kJ/kg

Energy balance in first FWH

$m_1{h}_2+(1-m_1)h_{11}=h_{12}$

$m_1=\frac{h_{12}-h_{11}}{h_2-h_{11}}=\frac{640.1-467.1}{2600-467.1}=0.081kJ/kg$

$m_1=0.081kJ/kg$ os entering stream

Energy balance in second FWH

$m_2{h}_3+(1-m_1-m_2)h_9=h_{10}$

$m_2=\frac{h_{10}-(1-m_1)h_9}{h_3-h_9}=\frac{467.1(1-0.081)\times 289.3}{2430-289.3}$

$m_2=0.09kJ/kg$ of entering steam

Enery balance in third FWH

$m_3{h}_4+(1-m_1-m_2-m_3)h_7=h_8$

$m_3=\frac{h_3-(1-m_1-m_2)h_7}{h_4-h_7}=\frac{289.3-(1-0.081-0.09)\times 137.8}{2210-137.8}$

$m_3=0.844kJ/kg$ of entering steam