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
Fig. 24.3. p-v, T-s and h-s diagrams
(a) Determine change in enthalpy
Formuls: Change in enthalpy = hsup,2 – hg,1
Answer: Change in enthalpy = hsup,2 – hg,1 = 2905.4 – 2801= 104.4 kJ/kg
(b) Determine change in internal energy
Formuls: The change in internal energy = usup,2 – ug,1
= (hsup,2 – p2 v2) - (hg,1 – p1 vg,1)
Answer: The change in internal energy = (hsup,2- p2 v2) – (hg,1 – p1 vg,1)
= (2905.4 – 2 x105 x 10-3 x 1.1215) – (2801 – 22 x105 x 10-3 x 0.09069)
= 79.62 kJ/kg.
(c) Determine change in entropy
Formuls: The Change in entropy = ssup,2 – sg,1
Answer: The Change in entropy = ssup,2 – sg,1
= 7.576 – 6.305 = 1.271 kJ/kg K
(d) Determine heat transfer, q1-2
Formuls: Heat transfer, q1-2 = T(ssup,2 – sg,1 )
Answer: q1-2 = T(ssup,2 – sg,1 )
= (217.2 + 273) x 1.271 = 623.04 kJ/kg K
(e) Work done, w1-2
Formuls: Work done = w1-2 = (ug,1 – usup,2) + q1-2
Answer: w1-2 = (ug,1 – usup,2) + q1-2
= –79.62 + 623.04 = 543.42 kJ/kg
Problem 24.3: Steam at a pressure of 5 bar and 0.8 dry expands in a cylinder according to the law pv1.35 = C to 2 bar. Find (a) the condition of steam at the end of expansion, (b) interchange of heat between the steam and the cylinder per kg of steam, (c) change in internal energy and (d) work done
Solution:
Given: The expansion of steam in the cylinder from state 1 to state 2 is a non-flow process according to the law pv1.35 = C.
Given: Steam at state ‘1’: p1= 5 bar and x1 = 0.8
By using steam table (for dry saturated steam):
For state 1, from steam tables for dry saturated steam, at p1 = 5 bar.
ts,1 = 151.8°C, hf,1 = 640 kJ/kg, hfg,1 =2109 kJ/kg, hg,1 = 2749 kJ/kg, vg,1 = 0.3748 m3/kg,
Given: Steam at state ‘2’: p2= 2 bar;
By using steam table (for dry saturated steam):
At state 2, from steam tables for dry saturated steam, at p1 = 2 bar.
ts,2 = 120°C, hf,2 = 505 kJ/kg, hg,2 = 2706.3 kJ/kg, hfg,2 =2202 kJ/kg, vg,2 = 0.8856 m3/kg
Fig. 24.4. p-v diagram
a) Determine the condition of steam at the end of expansion
Formula: The equation for the law of expansion is pv1.35 = C,
Hence, p1v11.35 = p2v21.35
or 
if v2 < vg,2 = 0.8856 m3/kg, then steam is wet.
if v2 = vg,2 = 0.8856 m3/kg, then steam is saturated.
if v2 > vg,2 = 0.8856 m3/kg, then steam is superheated.
Finding unknown, v1:
For wet steam, we have
v1 = x1 vg,1+(1-x1) vf,1
Neglecting volume of water, we have
v1 = x1 vg,1
= 0.8 x 0.3748 = 0.29984 m3/kg
Answer: 
m3/kg
Since, the actual specific volume of steam v2 = 0.59109 m3/kg is less than vg,2 = 0.8856 m3/kg,
Hence, the steam at state 2 is wet steam.
For wet steam at state ‘2’ we have
v2 = x2vg,2+(1-x2) vf,2
Neglecting volume of water, we have
v2 = x2vg,2
or 0.59109 = x2 (0.8856)
or 
Hence the final dryness fraction of steam = x2 = 0.6674
b) Determine the interchange of heat between the steam and the cylinder per kg of steam
Formula: During polytropic process, the heat transferred is given by
1q2 = (u2 − u1) + 
= (h2 − p2v2) − (h1 − p1v1 ) + kJ/kg
Finding unknown, h1 and h2;
For wet steam at state ‘1’ and state ‘2’, we have
h1 = hf,1 + x1. hfg,1 = 640 + 0.8 x 2109 = 2327 kJ/kg
h2 = hf,2 + x2. hfg,2 = 505 + 0.6674 x 2202 = 1974.61 kJ/kg
Answer: Therefore, 1q2 = (h2 − p2v2) − (h1 − p1v1 ) +
= (1974.61 − 2 x 105 x 10-3 x 0.59109) - (2327.2 − 5 x 105 x 10-3 x 0.29984) + 
= - 230.32 kJ/kg
-ve sign shows that heat is rejected from steam. Hence the heat transfer from steam to cylinder wall is = 230.32 kJ/kg
c) Determine change in internal energy, (u2-u1):
Formula: Change in internal energy = (u2-u1) = (h2 - p2v2) - (h1 - p1v1 )
Answer: (u2-u1) = (h2 - p2v2) - (h1 - p1v1 )
= (1974.61.- 2 x 105 x 10-3 x 0.59109) - (2327.2 - 5 x 105 x 10-3 x 0.29984)
= - 320.89 kJ/kg
d) Determine the work done
Formula: For polytropic process, the work done is given by
1w2 = 
Answer: Work done, 1w2 = = 
= 90.57 kJ/kg
Problem 24.4: Steam initially at a pressure of 15 bar and 0.95 dryness fraction expands isentropically to 7.5 bar and is then throttled until it is just dry. Determine per kg of steam:
(i) Change in entropy. (ii) Change in enthalpy. (iii) Change in internal energy.
Using: (a) Steam tables (b) Mollier chart.
Is the entire process reversible? Justify your statement.
Solution:
(A) USING STEAM TABLES
Given: Steam at state ‘1’: p1= 15 bar and x1 = 0.95
By using steam table (for dry saturated steam):
For state 1, from steam tables for dry saturated steam at p1 = 15 bar,
ts,1 = 198.3°C, hf,1 = 844.7 kJ/kg, hfg,1 =1945.2 kJ/kg, vg,1 = 0.132 m3/kg, sf,1 = 2.3145 kJ/kg K sfg,1 = 4.1261 kJ/kg K, sg,1 = 6.4406 kJ/kg K
Given: Steam at state ‘2’: p2= 7.5 bar;
By using steam table (for dry saturated steam):
For state 2, from steam tables for dry saturated steam at p2 = 7.5 bar.
ts,2 = 167.7°C, hf,2 = 709.3 kJ/kg, hfg,2 =2055.55 kJ/kg, vg,2 = 0.255 m3/kg, sf,2 = 2.0195 kJ/kg K sfg,2 = 4.6621 kJ/kg K, sg,2 = 6.6816 kJ/kg K
Given: Steam at state 3: p3 = 0.06 bar and x3 = 1
(i) Considering isentropic expansion 1-2
Determine per kg of steam: (i) Change in entropy (s2-s1), (ii) Change in enthalpy (h2 - h1) and (iii) Change in internal energy (u2-u1).
As the steam is wet at state ‘1’, therefore for wet steam, we have
h1= hf,1+ x1. hfg,1 = 844.7 + 0.95 x 1945.2 = 2692.64 kJ/kg
v1 = x1 vg,1 (neglecting volume of water)
= 0.95 x 0.132 = 0.1254 m3/kg
and s1= sf,1+ x1. sfg,1 = 2.3145 +0.95 x 4.1261 = 6.2343 kJ/kg K
Finding condition and properties of steam at state ‘2’
For isentropic or reversible adiabatic process ‘1-2’,
Change in entropy, (s2-s1) = 0
or s2 = s1
or s2 = 6.2343 kJ/kg K
Since, s2 = 6.2343 is less than the sg,2 = 6.6816 kJ/kg K
Therefore, condition of steam at point ‘2’ is wet
For wet steam, we have s2= sf,2+ x2. sfg,2
6.2343 = 2.0195 + x2. 4.6621
or x2 = 
= 0.9
As the steam is wet at state ‘2’, therefore for wet steam, we have
h2 = hf,2+ x2. hfg,2 = 709.3 +0.9 x 2055.55 = 2559.29 kJ/kg
v2 = x2 vg,2 (neglecting volume of water, vf,2)
= 0.9 x 0.255 = 0.2295 m3/kg
Answer:
Change in enthalpy = (h2 - h1) = 2559.29 – 2692.64 = - 133.35 kJ/kg (-ve sign indicates decrease)
Change in internal energy = (u2-u1) = (h2.- p2v2) - (h1.- p1v1 )
= (2559.29 –7.5 x 105 x 10-3 x 0.2295) - (2692.64 – 15 x 105 x 10-3 x 0.1254)
= - 117.38 kJ/kg (-ve sign indicates decrease)
Change in entropy, (s2-s1) = 0
(ii) Considering the throttling expansion 2-3
Determine per kg of steam: (i) Change in entropy (s3-s2), (ii) Change in enthalpy (h3 – h2) and (iii) Change in internal energy (u3-u2).
Finding condition and properties of steam at state ‘3’
For throttling process, ‘2-3’, we have change in enthalpy, (h3 - h2) = 0
or h3 = h2 = 2559.29 kJ/kg
Since the steam at state 3 is just dry (i.e. x3 = 1)
Hence hg,3 = h3
or hg,3 = 2559.29 kJ/kg
By using steam table (for dry saturated steam):
For state 3, From steam tables for dry saturated steam at p3 = 0.06 bar and hg,3= 2559.29 kJ/kg, we have
ts,3 = 36.18°C, hf,3 = 151.5 kJ/kg, hfg,3 =2416.0 kJ/kg, vg,3 = 23.741 m3/kg, sf,3 = 0.521 kJ/kg K sfg,3 = 7.809 kJ/kg K, sg,3 = 8.330 kJ/kg K
Answer: Change in entropy, sg,3– s2 = 8.330 - 6.2343 = 2.0957 kJ/kg K
Change in enthalpy, hg,3– h2 = 0 (throttling process)
Change in internal energy, (ug,3-u2) = (hg.3.- p3vg,3) - (h2- p2v2 )
= (hg,3- h2) - (p3vg,3- p2v2 )
= 0 - (0.06 x 105 x 10-3 x 23.741 – 7.5 x 105 x 10-3 x 0.2295)
= 29.68 kJ/kg
(iii) Combining the results obtained from isentropic ‘1-2’ and throttling expansion ‘2-3’ we get during the entire process
Determine per kg of steam: (i) Change in entropy (s3-s1), (ii) Change in enthalpy (h3 – h1) and (iii) Change in internal energy (u3-u1).
Answer:
(i) Change In entropy = (s2 – s1)+ (sg,3 – s2) = 0 +2.0957= 2.0957 kJ/kg K (Increase)
(ii) Change In enthalpy =(h2 – h1)+ (hg,3 – h2) = -133.35 + 0= -133.35 kJ/kg (decrease)
(iii) Change in Internal energy =(u2 – u1)+ (ug,3 – u2)= - 117.38 + 29.68 = -87.7 kJ/kg (decrease)
(B) BY USING MOLLIER CHART
Fig. 24.5. h-s diagram
(Refer Fig. 24.5)
Locate point 1 at an intersection of 15 bar pressure line and 0.95 dryness fraction line.
Draw vertical line from point ‘1’ intersecting 7.5 bar pressure line at point ‘2’. Line ‘1-2’ represents isentropic expansion.
From point 2, draw a horizontal line intersecting at the saturation line at point ‘3’. Line ‘2-3’ represents throttling expansion (constant enthalpy process).
From Mollier chart:
h1 = 2692 kJ/kg h2 = 2560 kJ/kg hg,3 = h2 = 2560 kJ/kg
s1 = s2 = 6.23 kJ/kg K sg,3 = 8.3 kJ/kg K
v1 = 0.125 m3/kg vg,3 = 24 m3/kg
Answer:
(i) Change in entropy .= (sg,3 – s2 ) + (s2 – s1 ) = (8.3 - 6.23) + (6.23 - 6.23) = 2.07 kJ/kg K
(ii) Change in enthalpy = (hg,3 – h2 ) + (h2 – h1 ) = (2560 – 2560) + (2560 – 2692) = -132 kJ/kg
(iii) Change in internal energy = (ug,3 – u2 ) + (u2 – u1 ) or
= [(hg,3 – p3 vg,3) – (h2 – p2 v2)] + [(h2 – p2 v2) – (h1 – p1 v1)]
= [(hg,3 – p3 vg,3) – (h1 – p1 v1)]
= [(2560 - 0.06 x 105 x 10-3 x 24) – (2692 – 15 x 105 x 10-3 x 0.125)]
= 2416 – 2504.5 = - 88.5 kJ/kg (decrease)
Answer: The entire process is not reversible as there is increase in entropy
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