Transcript HYDRO ELECTRIC POWER PLANT

AGUS HARYANTO
 Introduce
concept of energy and its various
forms.
 Discuss the nature of internal energy.
 Define concept of heat and terminology
associated
 Define concept of work and forms of mechanical
work.
 Define energy conversion efficiencies.
 Discuss relation of energy conversion and
environment.
 BENTUK
ENERGI:
1
2
1. Energi Kinetik (KE)  KE  mV
2
2. Energi Potensial (PE)  PE = mgh
3. Energi dakhil atau Internal Energy (U)
 ENERGI TOTAL:
E = U + KE + PE
e = u + ke + pe
(per satuan massa)
 Thermodynamics
deals only
with the change of the total
energy (E). Thus E of a system
can be assigned to zero (E = 0)
at some reference point.
 The change in total energy (E)
of a system is independent of
the reference point selected.
 For stationary systems, the E
is identical to the change of
internal energy U.
 The
macroscopic forms of energy are those a
system possesses as a whole with respect to
some outside reference frame, such as kinetic
and potential energies.
 The microscopic forms of energy are those
related to the molecular structure of a system
and the degree of the molecular activity, and
they are independent of outside reference
frames. The sum of all the microscopic forms of
energy is called the internal energy of a
system and is denoted by U.
SENSIBLE
and LATENT
energy
CHEMICAL
energy
NUCLEAR
energy
The internal
energy of a
system is the sum
of all forms of the
microscopic
energies.
The portion of the
internal energy of a
system associated with
the kinetic energies of
the molecules is called
the sensible energy
The various forms of
microscopic energies
that make up sensible
energy
 The
internal energy associated with the
phase of a system is called the latent
energy.
 The amount of energy absorbed or
released during a phase-change process
is called the latent heat coefficient.
 At 1 atm, the latent heat coefficient of
water vaporization is 2256.5 kJ/kg.
 The
internal energy associated with the
atomic bonds in a molecule is called
chemical energy.
 The
tremendous amount of energy
associated with the strong bonds within
the nucleus of the atom itself is called
nuclear energy.
Energy crosses
the boundaries in
the form of:
 Heat
 Work
 Mass flow*
*) 1,2 for Clossed System; 1,2,3 for Open System
Heat : the form of
energy that is
transferred
between two
systems (or a
system and its
surroundings) by
virtue of a
temperature
difference
 Mechanics: work
is the energy transfer
associated with a force acting through a
distance (W = F.s).
 Thermodynamics: work is an energy
interaction that is not caused by a
temperature difference between a
system and its surroundings.
+
 Qin
= + (positive)
 Qout = - (negative)
 Win = - (negative)
 Wout = + (positive)
+
1.
2.
3.
4.
Both are recognized at the boundaries of a
system as they cross the boundaries. That is,
both heat and work are boundary phenomena.
Systems possess energy, but not heat or work.
Both are associated with a process, not a state.
Unlike properties, heat or work has no
meaning at a state.
Both are path functions (i.e., their magnitudes
depend on the path followed during a process
as well as the end states), and not point
functions.
 Path
functions have inexact differentials
designated by (Q or W) NOT dQ or dW.
 Properties are point functions (i.e., they depend
on the state only, and not on how a system
reaches that state), and they have exact
differentials designated by d. A small change in
volume, for example, is represented by dV.
 Properties
are
point functions
 Heat
and Work
are path
functions
A candle is burning in
a well-insulated room.
Taking the room (the
air plus the candle) as
the system, determine
(a) if there is any heat
transfer during this
burning process and
(b) if there is any
change in the internal
energy of the system.
(a) The interior surfaces of the room form the system
boundary. As pointed out earlier, heat is recognized
as it crosses the boundaries. Since the room is well
insulated, we have an adiabatic system and no heat
will pass through the boundaries. Therefore, Q = 0
for this process.
(b) The internal energy involves energies that exist in
various forms. During the process just described,
part of the chemical energy is converted to sensible
energy. Since there is no increase or decrease in the
total internal energy of the system, U = 0 for this
process.
A potato initially at
room temperature
(25°C) is being
baked in an oven that
is maintained at
200°C, as shown in
Fig. 2–21. Is there any
heat transfer during
this baking process?
This is not a well-defined problem since the
system is not specified. Let us assume that
we are observing the potato, which will be
our system. Then the skin of the potato can
be viewed as the system boundary. Part of
the energy in the oven will pass through the
skin to the potato. Since the driving force for
this energy transfer is a temperature
difference, this is a heat transfer process.
Note: if the system is the oven, then Q = 0
A
well-insulated electric oven is being heated
through its heating element. If the entire oven,
including the heating element, is taken to be
the system, determine whether this is a heat or
work interaction.
 How if the system is taken as only the air in the
oven without the heating element.
Electrical Work: Wel = V.I.t = I.R.I.t
The energy content of the oven
obviously increases during this
process, as evidenced by a rise
in temperature. This energy
transfer to the oven is not
caused by a temperature
difference between the oven
and the surrounding air. Instead,
it is caused by electrons crossing
the system boundary and thus
doing work. Therefore, this is a
work interaction.
This time, no electrons will
be crossing the system
boundary at any point.
Instead, the energy
generated in the interior of
the heating element will be
transferred to the air
around it as a result of the
temperature difference
between the heating
element and the air in the
oven. Therefore, this is a
heat transfer process.
1.
Kinetical Work
Wk = F.s
Wb = P.A.ds = P.dV
Sebuah tangki kokoh berisi udara pada 500 kPa
dan 150oC. Akibat pertukaran panas dengan
lingkungannya, suhu dan tekanan di dalam tangki
berturut-turut turun menjadi 65oC dan 400 kPa.
Tentukan kerja lapisan batas selama proses ini.
Wb 

2
1
P.dV
Wb = 0 karena
dV = 0
2.
Shaft Work
Wsh = 2..n.
 = torsi = F.r
Daya Poros:
Wsh  2. .RPM.
Determine the power transmitted through
the shaft of a car when the torque applied
is 200 N.m and the shaft rotates at a rate of
4000 revolutions per minute (rpm).
The shaft power is determined directly
from:
Wsh  2. .RPM.
=
= 83.8 kW
(112 HP)
3.
Spring Work
Wsp = 0.5 k (x12 – x22)
k = spring constant
(kN/m)
F = kx
n = normal stress
n = F/A
Wa = 0.5 m.(V22-V12)
Wg = m.g.z
= m.g. h
Consider a 1200-kg car
cruising steadily on a level
road at 90 km/h. Now the
car starts climbing a hill
that is sloped 30° from the
horizontal (Fig. 2–35). If the
velocity of the car is to
remain constant during
climbing, determine the
additional power that must
be delivered by the
engine.
The additional power required is simply the
work that needs to be done per unit time to
raise the elevation of the car, which is equal to
the change in the potential energy of the car
per unit time:
Determine the power required to
accelerate a 900-kg car shown in Fig. 2–36
from rest to a velocity of 80 km/h in 20 s on
a level road.
The conservation of energy principle can be
expressed as follows: The net change (increase or
decrease) in the total energy of the system during a
process is equal to the difference between the total
energy entering and the total energy leaving the
system during that process.
Noting that energy can be
transferred in the forms of
heat, work, and mass, the
energy balance can be
written more explicitly as:
Esystem = Ein - Eout
= Qin - Qout + Win - Wout + Emass,in - Emass,out
 Energy
change = Energy at final state –
Energy at initial state
Esystem = Efinal – Einitial = E2 – E1
E = U + KE + PE
U = U2 – U1 = m(u2 – u1)
KE = KE2 – KE1 = 0.5 m(V22 – V12)
PE = PE2 – PE1 = mg(z2 – z1)
stationary systems, KE = 0 and PE =
0), and E = U
 For
 For
a closed system undergoing a cycle, the
initial and final states are identical, thus
Esystem = E2 – E1 = 0. The energy balance
simplifies to: Ein – Eout = 0 or Ein = Eout
 Noting
that a closed system
does not involve any mass
flow across its boundaries, the
energy balance for a cycle
can be expressed in terms of
heat and work interactions as:
Wnet,out = Qnet,in
A rigid tank contains a hot fluid that is
cooled while being stirred by a paddle
wheel. Initially, the internal energy of the
fluid is 800 kJ. During the cooling process,
the fluid loses 500 kJ of heat, and the
paddle wheel does 100 kJ of work on the
fluid. Determine the final internal energy
of the fluid. Neglect the energy stored in
the paddle wheel.
Assumptions :
 The tank is stationary
and thus the kinetic
and potential energy
changes are zero, KE
= PE = 0. Therefore,
E = U.
 Energy stored in the
paddle wheel is
negligible.
Applying the energy balance on the system gives
the final internal energy of the system is 400 kJ:
A fan that consumes 20
W of electric power
when operating is
claimed to discharge air
from a ventilated room
at a rate of 0.25 kg/s at a
discharge velocity of 8
m/s (Fig. 2–48).
Determine if this claim
is reasonable.
Performance = efficiency, is expressed in
desired output by the required input
A
pump or a fan receives shaft work
(from an electric motor) and transfers it
to the fluid as mechanical energy (less
frictional losses).
 A turbine, converts the mechanical
energy of a fluid to shaft work.
= useful pumping =
power
=
Vol( P2  P1 )
= Power rating x 
 mech, fan
E mech, fluid

W
shaft ,in
1
2


Emech, fluid  m(V2 )
2
is the rate of decrease in the mechanical energy of the
fluid, which is equivalent to the mechanical power
extracted from the fluid by the turbine
Qgeneration  Wel,in Wshaft ,out Wel,in 
Wshaft ,out
motor
1.

Polusi
Merusak lingkunan
dan kesehatan
Smog (asap kota metropolitan) dengan ciri
dark yellow or brown haze in stagnant air
mass and hangs over on calm hot summer
day. Komponen Smog:
 Ground Ozone (O3) : menyebabkan iritasi
mata, merusak paru-paru, dan merusak
jaringan daun tanaman.
 CO : racun mematikan
 VOCs (Benzene, butane, …)
Smog dapat dibawa angin melintasi perbatasan
 persoalan global.
2.
3.



Acid Rain
Fossil fuel contain
sulfur  SOx
SOx and NOx
+ Water + sunlight
 Sulfuric Acid +
Nitric Acid
The acid washed
out by rain water
 Acid Rain
4.
GHG: CO2, CH4, H2O, NOx
 Global Warming