Problem 23.1: A quantity of steam at 10 bar and 0.85 dryness occupies 0.15 m3. Determine the heat supplied to raise the temperature of the steam to 300°C at constant pressure and percentage of this heat which appears as external work.
Take specific heat of superheated steam as 2.2 kJ/kg K.
Solution: Refer Fig. 23.1.
Given: Steam at state ‘1’: Pressure of steam, p1 = 10 bar;
Dryness fraction, x1 = 0.85;
Volume of steam, V1 = 0.15 m3
By using steam table (for dry saturated steam):
For state 1, from steam tables for dry saturated steam, at p1 =10 bar, we have:
ts,1 = 179.9 °C ; hf,1 = 762.6 kJ/kg; hfg,1 = 2013.6 kJ/kg, vg,1 = 0.194 m3/kg
Fig. 23.1. T-s diagram
Given: At state ‘2’: Pressure of steam, p2 = 10 bar: Temperature of steam, t2 = 300°C
By using steam table (for dry saturated steam):
For state 2, from steam tables for dry saturated steam, at p2 = 10 bar, we have;
Saturated temperature of steam, ts,2 = 179.9 °C
Since the actual temperature of steam (t2 = 300°C) at state ‘2’ (at pressure p2= 10 bar) is more than its saturated temperature (ts,2 = 179.9 °C),
Hence the steam at state ‘2’ is superheated. Therefore, tsup,2 = t2 = 300°C
By using steam table (for superheated steam):
For state 2, From steam tables for superheated steam at p2 = 10 bar and tsup,2 = 300°C, we have:
hsup,2 = 3052.1 kJ/kg, vsup,2 = 0.258 m3/kg
Given: Specific heat of superheated steam, Cps = 2.2 kJ/kg K
(i) Determine the heat supplied to raise the temperature of the steam to 300°C at constant pressure, 1q2.
Formula: Heat supplied per kg of steam, q1-2 = (usup,2 – u1) + ∫pdv
= (usup,2 – u1) + (p2vsup,2 – p1v1)
= (hsup,2 – h1) kJ/kg
or Total heat supplied, 1Q2 = m x q1-2 kJ
Finding unknown, h1 and m:
h1 = hf,1 + x1. hfg,1 = 762.6 + 0.85 x 2013.6 = 2474.16 kJ/kg
Mass of steam, m = 
= 0.909 kg
Answer: Heat supplied per kg of steam, 1q2 = (hsup,2 – h1)
= 3052.1 - 2474.16 = 577.94 kJ/kg
Total heat supplied, 1Q2 = m x q1-2 = 0.909 x 577.94 = 525.35 kJ
(ii) Determine the percentage heat (1Q2) which appears as external work (w1-2):
Formula: Percentage of total heat supplied which appears as external work
= 
Finding unknown, w1-2:
External work done during this process,
w1-2 = 
kJ/kg
Finding unknown, v1:
v1 = (1 –x1).vf,1 +x1.vg,1
or v1 = x1.vg,1 =0.85 (0.194) = 0.1649 m3/kg
Therefore w1-2 = 
= 10 x 105 (0.258 – 0.1649) = 93 kJ/kg
Answer: Percentage of total heat supplied which appears as external work
= 
= 14.1%
Problem 23.2: Calculate the internal energy per kg of superheated steam at a pressure of 10 bar and a temperature of 300°C. Also find the change of internal energy if this steam is expanded to 1.4 bar and dryness fraction 0.8.
Solution: Refer Fig. 23.2.
Given: Steam at state ‘1’: Pressure of steam, p1 = 10 bar;
Temperature of superheated steam, tsup,1 = 300°C
By using steam table (for superheated steam):
For state 1, from steam tables for superheated steam at p1 = 10 bar and tsup,1 = 300°C, we have:
hsup,1 = 3052.1 kJ/kg, vsup,1 = 0.258 m3/kg Tsup,1 = 300+273 = 573 K
Given: At state ‘2’: Pressure of steam, p2= 1.4 bar: dryness fraction, x2 = 0.8.
By using steam table (for dry saturated steam):
For state 2, from steam tables for dry saturated steam at p2 = 1.4 bar, we have:
hf,2 = 458.4 kJ/kg; hg,2 = 1773.5 kJ/kg , hfg,2 = 2231.9 kJ/kg, vg,2 = 1.236 m3/kg
Fig. 23.2. T-s diagram
(i) Determine the internal energy per kg of superheated steam, usup,1:
Formula: Internal energy per kg of superheated steam at 10 bar,
usup,1 = hsup,1 – p1. vsup,1
Answer: usup,1 = hsup,1 – p1. vsup,1 = 3052.1 – 10 x 105 x 0.258 x 10-3
= 2806.2 kJ/kg
(ii) Determine the change in internal energy per kg if this steam is expanded to 1.4 bar and dryness fraction 0.8
Formula: change of internal energy = u2 – usup,1 ,kJ/kg
Finding unknown, u2:
Internal energy of wet steam after expansion,
u2= h2 – p. v2
Finding unknown, h2:
Specific enthalpy of wet steam, h2 = hf,2+ x. hfg,2
= 458.4 + 0.8 x 2231.9
= 2243.92 kJ/kg.
Finding unknown, v2:
Specific volume of wet steam, v2 = x2.vg,2+ (1 – x2).vf,2
or v2 = x2 .vg,2
= 0.8 x 1.236 = 0.9888 m3/kg
Therefore, u2= h2 – p. v2 = 2243.92 – 1.4 x 105 x 0.9888 x 10-3
= 2105.49 kJ/kg
Answer: Hence, change of internal energy = u2 – usup,1 = 2105.49 – 2806.2
= – 700.7 kJ/kg
Negative sign indicates decrease in internal energy.
Problem 23.3: A rigid cylinder of volume 0.028 m3 contains steam at 80 bar and 350°C. The cylinder is cooled until the pressure is 50 bar. Calculate:
(i) The state of steam before and after cooling.
(ii) The amount of heat rejected by the steam.
Solution: Refer Fig. 23.3.
Given:
Rigid cylinder volume V = 0.028 m3
Steam at state ‘1’: p1= 80 bar; t1 = 350°C
By using steam table(for dry saturated steam:
For state ‘1’: From steam tables for dry saturated steam at p1= 80 bar, we have
Temperature of saturated steam, ts,1 = 294.97°C
Given: Steam at state ‘2’: p2 = 50 bar
By using steam table (for dry saturated steam):
For state 2, From steam tables for dry saturated steam at p2 = 50 bar we have:
hf,2= 1154.5 kJ/kg, hg,2 = 2794.2 kJ/kg; hfg,2 = 1639.7 kJ/kg; vg,2 = 0.03943 m3/kg.
Fig. 23.3. T-s diagram
(i) Determine the state of steam before and after cooling i.e. at state ‘1’ and state ‘2’.
Condition of steam at state ‘1’ (before cooling):
Since the actual temperature of steam (t1 = 350°C) is more than the saturated temperature of steam at state ‘2’ (ts1 = 294.97°C),
Hence the steam at state ‘1’ is superheated. Therefore, tsup,1 = t1 = 350°C
By using steam table (for superheated steam):
For state ‘1’: From steam tables for superheated steam at p1= 80 bar and tsup,1 = 350°C, we have:
vsup,1 = 0.02995 m3/kg, hsup,1 = 2987.3 kJ/kg
Condition of steam at state ‘2’ (after cooling):
The condition of steam at state ‘2’ i.e. whether the steam is wet, dry saturated or superheated.
Because cylinder has constant volume (V = 0.028 m3) and fixed mass of steam,
Therefore, we have v2 = vsup,1
or v2 = 0.02995 m3/kg
Since the specific volume of steam (v2 = 0.02995 m3/kg) is less than the specific volume of saturated steam (vg,2 = 0.03943 m3/kg),
Hence the steam at state ‘2’ is wet.
Finding dryness fraction of wet steam
For wet steam, we have v2 = x2.vg,2+ (1 –x2).vf,2
or 0.02995 = x2.vg,2+ (1 –x2).vf,2
or x2.vg,2 = 0.02995 (Neglecting volume of water, vf)
Therefore, dryness fraction of wet steam is given by,
x2 = 
= 0.76
(ii) Determine the amount of heat rejected by the steam.
Formula: 1Q2 = m (u2 – usup,1) + 1W2 (Since dv = 0)
or 1Q2 = m (u2 – usup,1)
or = m [(h2 – p2v2) + (hsup,1- p1.vsup,1)]
Finding unknown, h2:
Sp. enthalpy of wet steam,
h2 = hf,2+ x. hfg,2 = 1154.5 + 0.76 x 1639.7
= 2400.67 kJ/kg.
Finding unknown, m:
Mass of steam in cylinder, m = 0.935 kg
Answer: 1Q2 = m [(h2 – p2v2) + (hsup,1 – p1.vsup,1)]
= 0.935[ (2400.67 – 50 x105 x 0.02995 x 10-3) – (2987.3 – 80 x 105 x 0.02995 x 10-3)]
= 0.935 [2250.92 – 2747.7]
= – 465.5 kJ i.e. Heat rejected
Problem 23.4: Steam at 10 bar and 200°C enters a convergent divergent nozzle with a velocity of 60 m/s and leaves at 1.5 bar and with a velocity of 650 m/s. Assuming that there is no heat loss, determine the quality of steam leaving the nozzle.
Solution:
Fig. 23.4. A Convergent divergent nozzle and T-s diagram
Given: Steam at state ‘1’: p1= 10 bar; t1 = 200°C; V1= 60 m/s
By using steam table (for superheated steam):
For state 1, from steam tables for superheated steam at p1 = 10 bar, we have:
ts,1 = 179.9 °C
Since the actual temperature of steam (t1 =200°C) is more than the saturated temperature of steam at state ‘1’ (ts,1 = 179.9 °C),
Hence the steam at state ‘1’ is superheated.
Therefore, tsup,1 = t1 = 200°C and Vsup,1 = V1 = 60 m/s
By using steam table (for superheated steam):
For state 1, from steam tables for superheated steam, at p1 = 10 bar and tsup,1 = 200°C, we have:
hsup,1= 2827.9 kJ/kg
Given: Steam at state ‘2’: p1= 1.5 bar; V2 = 650 m/s
By using steam table (for dry saturated steam):
For state 2, from steam tables for dry saturated steam at p2 = 1.5 bar, we have:
ts,1 = 111.4 °C, hf,2= 467.1 kJ/kg, hg,2 = 2693.4 kJ/kg, hfg,2 = 2226.2 kJ/kg.
Determine the quality of the steam leaving the nozzle
Formula: Apply the First law energy equation for steady flow process,
( hsup,1 + Vsup,12/2 + g Zsup,1) = 
( h2 + V22/2 + gZ2) + 
Since, 
Therefore, h2 = 
If h2 < hg,2, the steam is wet or
If h2 = hg,2, the steam is saturated or
If h2 > hg,2, the steam is superheated.
Answer: h2 = 
= 2618.45 kJ/kg
As h2 < hg,2, therefore the condition of steam is wet.
Dryness fraction of steam:
The enthalpy of wet steam is given by
h2 = hf,2+ x2. hfg,2
2618.45 = 467.1 + x2 2226.2
x2 = 
Hence the condition of steam leaving the nozzle is 96.6% dry.
Fig. 23.5. T-s diagram
Problem 23.5: Steam at 10 bar and 0.9 dryness fraction is throttled to a pressure of 2 bar Determine the exit condition of steam using Mollier chart.
Solution:
Given: Steam at state ‘1’: p1= 10 bar; x1 = 0.9
Steam at state ‘2’: p2= 2 bar;
By using Mollier chart
Locate point ‘1’ at an intersection of 10 bar pressure line and 0.9 dryness fraction line on the Mollier chart.
Draw horizontal linefrom point ‘1’ intersecting 2 bar pressure line at point ‘2’. Line ‘1-2’ represents throttling expansion (constant enthalpy process).
Fig. 23.6. Mollier chart
Read the value of dryness fraction corresponding to point ‘2’from chart.
Answer: x2 = 0.94
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