Question

1 kg of steam at 8 bar, entropy 6.55 kJ/ kg K is heated reversibly at constant pressure until the temperature is 200 °C. Calculate the heat supplied
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Answer 1

Answer:

Problem 24.1: Steam enters a steam turbine at a pressure of 15 bar and 350°C with a velocity of 60 m/s. The steam leaves the turbine at 1.2 bar and with a velocity of 180 m/s. Assuming the process to be reversible, adiabatic, determine the work done per kg of steam flowing through the turbine.

Solution:

Given: Steam at state ‘1’: p1= 15 bar; t1 = 350°C; V1=60 m/s

By using steam table (for dry saturated steam):

For state 1, from steam tables for dry saturated steam at p1 = 15 bar, we have

ts,1 = 198.2°C

As t1(350°C) > ts,1 (198.2°C), therefore steam at state ‘1’ is superheated steam

By using steam table (for superheated steam)

For state 1, from steam tables for superheated steam; at p1 = 15 bar; and tsup,1 =350°C.

hsup,1 = 3148.7 kJ/kg, ssup,1 = 7.102 kJ/kg Vsup,1 = V1 = 60 m/s

Fig. 24.1. Steam turbine

Given: Steam at state ‘2’: p2 = 1.2 bar; V2 =180 m/s

By using steam table (for dry saturated steam):

For state 2, from steam tables of dry saturated steam at p2 = 1.2 bar, we have

ts,2 = 104.8°C, hf,2 = 439.4 kJ/kg, hg,2 = 2683.4 kJ/kg, hfg,2 = 2244.1 kJ/kg, sf,2= 1.361 kJ/kg, sg,2= 7.2984 kJ/kg, sfg,2 = 5.937 kJ/kg

Fig. 24.2. T-s diagram

Determine the work done per kg of steam flowing through the turbine.

Apply the First law energy equation for steady flow process,

( hsup,1 + vsup,12/2 + g Zsup,1) = ( h2 + v22/2 + gZ2) +

( hsup,1 + Vsup,12/2) = ( h2 + V22/2) +

Therefore, = (hsup,1 - h2) +

Finding unknown, h2:

As the process is reversible adiabatic(constant entropy process), it will be represented by a vertical line on T-s diagram by 1-2.

The condition at point ‘2’ can be calculated by equating the entropy at point ‘1’ and point ‘2’

i.e. s2 = ssup,1 = 7.102

s2 = sf,2+ x2 .sfg,2

Therefore 7.102 = sf,2+ x2 .sfg,2

or 7.102 = 1.361 + x2 . 5.937

therefore, x2 =

h2 = hf,2+ x2. hfg,2 = 439.4 + 0.967 x 2244.1 = 2609.44 kJ/kg

Answer: Therefore, work done per kg of steam, 1w2 =

= (3148.7 - 2609.44) +

= 3147.5 – 2609.44 – 14.4

= 523.66 kJ/kg

Problem 24.2: In a steam engine cylinder, dry and saturated steam expands from 22 bar to 2 bar isothermally. Calculate (a) change in enthalpy, (b) change in internal energy, (c) change in entropy, (d) heat transfer, (e) work done. Assume the non-flow process in the cylinder.

Solution:

Given: Steam at state ‘1’: p1= 22 bar;

By using steam table (for dry saturated steam):

For state 1, From steam tables for dry saturated steam at p1 = 22 bar.

ts,1 = 217.2°C, hg,1 = 2801 kJ/kg, vg,1 = 0.09067 m3/kg, sg,1 = 6.305 kJ/kg K.

Given: Steam at state ‘2’: p2= 2 bar and t2 = ts,1 = 217.2°C (Isothermal process ‘1-2’)

By using steam table (for dry saturated steam):

For state 2, from steam tables for dry saturated steam at p1 = 2 bar, we have

Saturation temperature, ts,2 = 120.23°C.

Since the temperature of steam (t2 =217.2°C) is more than the saturated temperature of steam at state ‘2’ (ts,2 = 120.23°C), therefore condition of steam is superheated.

By using steam table (for superheated steam)

For state 2, From steam tables for superheated steam at p2 = 2 bar and tsup,2 = 217.2°C, we have

hsup,2 = 2905 kJ/kg, vsup,2 = 1.1215 m3/kg, ssup,2 = 7.576 kJ/kg K