Beirut Arab University Department of Mechanical Engineering MCHE-204 Foundation of Mechanical Engineering Spring 2015-2016
Chapter (4)
1ST Law of Thermodynamics (Open system)
Recall "Closed System"
constant (fixed amount) mass
No mass crosses its boundary Energy (heat and work) cross the boundary
Example
Qin M ass ass = constant
Remarks:
Qin
volume
, leads to output work
Beirut Arab University Department of Mechanical Engineering MCHE-204 Foundation of Mechanical Engineering Spring 2015-2016
Now
Open system, or a control volume (C.V): a properly selected region in space. It usually encloses a device that involves mass flow such as a compressor, turbine, or nozzle.
Mass can cross boundary
Energy can cross boundary
Mass does not have to be fixed
Example hot water water heater cold water
Qin
In this example: Volume = constant, Qin , min, mout
Applications of Open System A water heater A car radiator s Turbines and compressor
involve mass flow
Beirut Arab University Department of Mechanical Engineering MCHE-204 Foundation of Mechanical Engineering Spring 2015-2016
Boundary of an open system The boundaries of a control volume are called a control surface, and they can be real or imaginary. In the case of a nozzle, the inner surface of the nozzle forms the real part of the boundary, and the entrance and exit areas form the imaginary part, since there are no physical surfaces there: real imaginary
nozzle
A control volume can be fixed in size and shape, as in the case of a nozzle, or it may involve a moving boundary:
Most control volumes, however, have fixed boundaries and thus do not involve any moving boundaries. A control volume can also involve heat and work interactions just as a closed system, in addition to mass interaction.
Beirut Arab University Department of Mechanical Engineering MCHE-204 Foundation of Mechanical Engineering Spring 2015-2016
Important terms Steady: no change with time.
The opposite of steady is unsteady , or transient .
Under steady-state conditions, the mass and energy contents of a control volume remain constant.
Uniform: no change with location over a specified region.
Conservation of mass principle Mass is a conserved property that we cannot create or destroy (like energy),
Beirut Arab University Department of Mechanical Engineering MCHE-204 Foundation of Mechanical Engineering Spring 2015-2016
For an open system (control volume) Conservation of mass principle for a control volume:
The net mass transfer to or from a control volume during a time interval Δt is equal to the net change (increase or decrease) in the total mass within the control volume during Δt . mi me mCV
(kg )
me mcv
Where mCV
mi
m final minitial
In the rate form:
m i m e dmCV dt
dmCV
(kg/s)
dt
is the time rate of change of mass within the C.V. boundaries
Mass and Volume Flow Rates
) mass flowing per unit time ( m
Mass flow rate mass
density x volume
m ( ) ( dxA ) A dm dt
dx
A dt
m also
dx
VA
VA V
(1)
Mass flow rate in kg/s
(2)
Volumetric flow rate in m 3/s
From equations (1) & (2),
V m V v
Beirut Arab University Department of Mechanical Engineering MCHE-204 Foundation of Mechanical Engineering Spring 2015-2016
Conservation of Energy Principle
For a closed system W
Q-W = E Closed system
Q
For open system Mass leaving and mass entering contribute to the energy change
Total Energy of an Open System Recall that the total energy of a closed system (non-flowing fluid) is given by:
e u ke pe For a flowing fluid (for open systems): Work Flow (Wflow): It is the work required to push the mass into or out of the control volume.
The work flow per unit mass is given by: w
Pv [kJ/kg]
The total energy of a flowing fluid (for an open system) on a unit-mass basis becomes:
e Pv u Pv ke pe h
h ke pe
(kJ/kg)
where h = specific enthalpy = u + pv
Beirut Arab University Department of Mechanical Engineering MCHE-204 Foundation of Mechanical Engineering Spring 2015-2016
Energy Transport by Mass Amoun t of ener gy transpor t:
The total energy of a flowing fluid of mass m is simply mθ E mass
V 2 m m h gz 2
(kJ)
Rate of energy tr ansport:
When a fluid stream with uniform properties is flowing at a mass flow rate of m , the rate of energy flow with that stream is m
E
mass
V 2 m m h gz 2
(kJ/s
or kW)
Conservation of Mass Principle for an Open System During a steady flow process
_
Recall that we have previously shown that:
m m i
e
mc
v
But for steady flow process , no intensive or extensive properties within the
control volume change with time. Thus, the volume
V ,
the mass m, and the total
energy content E of the control volume remain constant. As a result, the total mass entering the control volume should be equal to the total mass leaving it (since mCV mc
v
=
constant)
0.0
V1A1/v1
m i m e For one inlet & one outlet
iV i Ai eV e Ae
mcv=const V2A2/v2
Beirut Arab University Department of Mechanical Engineering MCHE-204 Foundation of Mechanical Engineering Spring 2015-2016
Conservation of Energy Principle for an Open System During a steady flow process _ Qin Qout W in W out E in E out E CV Q W E in E out E CV
kJ
,
In the rate form:
dE CV
W E E Q in out
kW
,
dt
But for steady flow process, the total mass or energy entering the control
volume should be equal to the total mass or energy leaving it (since mCV = constant and E CV
=
constant). The amount of energy entering a control volume
in all forms (by heat, work, and mass) should be equal to the amount of energy leaving it. dmCV
mCV 0 min mout ;
dt
dE CV
E CV 0 ;
dt
0
0
| and it can easily be shown that
Q W m(h
In the rate form:
out
V 2 2
gz ) m(h
V 2 2
in
2
gz ),
2
m (h V gz ) m (h V gz ), Q W out
2
kJ
2
in
In the case of 1 inlet and 1 outlet: m in
out m
V o2 V i2 ho hi g z o z i Q W m 2
m
kW
kW
Beirut Arab University Department of Mechanical Engineering MCHE-204 Foundation of Mechanical Engineering Spring 2015-2016
Example In rural areas, water is often extracted from underground by pumps. Consider an underground water source whose free surface is 20 m below ground level. Water is to be raised 30 m above the ground by a pump at a rate of 1.5 m 3 /min. If the diameter of the pipe is 15 cm, determine a) the pumping speed b) the power input to the pump required for a steady flow of water. The local atmospheric pressure is 100 kPa and ambient temperature is 25°C. (2)
Given: z
2
V
d
z
1
50m
1.5m3 / min
(1.5 / 60)m3 / s
0.025 m3 / s
0.15m
Assumptions
No heat transfer
No change in pressure and temperature
(1)
between (1) and (2)
The pipe diameter is constant
Solution:
a) The water is in the subcooled phase. υ = A
0.001003 m3/kg (Table A-4)
4
2
d
2
0.15
4
0.0177 m
2
V 0.025 1.412m / s V AV V A
0.0177
Beirut Arab University Department of Mechanical Engineering MCHE-204 Foundation of Mechanical Engineering Spring 2015-2016
b) Energy balance for the CV:
V 22 V 12 h2 h1 g ( z 2 z 1 ) Q W m 2 No heat transfer is involved, Q
0
h1
h2
(same pressure and temperature)
V 1
V 2
(constant pipe area and constant density)
m 0 0 g ( z z ) 0 W 2 1 259.81 50 ) 12.26kW W 1000 12.26 kW W
The pump requires a power input of 12.26 kW in order to raise the water to 30 m above the ground.
Beirut Arab University Department of Mechanical Engineering MCHE-204 Foundation of Mechanical Engineering Spring 2015-2016
Problems set # 4 (1)
A tank of water of density 1000 kg/m3 sits on the roof of a building 85 m high. A 50 cm diameter pipe connects this tank to another tank situated on the ground floor. The lower tank is used for watering the lawns and is normally filled by rain-water. Where there is excessive rainfall, it is necessary to pump the water from the ground floor tank into the roof tank, and a pump is placed in the pipeline to do this. If the water is pumped up at a velocity of 0.6 m/s, what is the power input required at the pump if it has an energy transfer efficiency of 70%. Neglect the pipe frictional losses.
Beirut Arab University Department of Mechanical Engineering MCHE-204 Foundation of Mechanical Engineering Spring 2015-2016
(2)
In rural areas, water is often extracted from underground by pumps.
Consider
underground
water
an source
whose free surface is 60 m below ground level. The water is to be raised 5m above the ground
by
a
pump.
The
diameter of the pipe is 10 cm at the inlet and 15 cm at the exit. Neglecting any heat interaction with
the
frictional
surroundings heating
and
effects,
determine the power input to the pump required for a steady flow of water at a rate of 15 L/s (0.015 m3/s).
(3)
Water is to be pumped from a well to the top of a 200-m-tall building. There is a 15kW pump available in the basement, and the water surface level in the well is 40 m below ground level. Neglecting any heat transfer and frictional effects, determine the maximum flow rate of water that can be maintained by this pump.
Beirut Arab University Department of Mechanical Engineering MCHE-204 Foundation of Mechanical Engineering Spring 2015-2016
(4)
What is the maximum possible power output of the water turbine shown in the figure ?
(5) Steam expands in a turbine as shown below. Determine the power output and the turbine inlet diameter.