Introduction To Enthalpy

Introducing Enthalpy
Enthalpy and Internal Energy

Today’s main concepts:
• Define enthalpy, h
• Recognize that enthalpy and internal energy are both intensive
state properties that can also be determined on the steam
tables.
• Learn to pull values for enthalpy and internal energy from the
steam tables.
• Use steam table values to solve 1st Law balance problems.
3
Last time we learned that if you know two of

the three intensive properties:
Temperature, T
Specific Volume, v
and Pressure, p
...then you could find the other properties at
that state.
Other intensive properties include:

Enthalpy, h
Internal energy, u
and Entropy, s
(which are also found on the steam tables.
h  h(T , v) or u  u ( p, v ) or s  s (T , p)
Introducing Enthalpy 4
Internal Energy (U)

Specific Internal Energy (u) : energy / mass
Kinetic Energy
translational
rotational
vibrational
Potential Energy
electric energy of atoms, molecules, or crystals
Chemical Bonds
All at the microscale….
Enthalpy (H)
Specific Enthalpy (h) : energy / mass
Internal Energy + Work H  U  pV
h  u  pv
5
Internal Energy U
• Invincible microscopic energy on the atomic and molecular scale

• At microscopic scale, the glass is a seething mass of high speed
molecules travelling at hundreds of m/s.
6
Saturated Steam Table
Enthalpy, h
For the time being,

Internal Energy, u we’re going to still ignore
this last quantity.
Notice there are

additional columns on the
steam table that we
ignored last time.
7
What is enthalpy?
Enthalpy is defined in terms of internal energy and the work of
expansion as
Extensive Form Intensive Form
H  U  pV h  u  pv
This combination of terms will show up very frequently,
especially when we begin working with control volumes.
It’s a way to conveniently deal with the internal energy
in combination with the energy associated with
expansion or compression of the substance.
Enthalpy is another formulation of energy and is commonly expressed

in units of kJ or ft-lbf.
8
Example 1: Determine property values

Determine specific volume (v), internal energy (u), and enthalpy (h) of
the following state properties of H20.
a) p = 10 bar T = 179.9 deg C. x = 30%
b) p = 10 bar T = 320 deg. C.
c) p = 10 bar T = 450 deg C.

9

Determine specific volume, internal energy, and enthalpy of the
following state properties of H20.
a) p = 10 bar T = 179.9 deg C. x = 30%
from table A.3: at 10 bar the saturated temperature is 179.9 which

means the substance is a mixture of liquid and vapor.
10
From table A-3 the following values may be directly read:

quality spec. vol. int. energy enthalpy
[m3/kg] [kJ/kg] [kJ/kg]
x = 30% vf=0.001127 uf=761.68 hf=762.81
p=10 bar vg=0.1944 ug=2583.6 hg=2778.1
therefore:
v  v f  x (v g  v f ) v  0.0011  0.3(0.1944  0.0011)
v  0.0591 m3 / kg
u  u f  x(u g  u f ) u  761.7  0.3(2583.6  761.7)
u  1308.3 kJ / kg
h  h f  x(hg  h f ) h  762.8  0.3(2778.1  762.8)
h  1367.4 kJ / kg
11

Determine specific volume, internal energy, and enthalpy of the
following state properties of H20.
b) p = 10 bar T = 320 deg. C.
vapor
therefore:
v = 0.2678 m3/kg u = 2826.1 kJ/kg h = 3093.9 kJ/kg
12

c) p = 10 bar T = 450 deg C.
(v2  v1 )
This also is a gas or superheated v  v1  (T  T1 )
vapor and will need to be found (T2  T1 )
by interpolating from table A-4. (0.3541  0.3257)
 0.3257  (450  440)
(500  440)
 0.3304 m3 / kg
(u2  u1 )
u  u1  (T  T1 )
(T2  T1 )
(3124.4  3023.6)
 3023.6  (450  440)
(500  440)
 3040.4 kJ / kg
(h2  h1 )
h  h1  (T  T1 )
(T2  T1 )
(3478.5  3349.3)
 3349.3  (450  440)
(500  440)
 3370.8 kJ / kg
13
Linear Interpolation:
yunknown  y1 xknown  x1

y2  y1 x2  x1
 y2  y1 
yunknown  y1    ( xknown  x1 )
 x2  x1 
y2
*
yunknown
y1
*
x1 xknown x2
Your turn: Find the value for enthalpy at
T = 400 deg. C and p = 50 bar
at 400°C and p1=40 bar  h1 = 3213.6 kJ/kg

at 400°C and p2=60 bar  h2 = 3177.2 kJ/kg
 h2  h1 
hunknown  h1    ( pknown  p1 ) = 3195.4 kJ/kg
 p2  p1 
Saturation Tables
Example:
A cylinder-piston assembly initially contains water at 3 MPa and
300oC. The water is cooled at constant volume to 200 oC, then
compressed isothermally to a final pressure of 2.5 MPa. Find the
specific volume and enthalpy at each of these three states.
16
State 1:
p1 = 3 MPa= 30 bar and T1 = 300 deg C.
i) Start with superheated vapor table A-4
ii) Interpolate between 280 and 320 deg C.
vb  va
v1  va  (T1  Ta )
Tb  Ta
(0.0850  0.0771)
v1  0.0771  (300  280)
(320  280)
v1  0.0811 m3 / kg
17
State 2:
v2 = v1 = 0.0811 m3/kg and T2 = 200 deg C.
Recognize that this is in the liquid-vapor mixture range.

Refer to Table A-2
i) Locate vg and vf at T = 200 deg C.

v f  1.1565  103 vg  0.1274
v2  0.0811
ii) Determine quality, x2
v  v f  x (v g  v f )
v  vf
x
vg  v f
0.0811  0.0011565
  0.633  63.3%
0.1274  0.0011565
18
State 3:
p3 = 2.5 MPa = 25 bar and T2 = T3= 200 deg C.
Recognize that this is in the compressed liquid.

Refer to Table A-5
i) Locate p3=25 bar and T3=200 deg C
ii) Read the values directly.
v3  1.1555  103 m3 / kg
u3  849 kJ / kg
h3  852.8 kJ / kg
Retrieving u and h Data
Workout Problem
Given Ammonia at 12°C and v = 0.1217 m3/kg
determine P, u and h.
a) Using Tables in Appendix:
b) Using IT:
Workout Problem
Given Ammonia at 12°C and v = 0.1217 m-3/kg
a) Using Tables in Appendix:

Example:
Given Ammonia at 12°C and v=0.1217 m-3/kg
Notice that
vf = 0.001608 and vg = 0.1923
Since: vf < v < vg

the substance is a liquid-gas mixture
Saturation Properties are:
Tsat = 12 deg C. Psat= 658.9 kPa
Quality is found using:

v  vf
v  v f  x  vg  v f   x
vg  v f
Example:
v  vf
x
vg  v f
with
vf = 1.608x10-3,
vg = 0.1923 m3/kg
0.1217  0.001608
x
0.1923  0.001608
x  0.63  63%
Example:
Internal Energy:
u  1  x u f  xug  1  0.63255.16  kJ
kg  0.631346.8  kJ
kg
u  942.9 kJ
kg
Enthalpy:
h  1  x h f  xhg  1  0.63256.2  kJ
kg  0.631473. 6  kJ
kg
h  1023.2 kJ
kg
Check:
h  u  Pv  942.9 kJ
kg  658.9kPa0.1217 m3 kJ
kg 103 Nm
103 N m 2
kPa
Appling the Energy Balance using the Property Tables 24
Example: (3.77) A system consisting of 1 kg or water undergoes a

power cycle composed of the following processes.
Process 1-2 : Constant-pressure heating at 10 bar from saturated vapor.
Process 2-3: Constant-volume cooling to P3 = 5 bar, T3 = 160°C
Process 3-4: Isothermal compression with Q34 = -815.8 kJ
Process 4-1: Constant-volume heating
Sketch the cycle on T-v and P-v diagrams. Neglecting kinetic and
potential energy effects, determine the thermal efficiency.


state 1 2 3 4
P (bar) 10 10 5
T (°C) Tsat=179.9 160 160
v (m3/kg) v1=0.1944 v2= v3=0.3835 v4=v1
u (kJ/kg) 2583.6 3231.8 2575.2

W12   PdV  Pm dv  Pm v2  v1 

V2 v2
V1 v1
105 N m 2 m3
W12  10 bar 1kg0.3835  0.1944 kJ
bar kg 103 Nm
W12  189.1kJ
U12  Q12  W12  Q12  U12  W12  m  u2  u1   W12
Q12  1 kg  3231.8  2583.6  kJkg  189.1kJ  837.3kJ
Q23  U 23  W23  m  u3  u2   0
Q23  1 kg  2575.2  3231.8  kJkg  656.6kJ

W34  Q34  U 34  Q34  m  u3  u4 
v4  v f 0.1944  1.102 10 3

x4   3
 0.6317
vg  v f 0.3071  1.102 10
u4  u f  x4 u g  u f   674.86  0.6317 2568.4  674.86  1871 kJ

kg
W34  Q34  mu3  u4 

 815.8kJ  1kg1871  2575.2 kJ
kg  111.6kJ
Q41  U 41  W41  m  u1  u4   0
Q41  1 kg  2583.6  1871 kJkg  712.6kJ

Q (kJ) W (kJ) Qin  Q12  Q41

Process 1-2 837.3 189.1
 837.3  712.6  1549.9kJ
Process 2-3 -656.6 0
Process 3-4 -815.8 -111.6 Wcycle
Process 4-1 712.6 0

Qin
Total 77.5 77.5
77.5kJ
  0.05  5%
1549.9kJ
Workout Problem: (3.38) A rigid, insulated tank fitted with a paddle

wheel is filled with water, initially a two phase liquid-vapor mixture at
200 kPa, consisting of 0.04 kg of saturated liquid and 0.04 kg of
saturated vapor. The tank contents are stirred by the paddle wheel
until all of the water is saturated vapor at a pressure greater than 138
kPa. Kinetic and potential energy effects are negligible. For the water,
determine
(a) Volume occupied, in m3, Pi = 200 kPa
(b) Initial temperature, in °C, mf = 0.04 kg
(c) Final pressure, in kPa, ml = 0.04 kg
(d) Work, in kJ
mvapor 0.04
x   0.5
mliquid  mvapor 0.04  0.04

(d) Work, in kJ
mvapor 0.04
x   0.5
mliquid  mvapor 0.04  0.04

(d) Work, in kJ xi = 0.5
a) V  mfv f  mgvg
b) T = Tsat

(d) Work, in kJ xi = 0.5
V
c) v
m
Properties: Specific volume & that it is saturated vapor,
Interpolate values to find P2 & u2
d)
U  KE  PE  Q  W  W   U  mu 2  u1 

u1  u f  x  u g  u f 
U  W
34
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Introduction To Enthalpy

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Introduction To Enthalpy

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What are enthalpy and internal energy?

What are enthalpy and internal energy?

How are enthalpy and internal energy related?

How are enthalpy and internal energy related?

Introducing Enthalpy

Enthalpy and Internal Energy

Last time we learned that if you know two of

Other intensive properties include:

Internal Energy (U)

• Invincible microscopic energy on the atomic and molecular scale

Saturated Steam Table

For the time being,

Notice there are

Enthalpy is another formulation of energy and is commonly expressed

Example 1: Determine property values

a) p = 10 bar T = 179.9 deg C. x = 30%

b) p = 10 bar T = 320 deg. C.

c) p = 10 bar T = 450 deg C.

Example 1: Determine property values

a) p = 10 bar T = 179.9 deg C. x = 30%

from table A.3: at 10 bar the saturated temperature is 179.9 which

From table A-3 the following values may be directly read:

Example 1: Determine property values

b) p = 10 bar T = 320 deg. C.

Example 1: Determine property values

at 400°C and p1=40 bar  h1 = 3213.6 kJ/kg

i) Start with superheated vapor table A-4

ii) Interpolate between 280 and 320 deg C.

Recognize that this is in the liquid-vapor mixture range.

i) Locate vg and vf at T = 200 deg C.

Recognize that this is in the compressed liquid.

i) Locate p3=25 bar and T3=200 deg C

ii) Read the values directly.

a) Using Tables in Appendix:

a) Using Tables in Appendix:

Since: vf < v < vg

Quality is found using:

Example: (3.77) A system consisting of 1 kg or water undergoes a

Example: (3.77) A system consisting of 1 kg or water undergoes a

Example: (3.77) A system consisting of 1 kg or water undergoes a

Example: (3.77) A system consisting of 1 kg or water undergoes a

W12   PdV  Pm dv  Pm v2  v1 

Example: (3.77) A system consisting of 1 kg or water undergoes a

v4  v f 0.1944  1.102 10 3

u4  u f  x4 u g  u f   674.86  0.6317 2568.4  674.86  1871 kJ

W34  Q34  mu3  u4 

Example: (3.77) A system consisting of 1 kg or water undergoes a

Q (kJ) W (kJ) Qin  Q12  Q41

Workout Problem: (3.38) A rigid, insulated tank fitted with a paddle

(a) Volume occupied, in m3, Pi = 200 kPa

(a) Volume occupied, in m3, Pi = 200 kPa

(a) Volume occupied, in m3, Pi = 200 kPa

Interpolate values to find P2 & u2

U  KE  PE  Q  W  W   U  mu 2  u1 

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