Thermodynamics
ME-112
Lecture # 18
THERMODYNAMICS ME:112
LECTURE: 14
MASS & ENERGY ANALYSIS
SYSTEMS
OF
OPEN
Dr.-Ing. Usman Allauddin,
Assistant Professor, Mechanical Department,
Director NED-DICE Energy Innovation Centre,
NED University of Engineering & Technology, Karachi.
EMAIL:
usman.allauddin@neduet.edu.pk
TEL:
+92 21 99261261-8 Ext: 2315
MOBILE #:
+92-345-2127526
Office:
NED-DICE Energy Innovation Centre
LECTURE LEARNING OUTCOMES
 Energy analysis of steady flow devices:
 Turbines
 Compressors
 Throttling valves
Prepared by: Dr.-Ing. Usman Allauddin
Sunday, June 7, 2020
TURBINES & COMPRESSORS
 Turbine drives the electric generator in
steam, gas, or hydroelectric power
plants.
 As the fluid passes through the turbine,
work is done against the blades, which
are attached to the shaft. As a result,
the shaft rotates, and the turbine
produces work.
 Compressors, as well as pumps and
fans, are devices used to increase the
pressure of a fluid. Work is supplied to
these devices from an external source
through a rotating shaft.
Prepared by: Dr.-Ing. Usman Allauddin
Sunday, June 7, 2020
TURBINES & COMPRESSORS
Prepared by: Dr.-Ing. Usman Allauddin
Sunday, June 7, 2020
TURBINES & COMPRESSORS
 A fan increases the pressure of a gas slightly and is mainly used to
mobilize a gas.
 A compressor is capable of compressing the gas to very high pressures.
 Pumps work very much like compressors except that they handle liquids
instead of gases.
 Note that turbines produce power output whereas compressors, pumps and
fans require power input.
 Heat transfer from turbines and compressors is usually negligible (𝑸 ≈ 𝟎)
since they are typically well insulated.
 Potential energy changes are negligible for all of these devices (∆𝑷𝑬 ≈ 𝟎).
 The fluid velocities encountered in most turbines are very high and the fluid
experiences a significant change in its kinetic energy. However, this change
is usually very small relative to the change in enthalpy and thus it is often
disregarded (∆𝑲𝑬 ≈ 0).
Prepared by: Dr.-Ing. Usman Allauddin
Sunday, June 7, 2020
5-51 Steam enters an adiabatic turbine at 10 MPa and 500°C and leaves at
10 kPa with a quality of 90 percent. Neglecting the changes in kinetic and
potential energies, determine (a) the mass flow rate required for a power
output of 5 MW and (b) the exit temperature.
Sol.:
Inlet state = Superheated vapor
Exit state = Saturated mixture
(a) 𝑚 = ? (b) 𝑇2 = ?
For 𝒉𝟏 :
From Table A-6 at P1 = 10 MPa & T1 = 500 C
ℎ1 = 3375.1 kJ/kg
For 𝒉𝟐 :
From Table A-5 at P2 = 10 kPa & x2 = 0.9
ℎ𝑓 = 191.81 kJ/kg
ℎ𝑓𝑔 = 2392.1 kJ/kg
ℎ2 = ℎ𝑓 + 𝑥2 ℎ𝑓𝑔 = 2344.7 kJ/kg
For 𝑻𝟐 :
Since phase is changing at the final state, therefore T2 = Tsat @ P2 = 10 kPa
From Table A-5 at P2 = 10 kPa
T2 = 45.81 C
Ans
Prepared by: Dr.-Ing. Usman Allauddin
Sunday, June 7, 2020
For 𝒎:
Since 𝑄 = 0, ∆𝐾𝐸 ≈ 0and ∆𝑃𝐸 ≈ 0
𝑚 ℎ1 = 𝑊𝑜𝑢𝑡 + 𝑚 ℎ2
𝑚 ∗ 3375.1 = 5000 + 𝑚 ∗ 2344.7
𝑚 = 4.852 𝑘𝑔/𝑠
Prepared by: Dr.-Ing. Usman Allauddin
Ans
Sunday, June 7, 2020
5–56 Refrigerant-134a enters an adiabatic compressor as saturated vapor at
24°C and leaves at 0.8 MPa and 60°C. The mass flow rate of the refrigerant
is 1.2 kg/s. Determine (a) the power input to the compressor and (b) the
volume flow rate of the refrigerant at the compressor inlet.
Sol.:
Inlet state = Saturated vapor
Exit state = Superheated vapor
(a) 𝑊𝑖𝑛 = ? (b) 𝑉1 = ?
For 𝒉𝟏 & 𝒗𝟏 :
From Table A-11 at T1 = - 24 C
ℎ1 = ℎ𝑔 = 235.92 kJ/kg
𝑣1 = 𝑣𝑔 = 0.17395 m3/kg
For 𝒉𝟐 :
From Table A-13 at P2 = 0.8 MPa & T2 = 60 C
ℎ2 = 296.81 kJ/kg
For 𝑽𝟏 :
𝑚1 =𝜌1 𝑉1 = 𝑉1 /𝑣1 → 1.2 = 𝑉1 /0.17395
𝑉1 = 0.209 m3/s
Ans
Prepared by: Dr.-Ing. Usman Allauddin
Sunday, June 7, 2020
For 𝒎:
Since 𝑄 = 0, ∆𝐾𝐸 ≈ 0and ∆𝑃𝐸 ≈ 0
𝑊𝑖𝑛 + 𝑚 ℎ1 = 𝑚 ℎ2
𝑊𝑖𝑛 + 1.2 ∗ 235.92 = 1.2 ∗ 296.81
𝑊𝑖𝑛 = 73.06 𝑘𝐽/𝑠
Prepared by: Dr.-Ing. Usman Allauddin
Ans
Sunday, June 7, 2020
TROTTLING VALVES
 Throttling valves are any kind of flow-restricting devices that cause a
significant pressure drop in the fluid.
0
 The pressure drop in the fluid is often accompanied by a large drop in
temperature, and for that reason throttling devices are commonly used
in refrigeration and air-conditioning applications.
Prepared by: Dr.-Ing. Usman Allauddin
Sunday, June 7, 2020
TROTTLING VALVES
 Throttling valves are usually small devices, and the flow through them
may be assumed to be adiabatic (𝑸 ≈ 𝟎) since there is neither sufficient
time nor large enough area for any effective heat transfer to take place.
 Also, there is no work done (𝑾 = 0) and the change in potential energy, if
any, is very small (∆𝑷𝑬 ≈ 𝟎).
 Even though the exit velocity is often considerably higher than the inlet
velocity, in many cases, the increase in kinetic energy is insignificant
(∆𝑲𝑬 ≈ 𝟎). Then the conservation of energy equation for this singlestream steady-flow device reduces to
𝒉𝟏 ≈ 𝒉𝟐
Prepared by: Dr.-Ing. Usman Allauddin
1
 That is, enthalpy values at the inlet and exit of a throttling valve are the
same. For this reason, a throttling valve is sometimes called an
isenthalpic device.
Sunday, June 7, 2020
2
TROTTLING VALVES
Prepared by: Dr.-Ing. Usman Allauddin
Sunday, June 7, 2020
5–66 Refrigerant-134a is throttled from the saturated liquid state at 700 kPa
to a pressure of 160 kPa. Determine the temperature drop during this
process and the final specific volume of the refrigerant.
Sol.:
For T1 & h1:
Since at the inlet the state is saturated liquid, so phase is changing.
Thus, T1 = Tsat at P1 = 700 kPa
From Table A-12 at P1 = 700 kPa
T1 = Tsat = 26.69 C
h1 = hf = 88.82 kJ/kg
For h2:
We have studied earlier that when 𝑄 ≈ 0, 𝑊 = 0, ∆𝑘𝑒 ≈ 0and ∆𝑝𝑒 ≈ 0
the enthalpy remains constant in throttling valves, so:
h2 = h1 = 88.82 kJ/kg
3
For Final State:
From Table A-12 at P2 = 160 kPa
hf = 31.21 kJ/kg
hg = 241.11 kJ/kg
Since hf < h2 < hg → 31.21 < 88.81 < 241.11
Therefore, the final state is saturated mixture
Prepared by: Dr.-Ing. Usman Allauddin
Sunday, June 7, 2020
Sol.:
For T2:
Since at the outlet the state is saturated mixture, so phase is changing.
Thus, T2 = Tsat at P2 = 160 kPa
From Table A-12 at P2 = 160 kPa
T2 = Tsat = -15.60 C
For ∆T:
∆T = T2 – T1 = -15.60 – 26.69 = - 42.3 C
∆T = - 42.3 C
Ans
For x2:
From Table A-12 at P2 = 160 kPa
hf = 31.21 kJ/kg
hg = 241.11 kJ/kg
h2 = hf +x2 hfg → 88.81 = 31.21 + x2 * 241.11
x2 = 0.2745
4
For v2:
From Table A-12 at P2 = 160 kPa
vf = 0.0007437 m3/kg
vg = 0.12348 m3/kg
v2 = vf +x2 vfg → v2 = 0.0007437 + x2 * (0.12348 - 0.0007437)
v2 = 0.0344 m3/kg
Ans
Prepared by: Dr.-Ing. Usman Allauddin
Sunday, June 7, 2020
5–68 A well-insulated valve is used to throttle steam from 8 MPa and 500°C
to 6 MPa. Determine the final temperature of the steam.
Sol.:
For h1:
At the inlet at P1 = 8 MPa & T1 = 500 C the state must be superheated vapor.
From Table A-6 at P1 = 8 MPa & T1 = 500 C
h1 = 3399.5 kJ/kg
For h2:
We have studied earlier that when 𝑄 ≈ 0, 𝑊 = 0, ∆𝑘𝑒 ≈ 0 and ∆𝑝𝑒 ≈ 0
the enthalpy remains constant in throttling valves, so:
h2 = h1 = 3399.5 kJ/kg
5
For Final State:
P2 = 6 MPa is not present in Table A-5, it means the final state is also superheated
vapor.
From Table A-6 at P2 = 6 MPa & h2 = 3399.5 kJ/kg
T2 = 490.1 C
Ans
(you need to do interpolation)
T2
h2
450
3302.9
T2=?
3399.5
500
3423.1
Prepared by: Dr.-Ing. Usman Allauddin
Sunday, June 7, 2020
6
Thank you!!
Prepared by: Dr.-Ing. Usman Allauddin
Sunday, June 7, 2020