Adiaba c Process of Vapor

- A process during which there is no heat transfer. The word adiaba c comes from the Greek word
adiabatos, which means not to be passed.
1. Reversible
a. Isentropic process
𝑄=0
∆𝑠 = 0

Non-flow work: 𝑊 = 𝑢 − 𝑢
Steady flow work: 𝑊 = ℎ − ℎ
2. Irreversible

Non-flow work: 𝑊 = 𝑢 − 𝑢 ′
Steady flow work: 𝑊 = ℎ − ℎ ′
Efficiency of the process (expansion):
𝐴𝑐𝑡𝑢𝑎𝑙 𝑤𝑜𝑟𝑘 ℎ − ℎ ′
𝜂= =
𝐼𝑑𝑒𝑎𝑙 𝑤𝑜𝑟𝑘 ℎ −ℎ
Efficiency of the process (compression):
𝐼𝑑𝑒𝑎𝑙 𝑤𝑜𝑟𝑘 ℎ −ℎ
𝜂= =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑤𝑜𝑟𝑘 ℎ − ℎ ′
Process internal reheat: ℎ ′ − ℎ
Irreversibility of the process: 𝑠 ′ − 𝑠
Example 1

A steam engine isentropically expands 5 kg/sec of steam from 0.68 MPa, 220°C. The exhaust is dry and
saturated.

Determine:

(a) The work of a nonflow process

(b) The final temperature
(c) The work of a steady flow process

Given:

Process: Isentropic

ṁ = 5 kg/s
Ini al state (Point 1):

𝑃 = 0.68 𝑀𝑃𝑎
𝑡 = 220°𝐶
Check with steam tables if the given condi on of steam is dry saturated (saturated vapor), superheated
vapor, or a saturated liquid-vapor mixture (wet steam).

Using Table 2, get the satura on temperature of 𝑃 = 0.68 𝑀𝑃𝑎:

𝑡 = 163.81°𝐶
Since 𝑡 > 𝑡 , the ini al state is superheated vapor.

Refer to table 3 to get 𝑠 , 𝑢 , 𝑎𝑛𝑑 ℎ .

𝑘𝐽
𝑠 = 6.9929
𝑘𝑔 𝐾
𝑘𝐽
𝑢 = 2669.4
𝑘𝑔
𝑘𝐽
ℎ = 2889.9
𝑘𝑔
Final state (Point 2):

Since the process is isentropic, the final entropy is equal to ini al entropy.
𝑘𝐽
𝑠 = 𝑠 = 6.9929
𝑘𝑔 𝐾
According to the problem, the exhaust is dry and saturated. Thus, the final state is saturated vapor.

𝑠 = 𝑠 , 𝑢 = 𝑢 , 𝑎𝑛𝑑 ℎ = ℎ .
Find the pressure using the value of entropy on table 2 or find the temperature using table 1.

From Table 2:

P, MPa t, °C s, kJ/kg K u, kJ/kg h, kJ/k/g

0.295 132.97 6.9975 2543.0 2724.5
6.9929
0.300 133.55 6.9919 2543.6 2725.3

By interpola on:

𝑃 = 0.2991071429 𝑀𝑃𝑎 𝑜𝑟 0.2991 𝑀𝑃𝑎

𝑡 = 133.4464286°𝐶
𝑘𝐽
𝑢 = 2543.492857
𝑘𝑔
𝑘𝐽
ℎ = 2725.157143
𝑘𝑔
(a) The work of a nonflow process

𝑊 =𝑢 −𝑢
𝑘𝐽 𝑘𝐽
𝑊 = 2669.4 − 2543.492857
𝑘𝑔 𝑘𝑔
𝑘𝐽
𝑊 = 125.907143
𝑘𝑔
Mul ply with the mass flow rate:
𝑘𝐽 𝑘𝑔
𝑊 = 125.907143 5
𝑘𝑔 𝑠
𝑾𝒏 = 𝟔𝟐𝟗. 𝟓𝟑𝟓𝟕𝟏𝟓 𝒌𝑾

(b) The final temperature

𝒕𝟐 = 𝟏𝟑𝟑. 𝟒𝟒𝟔𝟒𝟐𝟖𝟔°𝑪

(c) The work of a steady flow process

𝑊 =ℎ −ℎ
𝑘𝐽 𝑘𝐽
𝑊 = 2889.9 − 2725.157143
𝑘𝑔 𝑘𝑔
𝑘𝐽
𝑊 = 164.742857
𝑘𝑔
Mul ply with the mass flow rate:
𝑘𝐽 𝑘𝑔
𝑊 = 164.742857 5
𝑘𝑔 𝑠
𝑾𝒔 = 𝟖𝟐𝟑. 𝟕𝟏𝟒𝟐𝟖𝟓 𝒌𝑾
Example 2

Saturated steam at 0.16 MPa is compressed in an irreversible adiaba c process with an efficiency of
80.24%. For a final pressure of 0.45 MPa, determine

(a) The final and ini al entropies

(b) The ideal and actual temperatures of compressed steam

Given:

Process: Irreversible adiaba c

𝜂 = 80.24%

Ini al state (Point 1): Saturated vapor

𝑃 = 0.16 𝑀𝑃𝑎
From Table 2 at 0.16 MPa:
𝒌𝑱
𝒔𝟏 = 𝒔𝒈 = 𝟕. 𝟐𝟎𝟏𝟕
𝒌𝒈 𝑲
𝑘𝐽
ℎ = ℎ = 2696,5
𝑘𝑔

Final state (Point 2):

𝑃 = 0.45 𝑀𝑃𝑎

Required:

(a) 𝑠 ′ and 𝑠
(b) 𝑡 and 𝑡 ′

Solu on:

𝑠 = 𝑠 for reversible adiaba c process.

𝑠 must be known before Ideal ℎ can be determined.

Using Table 2, get the entropy of saturated vapor at 𝑃 = 0.45 𝑀𝑃𝑎:

𝑘𝐽
𝑠 = 6.8565
𝑘𝑔 𝐾
Since 𝑠 > 𝑠 , the final state is superheated vapor.
From Table 3 at 0.45 MPa and 7.2017 kJ/kg K:

t, °C 0.44 MPa 0.45 MPa 0.46 MPa 0.44 MPa 0.45 MPa 0.46 MPa
ideal
s, kJ/kg K h, kJ/kg
ideal ideal
220 7.2103 7.1994 7.1885 2900.5 2900.1 2899.7
𝑡 7.2017 ℎ
230 7.2521 7.24125 7.2304 2921.4 2921 2920.6

By interpola on:

𝒕𝟐 = 𝟐𝟐𝟎. 𝟓𝟒𝟗𝟓𝟖𝟏𝟖°𝑪
𝑘𝐽
ℎ = 2901.248626
𝑘𝑔
Solving for ℎ :
ℎ −ℎ
𝑒=
ℎ −ℎ ′
ℎ −ℎ
ℎ =ℎ −
𝑒
𝑘𝐽
𝑘𝐽 2696.5 𝑘𝑔 − 2901.248626
ℎ = 2696.5 −
𝑘𝑔 0.8024
𝑘𝐽
ℎ = 2951.670272
𝑘𝑔

t, °C 0.44 MPa 0.45 MPa 0.46 MPa 0.44 MPa 0.45 MPa 0.46 MPa
actual
h, kJ/kg K s, kJ/kg K
actual actual
240 2942.1 2941.75 2941.4 7.2930 7.2822 7.2714
𝑡 ′ 2951.670272 𝑠 ′
250 2962.8 2962.45 2962.1 7.3329 7.32215 7.3114

By interpola on:

𝒕𝟐 = 𝟐𝟒𝟒. 𝟕𝟗𝟐𝟒𝟎𝟏𝟖°𝑪
𝒌𝑱
𝒔𝟐 = 𝟕. 𝟑𝟎𝟏𝟑𝟒𝟓𝟔𝟒𝟓
𝒌𝒈 𝑲
Review problems:

1. Three kg of steam has an ini al enthalpy of 7200 kJ at 0.145 MPa. Isentropic compression gives a
final pressure of 0.29 MPa. Find
(a) the ini al and final temperatures
(b) the work of a steady and nonflow processes
2. Wet steam at 200°C and 20% moisture is compressed in an irreversible adiaba c process to a
pressure of 2.30 MPa and a temperature of 225°C. Determine
(a) The work of compression
(b) The compression efficiency
(c) The irreversibility of the process
(d) The compression internal reheat
Answers to review problems:

1. Three kg of steam has an ini al enthalpy of 7200 kJ at 0.145 MPa. Isentropic compression gives a
final pressure of 0.29 MPa. Find
(a) the ini al and final temperatures
(b) the work of a steady and nonflow processes

Given:
Process: Isentropic
m = 3 kg
H1 = 7200 kJ
P1 = 0.145 MPa
P2 = 0.29 MPa
Required: t1, t2, Ws, Wn

Solu on:
𝐻 7200 𝑘𝐽 𝑘𝐽
ℎ = = = 2400
𝑚 3 𝑘𝑔 𝑘𝑔
From Table 2 At 0.145 MPa:
𝑘𝐽
ℎ = 462.80
𝑘𝑔
𝑘𝐽
ℎ = 2229.2
𝑘𝑔
𝑘𝐽
𝑢 = 462.65
𝑘𝑔
𝑘𝐽
𝑢 = 2055.9
𝑘𝑔
𝑘𝐽
𝑠 = 1.4224
𝑘𝑔 𝐾
𝑘𝐽
𝑠 = 5.8123
𝑘𝑔 𝐾
Ini al state is saturated liquid-vapor mixture. (point 1)
𝑡 = 𝑡 𝑎𝑡 0.145 𝑀𝑃𝑎
𝒕𝟏 = 𝟏𝟏𝟎. 𝟑𝟔°𝑪

𝑘𝐽 𝑘𝐽
ℎ −ℎ 2400 − 462.80
𝑘𝑔 𝑘𝑔
𝑥 = =
ℎ 𝑘𝐽
2229.2
𝑘𝑔
𝑥 = 0.8690113045

𝑢 =𝑢 +𝑥 𝑢
𝑘𝐽 𝑘𝐽
𝑢 = 462.65 + (0.8690113045) 2055.9
𝑘𝑔 𝑘𝑔
𝑘𝐽
𝑢 = 2249.250341
𝑘𝑔
 𝑠 =𝑠 +𝑥 𝑠
𝑘𝐽 𝑘𝐽
𝑠 = 1.4224 + (0.8690113045) 5.8123
𝑘𝑔 𝐾 𝑘𝑔 𝐾
𝑘𝐽
𝑠 = 6.473354405 = 𝑠 (𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐)
𝑘𝑔 𝐾

From Table 2 At 0.29 MPa:

𝑘𝐽
ℎ = 556.54
𝑘𝑔
𝑘𝐽
ℎ = 2167.2
𝑘𝑔
𝑘𝐽
𝑢 = 556.23
𝑘𝑔
𝑘𝐽
𝑢 = 1986.2
𝑘𝑔
𝑘𝐽
𝑠 = 1.6597
𝑘𝑔 𝐾
𝑘𝐽
𝑠 = 5.3435
𝑘𝑔 𝐾

Final state is saturated liquid-vapor mixture. (point 2)

𝑡 = 𝑡 𝑎𝑡 0.29 𝑀𝑃𝑎
𝒕𝟐 = 𝟏𝟑𝟐. 𝟑𝟗°𝑪

𝑘𝐽 𝑘𝐽
𝑠 −𝑠 6.473354405 − 1.6597
𝑘𝑔 𝐾 𝑘𝑔 𝐾
𝑥 = =
𝑠 𝑘𝐽
5.3435
𝑘𝑔 𝐾
𝑥 = 0.900842969

ℎ =ℎ +𝑥 ℎ
𝑘𝐽 𝑘𝐽
ℎ = 556.54 + (0.900842969) 2167.2
𝑘𝑔 𝑘𝑔
𝑘𝐽
ℎ = 2508.846882
𝑘𝑔

𝑢 =𝑢 +𝑥 𝑢
𝑘𝐽 𝑘𝐽
𝑢 = 556.23 + (0.900842969) 1986.2
𝑘𝑔 𝑘𝑔
𝑘𝐽
𝑢 = 2345.484305
𝑘𝑔
 𝑊 =ℎ −ℎ
𝑘𝐽 𝑘𝐽
𝑊 = 2400 − 2508.846882
𝑘𝑔 𝑘𝑔
𝑘𝐽
𝑊 = −108.8468825 (3 𝑘𝑔)
𝑘𝑔
𝑾𝒔 = −𝟑𝟐𝟔. 𝟓𝟒𝟎𝟔𝟒𝟕𝟒 𝒌𝑱

𝑊 =𝑢 −𝑢
𝑘𝐽 𝑘𝐽
𝑊 = 2249.250341 − 2345.484305
𝑘𝑔 𝑘𝑔
𝑘𝐽
𝑊 = 1 − 96.23396403 (3 𝑘𝑔)
𝑘𝑔
𝑾𝒏 = −𝟐𝟖𝟖. 𝟕𝟎𝟏𝟖𝟗𝟐𝟏 𝒌𝑱

2. Wet steam at 200°C and 20% moisture is compressed in an irreversible adiaba c process to a
pressure of 2.30 MPa and a temperature of 225°C. Determine
(a) The work of compression
(b) The compression efficiency
(c) The irreversibility of the process
(d) The compression internal reheat

Given:

Process: Irreversible adiaba c

Ini al state: wet steam

t1 = 200°C and y = 20% or 0.20

P2 = 2.30 MPa and t2 = 225°C

Tsat at 2.30 MPa = 219.60°C

Thus, final state is superheated vapor

Solu on:

From table 1 at 𝑡 = 200°𝐶

𝑘𝐽
ℎ = 852.45
𝑘𝑔
𝑘𝐽
ℎ = 1940.7
𝑘𝑔
𝑘𝐽
𝑠 = 2.3309
𝑘𝑔 𝐾
 𝑘𝐽
𝑠 = 4.1014
𝑘𝑔 𝐾
𝑘𝐽 𝑘𝐽
ℎ =ℎ +𝑥 ℎ = 852.45 + (1 − 0.20) 1940.7
𝑘𝑔 𝑘𝑔
𝑘𝐽
ℎ = 2405.01
𝑘𝑔
𝑘𝐽 𝑘𝐽
𝑠 =𝑠 +𝑥 𝑠 = 2.3309 + (1 − 0.20) 4.1014
𝑘𝑔 𝐾 𝑘𝑔 𝐾
𝑘𝐽
𝑠 = 5.61202 =𝑠
𝑘𝑔 𝐾
𝑘𝐽
ℎ = ℎ 𝑎𝑡 2.30 𝑀𝑃𝑎 𝑎𝑛𝑑 5.61202
𝑘𝑔 𝐾
From Table 2 at 2.30 MPa:
𝑘𝐽
𝑠 = 2.5141
𝑘𝑔 𝐾
𝑘𝐽
𝑠 = 3.7749
𝑘𝑔 𝐾
𝑠 =𝑠 +𝑥 𝑠
𝑠 −𝑠 5.61202 − 2.5141
𝑥 = = = 0.820662799
𝑠 3.7749
𝑘𝐽
ℎ = 941.77
𝑘𝑔
𝑘𝐽
ℎ = 1860.2
𝑘𝑔
𝑘𝐽 𝑘𝐽
ℎ =ℎ +𝑥 ℎ = 941.77 + (0.820662799) 1860.2
𝑘𝑔 𝑘𝑔
𝑘𝐽
ℎ = 2468.366939
𝑘𝑔
From Table 3 at 𝑃 = 2.30 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑡 = 225°𝐶:
𝑘𝐽
ℎ ′ = 2818.5
𝑘𝑔
𝑘𝐽
𝑠 ′ = 6.3222
𝑘𝑔 𝐾
(a) Work of compression
𝑘𝐽 𝑘𝐽
W = ℎ − ℎ = 2405.01 − 2818.5
𝑘𝑔 𝑘𝑔
𝒌𝑱
𝑾𝒔 = −𝟒𝟏𝟑. 𝟒𝟗
𝒌𝒈

(b) Efficiency of the process (compression):

𝐼𝑑𝑒𝑎𝑙 𝑤𝑜𝑟𝑘 ℎ −ℎ
𝜂= =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑤𝑜𝑟𝑘 ℎ − ℎ ′
𝑘𝐽 𝑘𝐽
2405.01 − 2468.366939
𝑘𝑔 𝑘𝑔
𝜂=
𝑘𝐽 𝑘𝐽
2405.01 − 2818.5
𝑘𝑔 𝑘𝑔
𝜼 = 𝟎. 𝟏𝟓𝟑𝟐𝟐𝟒𝟖𝟑𝟗𝟖 𝒐𝒓 𝟏𝟓. 𝟑𝟐%

𝒌𝑱
(c) Irreversibility of the process = 𝑠 − 𝑠 = 6.3222 − 5.61202 = 𝟎. 𝟕𝟏𝟎𝟏𝟖
𝒌𝒈 𝑲

𝒌𝑱
(d) Process internal reheat = ℎ − ℎ = 2818.5 − 2468.366939 = 𝟑𝟓𝟎. 𝟏𝟑𝟑𝟎𝟔𝟏
𝒌𝒈
Polytropic Process

 Nonflow work, Wn
𝑃 v −𝑃 v
𝑊 =
1−𝑛
 Heat, Q

𝑄 = ∆𝑢 + 𝑊
𝑃 v −𝑃 v
𝑄 =𝑢 −𝑢 +
1−𝑛
 Steady flow work, Ws

𝑊 = 𝑄 − ∆ℎ − ∆𝐾𝐸
𝑛(𝑃 v − 𝑃 v )
𝑊 = 𝑛(𝑊 ) =
1−𝑛
Problems
Thro ling Process

Thro ling is a flow process during which pressure drops but enthalpy remains constant. It reduces
the temperature of the steam but increases the dryness frac on of wet steam which may become dry
saturated or even superheated. This forms the principle of thro ling calorimeter, a device used to measure
the quality of wet steam.

Thro ling Calorimeter

In the thro ling calorimeter, a sample of wet steam of mass m and at pressure P 1 is taken from
the steam main through a perforated sampling tube. Then it is thro led by the par ally-opened valve (or
orifice) to a pressure P2, measured by mercury manometer, and temperature t2, so that a er thro ling the
steam is in the superheated region. The steady flow energy equa on gives the enthalpy a er thro ling as
equal to enthalpy before thro ling. The ini al and final equilibrium states 1 and 2 are joined by a do ed
line since thro ling is irreversible (adiaba c but not isentropic) and the intermediate states are non-
equilibrium states not describable by thermodynamic coordinates.
Thro ling Valves

Thro ling valves are any kind of flow-restric ng devices that cause a significant pressure drop in
a flowing fluid. Some familiar examples are ordinary adjustable valves, capillary tubes, and porous plugs.
Unlike turbines, they produce a pressure drop without involving any work. The pressure drop in the fluid
is o en accompanied by a large drop in temperature, and for that reason thro ling devices are commonly
used in refrigera on and air-condi oning applica ons.

Thro ling valves are usually small devices, and the flow through them may be assumed to be
adiaba c (q ~= 0) since there is neither sufficient me nor large enough area for any effec ve heat transfer
to take place. Also, there is no work done (w = 0), and the change in poten al energy, if any, is very small
(Δpe ~= 0). Even though the exit velocity is o en considerably higher than the inlet velocity, in many cases,
the increase in kine c energy is insignificant (Δke ~= 0). Then the conserva on of energy equa on for this
single-stream steady-flow device reduces to

ℎ ≅ℎ
Joule-Thomson Coefficient

The magnitude of the temperature drop (or, some mes, the temperature rise) during a thro ling
process is governed by a property called the Joule-Thomson coefficient.
∆𝑇
𝜇=
∆𝑃

Separa ng and Thro ling Calorimeter

When the steam is very wet and the pressure a er thro ling is not low enough to take the steam
to the superheated region, then a combined separa ng and thro ling calorimeter is used for the
measurement of quality. Steam from the main is first passed through a separator where some part of the
moisture separates out due to the sudden change in direc on and falls by gravity, and the par ally dry
vapor is then thro led and taken to the superheated region.

ℎ =ℎ
ℎ =ℎ +𝑥 ℎ

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑑𝑟𝑦 𝑣𝑎𝑝𝑜𝑟 𝑎𝑡 𝑠𝑡𝑎𝑡𝑒 1

𝑥 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 − 𝑣𝑎𝑝𝑜𝑟 𝑚𝑖𝑥𝑡𝑢𝑟𝑒 𝑎𝑡 𝑠𝑡𝑎𝑡𝑒 1
𝑥 𝑚
𝑥 =
𝑚 +𝑚
Example 1:

Steam at 800 kPa, 300°C is thro led to 200 kPa. Changes in kine c energy are negligible for this
process. Determine the final temperature of the steam, and the average Joule-Thomson coefficient.

Given:

𝑃 = 800 𝑘𝑃𝑎 = 0.8 𝑀𝑃𝑎

𝑡 = 300°𝐶
𝑃 = 200 𝑘𝑃𝑎 = 0.2 𝑀𝑃𝑎
From table 2 at 0.8 MPa:

𝑡 = 170.43°𝐶
Since 𝑡 > 𝑡 , then the ini al state is superheated vapor.

From table 3 at 0.8 MPa and 300°C:

𝑘𝐽
ℎ = ℎ = 3056.5
𝑘𝑔

From table 2 at 0.2 MPa:

𝑘𝐽
ℎ = 2706.7
𝑘𝑔
Since ℎ > ℎ𝑔 , then the final state is superheated vapor.
From table 3 at 0.2 MPa and ℎ = 3056.5 :

𝑡 ℎ
290 3051.6
𝑡 3056.5
300 3071.8

By interpola on:

𝑡 = 292.4257426°𝐶
∆𝑇 (300 + 273)𝐾 − (292.4257426 + 273)𝐾 7.5742574 𝐾 𝐾
𝜇 = = = = 0.01262376233
∆𝑃 800 𝑘𝑃𝑎 − 200 𝑘𝑃𝑎 600 𝑘𝑃𝑎 𝐾𝑃𝑎

Example 2

Steam flows in a pipeline at 1.5 MPa. A er expanding to 0.1 MPa in a thro ling calorimeter, the
temperature is found to be 120°C. Find the quality of steam in the pipeline. What is the maximum moisture
at 1.5 MPa that can be determined with this setup if at least 5°C of superheat is required a er thro ling
for accurate readings?

Given:

Ini al state 𝑃 = 1.5 𝑀𝑃𝑎

Final state 𝑃 = 0.1 𝑀𝑃𝑎 and 𝑡 = 120°𝐶

𝑥 =?
𝑦 =?
𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 2 𝑎𝑡 𝑃 = 0.1 𝑀𝑃𝑎
𝑡 = 99.63°𝐶
Therefore point 2 is superheated vapor while point 1 is a wet vapor mixture.

𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 3 𝑎𝑡 𝑃 = 0.1 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑡 = 120°𝐶

𝑘𝐽
ℎ = 2716.6 =ℎ
𝑘𝑔
ℎ =ℎ
ℎ =ℎ +𝑥 ℎ

ℎ −ℎ
𝑥 =
ℎ
𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 2 𝑎𝑡 𝑃 = 1.5 𝑀𝑃𝑎
𝑘𝐽
ℎ = 844.89
𝑘𝑔
𝑘𝐽
ℎ = 1947.3
𝑘𝑔
𝑘𝐽 𝑘𝐽
ℎ −ℎ 2716.6 − 844.89
𝑘𝑔 𝑘𝑔
𝑥 = =
ℎ 𝑘𝐽
1947.3
𝑘𝑔
𝒙𝟏 = 𝟎. 𝟗𝟔𝟏𝟐 𝒐𝒕 𝟗𝟔. 𝟏𝟐%
The quality of steam in the pipeline

𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 2 𝑎𝑡 𝑃 = 0.1 𝑀𝑃𝑎

𝑡 = 99.63°𝐶
if at least 5°C of superheat is required a er thro ling:

𝑡 =𝑡 + 5°𝐶 𝑆𝐻

𝑡 = 99.63°𝐶 + 5°𝐶 = 104.63°𝐶

𝐹𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 3 𝑎𝑡 𝑃 = 0.1 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑡 = 104.63°𝐶

t h
100 2676.2
104.63
110 2696.5

By interpola on:
𝑘𝐽
ℎ = 2685.5989 =ℎ
𝑘𝑔
ℎ =ℎ +𝑥 ℎ

𝑘𝐽 𝑘𝐽 𝑘𝐽
2685.5989 = 844.89 +𝑥 1947.3
𝑘𝑔 𝑘𝑔 𝑘𝑔
𝑥 = 0.9452621065

𝑦 = 1 − 𝑥 = 1 − 0.9452621065

𝒚𝟏 = 𝟎. 𝟎𝟓𝟒𝟕𝟒 𝒐𝒓 𝟓. 𝟒𝟕𝟒%
The maximum moisture at 1.5 MPa that can be determined with this setup if at least 5°C of superheat
is required a er thro ling for accurate readings
Problems
The Carnot Vapor Cycle

Carnot cycle is the most efficient cycle opera ng between two specified temperature limits, that
is why we look at Carnot cycle first as a prospec ve ideal cycle for vapor power plants. However, the Carnot
cycle is not a suitable model for power cycles.

Consider a steady-flow Carnot cycle executed within the satura on dome of a pure substance.

Process 1-2: The fluid is heated reversibly and isothermally in a boiler

Process 2-3: Steam is expanded isentropically in a turbine
Process 3-4: Steam is condensed reversibly and isothermally in a condenser
Process 4-1: Isentropical compression by a compressor to the ini al state (Process 4-1).

There are several imprac cali es associated with this cycle:

1. Limi ng the heat transfer process to two-phase systems severely limits the maximum
temperature that can be used in the cycle. Limi ng the maximum temperature in the cycle
also limits the thermal efficiency. Any a empt to raise the maximum temperature in the cycle
involves heat transfer to the working fluid in a single phase, which is not easy to accomplish
isothermally.
2. The isentropic expansion process cannot be approximated closely by a well-designed turbine.
The quality of the steam decreases during this process. Thus, the turbine has to handle steam
with low quality, that is, steam with a high moisture content.
3. The impingement of liquid droplets on the turbine blades causes erosion and is a major
source of wear. Thus, steam with quali es less than about 90 percent cannot be tolerated in
the opera on of power plants. This problem could be eliminated by using a working fluid
with a very steep saturated vapor line.
4. The isentropic compression process (4-1) involves the compression of a liquid-vapor mixture
to a saturated liquid. There are two difficul es associated with this process.
a. First, it is not easy to control the condensa on process so precisely as to end up with
the desired quality at state 4.
b. Second, it is imprac cal to design a compressor that handles two phases.
 Some of these problems could be eliminated by execu ng the Carnot cycle in a different way. This
cycle, however, presents other problems such as isentropic compression to extremely high pressures and
isothermal heat transfer at variable pressures.

Thus, we conclude that the Carnot cycle cannot be approximated in actual devices and is not a
realis c model for vapor power cycles.

Rankine Cycle

Rankine cycle is the ideal cycle for vapor power plants. The ideal Rankine cycle does not involve
any internal irreversibili es and consists of the following four processes:

Process 1-2 Isentropic compression in a pump,

Process 2-3 Constant pressure heat addi on in a boiler,
Process 3-4 Isentropic expansion in a turbine,
Process 4-1 Constant pressure heat rejec on in a condenser.
Process 1-2 (Pump)

Water enters the pump at state 1 as saturated liquid and is compressed isentropically to the
opera ng pressure of the boiler. The water temperature increases somewhat during this isentropic
compression process due to a slight decrease in the specific volume of water.

Process 2-3 (Boiler)

Water enters the boiler as a compressed liquid at state 2 and leaves as a superheated vapor at
state 3. The boiler is a large heat exchanger where the heat origina ng from combus on gases, nuclear
reactors, or other sources is transferred to the water essen ally at constant pressure. The boiler, together
with the sec on where the steam is superheated (the superheater), is o en called the steam generator.

Process 3-4 (Turbine)

The superheated vapor at state 3 enters the turbine, where it expands isentropically and produces
work by rota ng the sha connected to an electric generator. The pressure and temperature of steam
drop during this process to the values at state 4, where steam enters the condenser.

Turbine

A turbine is a rotary machine whose purpose is the produc on of sha work (power, on a rate
basis) at the expense of the pressure of the working fluid. Two general classes of turbines are steam (or
other working fluid) turbines, in which the steam exi ng the turbine passes to a condenser, where it is
condensed to liquid; and gas turbines, in which the gas usually exhausts to the atmosphere from the
turbine.

Process 4-1 (Condenser)

At this state, steam is usually a saturated liquid-vapor mixture with a high quality. Steam is
condensed at constant pressure in the condenser by rejec ng heat to a cooling medium such as a lake, a
river, or the atmosphere. Steam leaves the condenser as saturated liquid and enters the pump, comple ng
the cycle.

In the Rankine cycle, which is the ideal cycle for steam power plants, the working fluid flows
through a series of steady-flow devices such as the turbine and condenser. Thus, equa ons pertaining to
steady-flow systems should be used in the analysis of the Rankine cycle. All four components associated
with the Rankine cycle (pump, boiler, turbine, and condenser) are steady-flow devices, and thus all four
processes that make up the Rankine cycle can be analyzed as steady-flow processes. The kine c and
poten al energy changes of the steam are usually small rela ve to the work and heat transfer terms and
are therefore usually neglected.

The steady-flow energy equa on per unit mass of steam reduces to:

𝑄 −𝑄 +𝑊 −𝑊 =ℎ −ℎ
The boiler and the condenser do not involve any work, and the pump and the turbine are
assumed to be isentropic.
Pump work

 𝑊 =ℎ −ℎ

 𝑊 = v (𝑃 − 𝑃 )

 ℎ = ℎ 𝑎𝑡 𝑃

 v = v 𝑎𝑡 𝑃

Boiler

 𝑄 =ℎ −ℎ
Turbine

 𝑊 =ℎ −ℎ
Condenser

 𝑄 =ℎ −ℎ
Thermal efficiency of Rankine cycle

 𝑒= = =

Example 1

Consider a steam power plant opera ng on the simple ideal Rankine cycle. Steam enters the
turbine at 3 MPa and 350°C and is condensed in the condenser at a pressure of 75 kPa. Determine the
thermal efficiency of this cycle.
Given:

Turbine (Process 3-4):

𝑃 = 3 𝑀𝑃𝑎 = 𝑃 (𝑖𝑠𝑜𝑏𝑎𝑟𝑖𝑐)
𝑡 = 350°𝐶
Condenser (Process 4-1):

𝑃 = 75 𝑘𝑃𝑎 = 0.075 𝑀𝑃𝑎 = 𝑃 (𝑖𝑠𝑜𝑏𝑎𝑟𝑖𝑐)

Required:

e=?
𝑊 𝑄 −𝑄 𝑄 ℎ −ℎ
𝑒= = =1− =1−
𝑄 𝑄 𝑄 ℎ −ℎ
(Get all enthalpies)

Solu on:

𝑡 𝑎𝑡 3 𝑀𝑃𝑎 = 233.90°𝐶
Therefore, point 3 is superheated vapor.

ℎ = ℎ 𝑎𝑡 3 𝑀𝑃𝑎 𝑎𝑛𝑑 350°𝐶 from Table 3

𝑘𝐽
ℎ = 3115.3
𝑘𝑔
𝑠 = 𝑠 𝑎𝑡 3 𝑀𝑃𝑎 𝑎𝑛𝑑 350°𝐶 from Table 3
𝑘𝐽
𝑠 = 6.7428 = 𝑠 (𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐)
𝑘𝑔 𝐾

Point 1 is saturated liquid.

ℎ = ℎ 𝑎𝑡 0.075 𝑀𝑃𝑎 from Table 2

𝑘𝐽
ℎ = 384.39
𝑘𝑔
𝑠 = 𝑠 𝑎𝑡 0.075 𝑀𝑃𝑎 from Table 2

𝑘𝐽
𝑠 = 1.2130 = 𝑠 (𝑖𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐)
𝑘𝑔 𝐾
Point 4 is saturated liquid-vapor mixture (wet vapor).

ℎ =ℎ +𝑥 ℎ

From Table 2:
𝑘𝐽
ℎ = 384.39
𝑘𝑔
𝑘𝐽
ℎ = 2278.6
𝑘𝑔
𝑥 can be solved using the entropy values:

𝑠 =𝑠 +𝑥 𝑠

𝑘𝐽
𝑠 = 1.2130
𝑘𝑔 𝐾
𝑘𝐽
𝑠 = 6.2434
𝑘𝑔 𝐾
𝑘𝐽 𝑘𝐽
𝑠 −𝑠 6.7428 − 1.2130
𝑘𝑔 𝐾 𝑘𝑔 𝐾
𝑥 = =
𝑠 𝑘𝐽
6.2434
𝑘𝑔 𝐾
𝑥 = 0.8857033027 𝑜𝑟 88.57%
𝑘𝐽 𝑘𝐽
ℎ =ℎ +𝑥 ℎ = 384.39 + (0.8857033027) 2278.6
𝑘𝑔 𝑘𝑔
𝑘𝐽
ℎ = 2402.553546
𝑘𝑔

Point 2 is a subcooled liquid.

ℎ = ℎ 𝑎𝑡 3 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑠 = 1.2130 from Table 4

t, °C 2.5 MPa 3.0 MPa 5.0 MPa 2.5 MPa 3.0 MPa 5.0 MPa
s, kJ/kg K h, kJ/kg
80 1.0737 1.07336 1.0720 336.86 337.258 338.85
𝑡 1.2130 ℎ
100 1.3050 1.3046 1.3030 420.85 421.224 422.72
By interpola on:

𝑡 = 92.07749524°𝐶 (not required, for reference only)

𝑘𝐽
ℎ = 387.9629483
𝑘𝑔
 𝑊 𝑄 −𝑄 𝑄 ℎ −ℎ
𝑒= = =1− =1−
𝑄 𝑄 𝑄 ℎ −ℎ
𝑘𝐽 𝑘𝐽
2402.553546 − 384.39
𝑘𝑔 𝑘𝑔
𝑒 =1−
𝑘𝐽 𝑘𝐽
3115.3 − 387.9629483
𝑘𝑔 𝑘𝑔
𝒆 = 𝟎. 𝟐𝟔𝟎𝟎 𝒐𝒓 𝟐𝟔%

Other method:

Solve for ℎ using the approximate pump work value:

v = v 𝑎𝑡 0.075 𝑀𝑃𝑎 from Table 2

𝑚
v = 1.0373 𝑥10
𝑘𝑔

𝑊 = v (𝑃 − 𝑃 )

𝑚
= 1.0373 𝑥10 (3000 𝐾𝑃𝑎 − 75 𝑘𝑃𝑎)
𝑘𝑔
𝑚 𝐾𝑁 𝑘𝑁
= 1.0373 𝑥10 3000 − 75
𝑘𝑔 𝑚 𝑚
𝑘𝐽
𝑊 = 3.0341025
𝑘𝑔

𝑊 =ℎ −ℎ

ℎ =𝑊 +ℎ
𝑘𝐽 𝑘𝐽
ℎ = 3.0341025 + 384.39
𝑘𝑔 𝑘𝑔
𝑘𝐽
ℎ = 387.4241025
𝑘𝑔

𝑘𝐽 𝑘𝐽
ℎ −ℎ 2402.553546 − 384.39
𝑘𝑔 𝑘𝑔
𝑒 =1− =1−
ℎ −ℎ 𝑘𝐽 𝑘𝐽
3115.3 − 387.4241025
𝑘𝑔 𝑘𝑔
𝒆 = 𝟎. 𝟐𝟔𝟎𝟏𝟕𝟎𝟑𝟐𝟐𝟑 𝒐𝒓 𝟐𝟔. 𝟎𝟐%
Problems

1. In an ideal Rankine cycle, steam is generated at 4.1 MPa and 480°C. The condenser is at 32°C.
Determine
(a) The ideal pump work
(b) The cycle thermal efficiency

Ans. (a) 4.12 kJ/kg; (b) 38.65%

2. Consider a 210-MW steam power plant that operates on a simple ideal Rankine cycle. Steam
enters the turbine at 10 MPa and 500°C and is cooled in the condenser at a pressure of 10 kPa.
Show the cycle on a T-s diagram with respect to satura on lines, and determine (a) the quality of
the steam at the turbine exit, (b) the thermal efficiency of the cycle, and (c) the mass flow rate of
the steam.
Answers: (a) 0.793, (b) 40.2 percent, (c) 163.47 kg/s
3. Consider a steam power plant that operates on a simple ideal Rankine cycle and has a net power
output of 45 MW. Steam enters the turbine at 7 MPa and 500°C and is cooled in the condenser
at a pressure of 10 kPa by running cooling water from a lake through the tubes of the condenser
at a rate of 2000 kg/s. Show the cycle on a T-s diagram with respect to satura on lines, and
determine (a) the thermal efficiency of the cycle, (b) the mass flow rate of the steam, and
(c) the temperature rise of the cooling water.
Answers: (a) 38.9 percent, (b) 36 kg/s, (c) 8.4°C