A well-insulated turbine operating at steady state with seam enters at 3 MPa, 40

Question

A well-insulated turbine operating at steady state with seam enters at 3 MPa, 400°C, with a volumetric flow rate of 85 m3/min. Some steam is extracted from the turbine at a pressure of 0.5 MPa and a temperature of 180°C. The rest expands to a pressure of 6 kPa and a quality of 90%. The total power developed by the turbine is 11,400 kW. Kinetic and potential energy effects can be neglected. Determine.

(a) the mass flow rate of the steam at each of the two exits, in kg/h.

(b) the diameter, in m, of the duct through which steam is extracted, if the velocity there is 20 m/s.

Explanation / Answer

At P1 = 3 Mpa and T1 = 400 C, from mollier diagram:

h1 = 3231.57 KJ/Kg.K

and mass flow rate = m1

density at 400 C = = 10.0627 kg/m^3

Q = volume flow rate = 85 m^3/min = 85/60 = 1.4167 m3/s

m1 = .Q = 10.0627 x 1.4167 = 14.255 Kg/s

At the extraction point , at P2 = 0.5 Mpa and T2 = 180 C, from mollier diagram:

h2 = 2812.45 KJ/Kg

and mass flow rate = h2

and at P3 = ^ Kpa and quality = 90 %,

h3 = h f@6 Kpa + 0.9 x h fg@6KPa

   = 151.6 + 0.9 x 2415.8

   = 2325.82 Kj/kg.K

and mass flow rate = m3

neglecting kinetic energy and potential energy chang:

Q - W = m3.h3 + m2.h2 - m1.h1

Q = 0, as turbine is adiabatic

W = m1.h1 - m2.h2 - m3.h3 = 11,400 KW (given)

a) 11,400 = 14.255 x 3231.57 - m2 x 2812.45 - m3 x 2325.82

2812.45 .m2 + 2325.82.m3 = 34666.03 (i)

Also by mass conservation:

m1 = m2 + m3

14.255 = m2 + m3 (ii)

on solving (i) and (ii)

m2 = 3.106 kg/s = 0.0008627 kg/h and m3 = 11.15 = 0.00309 kg/h

b) Q = A.V

1.4167 = (/4 . d2) x 20

d = 0.3 m

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