Download CHAPTER 2: MASS, ENERGY, AND MOMENTUM BALANCES

CHE 204: TRANSPORT PHENOMENA I
TERM 102
ALSHAMI
CHAPTER 2: MASS, ENERGY, AND MOMENTUM BALANCES
Energy: ability to do work
Work: Force acting through a distance
ܹ =‫߱∗ܨ= ݀∗ܨ‬
Dimensions of Work:
ܹ =‫∗ܯ= ݀∗ܽ∗ܯ= ݀∗ܨ‬
Units of work = Joules, (J = N.m)
‫ܮ‬
‫ܮ‬ଶ
∗
‫ܮ‬
=
‫ܯ‬
ܶଶ
ܶଶ
Power = rate of expanding energy in order to perform work
ܲ‫= )ܲ( ݎ݁ݓ݋‬
ܳ‫ܹ∆ )ݕ݃ݎ݁݊݁( ݇ݎ݋ݓ ݂݋ ݕݐ݅ݐ݊ܽݑ‬
=
= ܹሶ
݅݊‫݁݉݅ݐ ݂݋ ݈ܽݒݎ݁ݐ‬
∆ܶ
Dimensions of Power:
ܹ ‫݀ܨ‬
‫ܮܯ‬ଶ
ܲ=
=
= ‫ = ܸܨ‬ଷ
ܶ
ܶ
ܶ
௃
ே.௠
Units of power = Watt, (1ܹ = ௦ = ௦ )
For flowing stream, the work is done by applying pressure (∆p), therefore:
ܲ=
ܹ ‫݀ܣ݌∆ ݀ܨ‬
=
=
= ∆‫ܳ݌‬ሶ
ܶ
ܶ
ܶ
------------------------------------------------------Energy Equation
dW
Systems
dMin
dMout
dQ
1
CHE 204: TRANSPORT PHENOMENA I
Conservation law:
TERM 102
ALSHAMI
ࢄ࢏࢔ − ࢄ࢕࢛࢚ = ∆ࢄ࢙࢙࢚࢟ࢋ࢓
(1)
X in eqn. (1) is now a ‘quantity of energy’
Think of a very small amount of fluid (ie., differential element) entering and leaving your defined
system. This amount of matter naturally contains (carries) energy with it in the form of Internal,
Potential, and Kinetic energy. In addition to these ‘naturally’ existing amounts of energy, the flow can
also be under an external force ‘pushing or pulling it’ between two points; for example a pump or a
turbine. This force is generally results from applying Pressure (positive or negative) and commonly
termed FLOW work. Applying the conservation of energy principle (per unit time) results with the
following general equation, ie the ENERGY EQUATION (Eqn. 2):
ܲ
‫ݑ‬ଶ
ܲ
‫ݑ‬ଶ
ܲ
‫ݑ‬ଶ
݀‫ܯ‬௜௡ ቆ + ݁ + + ݃‫ݖ‬ቇ − ݀‫ܯ‬௢௨௧ ቆ + ݁ + + ݃‫ݖ‬ቇ + ݀ܳ − ܹ݀ = ݀ ቈ‫ ܯ‬ቆ + ݁ + + ݃‫ݖ‬ቇ቉
ߩ
2
ߩ
2
ߩ
2
௜௡
௢௨௧
(2)
ࡸ૛
Note that the dimensions of each term in the above equation is ࢀ૛ ; which is
Where,
dM_in
dM_out
dQ
dW
‘e’
‘½ u2’
‘gz’
P/ρ
ࡱ࢔ࢋ࢘ࢍ࢟
࢛࢔࢏࢚ ࢓ࢇ࢙࢙
.
Differential amount of MASS entering the system
Differential amount of MASS leaving the system
Differential amount of HEAT added to the system
Differential amount of WORK done by the system
INTERNAL energy
KINETIC energy
POTENTIAL energy
FLOW (INJECTION) work/energy
Now, if the process is in a ‘Steady-State’ conditions (meaning no accumulation/disappearance of matter
within the system); then the Right Hand Side (RHS) of eqn. (2) vanishes. Also, steady-state results with
dMin = dMout = dM. Dividing through by dM, yields:
ܲ
‫ݑ‬ଶ
ܲ
‫ݑ‬ଶ
ቆ + ݁ + + ݃‫ݖ‬ቇ + ‫ = ݍ‬ቆ + ݁ + + ݃‫ݖ‬ቇ + ‫ݓ‬
ߩ
2
ߩ
2
௜௡
௢௨௧
(3)
2
CHE 204: TRANSPORT PHENOMENA I
TERM 102
ALSHAMI
(Not: in the textbook, the author uses CAPITAL letters to denote total quantities and small letters
denotes ‘specific quantities’ (QUANTITY/unit MASS))
Rearrangement of eqn. (3), yields:
ܲ
ܲ
‫ݑ‬ଶ
‫ݑ‬ଶ
ቈ൬ ൰ − ൬ ൰ ቉ + ቈቆ ቇ − ቆ ቇ ቉ + ሾ(݃‫)ݖ‬௢௨௧ − (݃‫)ݖ‬௜௡ ሿ + ሾ(݁)௢௨௧ − (݁)௜௡ ሿ − ‫ = ݍ‬−‫ݓ‬
ߩ ௢௨௧
ߩ ௜௡
2 ௢௨௧
2 ௜௡
(4)
Now, we desire to eliminate the internal energy (e) and replace it with another quantity that is more
common and practical for engineering applications. The most obvious quantity for fluid flows is Ƒ
(Frictional forces) which would result in energy loss and /or heat generation. This new term involves
energy and heat, so the last two terms on the LHS of eqn. (4) can, therefore, be defined as:
Substituting (5) into (4) gives:
Ƒ = ሾ(݁)௢௨௧ − (݁)௜௡ ሿ − ‫ݍ‬
(5)
∆ܲ
‫ݑ‬ଶ
+ ∆ ቆ ቇ + ݃∆‫ ݖ‬+ Ƒ + w = 0
2
ߩ
(6)
Eqn. (6) is the generalized energy equation, commonly referred to as the Bernoulli equation.
Examining this equation reveals that if the inlet pressure is higher than the outlet (as in the case of a
turbine), the change in kinetic energy from inlet to outlet will yield negative value and the potential
energy change will have to be either negative or less than the precious terms in order to have flow, and
finally Ƒ (positive value) which is always minimized as much as possible, then the work (w) will be
positive. In other words, the fluid flow produced work!
The opposite is for a pump: outlet quantities are always higher resulting with positive terms in LHS of
eqn. (6), and consequently negative work (w)…meaning that we have had to provide work in the form
of ‘electricalmechanical’ energy to the system.
Example 2.2: Pumping n-Pentane\
The figure below shows an arrangement for pumping n-pentane (density = 39.3 lbm/cft) at 25 C from
one tank to another, through a vertical distance of 40 ft. All piping is 3-in ID. Assume that the overall
frictional losses in the pipes are given by:
Ƒ = 2.5
uଶ୫
݂‫ ݐ‬ଶ
2.5uଶ୫ ݈ܾ௙ ݂‫ݐ‬
=
‫ݏ‬ଶ
݃௖
݈ܾ௠
3
CHE 204: TRANSPORT PHENOMENA I
TERM 102
ALSHAMI
4
40 ft
1
3
4.5 ft
4 ft
2
(you may ignore the friction in the short length of pipe leading to the pump inlet). The pump and its
motor have a combined efficiency of 75%. If the mean velocity um is 25 ft/s, determine the following:
a) The power required to derive the pump.
b) The pressure at the inlet of the pump, and compare it with 10.3 psia, which is the vapor pressure
of n-pentane at 25 C.
c) The pressure at the pump exit.
Solution
This is a system that involves the transport of MASS and ENERGY. Therefore, the conservation
principles of mass and energy MUST be satisfied.
1st, conservation of mass - continuity equation:
(݉ሶ)௜௡ = (݉ሶ)௜௡
(ߩ‫)ܣݑ‬௜௡ = (ߩ‫)ܣݑ‬௢௨௧
Incompressible fluid (constant density), with no change in velocity and cross-sectional area, then:
3 ଶ
݈ܾ௠
݂‫ ߨ ݐ‬ቀ ቁ
݈ܾ௠
(ߩ‫)ܣݑ‬௜௡ = (ߩ‫)ܣݑ‬௢௨௧ = ݉ሶ = ൬39.3 ଷ ൰ ൬25 ൰ ൮ 12 ݂‫ ݐ‬ଶ ൲ = 48.2
݂‫ݐ‬
‫ݏ‬
4
‫ݏ‬
2nd, conservation of energy – energy equation:
∆ܲ
‫ݑ‬ଶ
+ ∆ ቆ ቇ + ݃∆‫ ݖ‬+ Ƒ + ‫ = ݓ‬0
ߩ
2
4
CHE 204: TRANSPORT PHENOMENA I
TERM 102
ALSHAMI
applying the energy equation correctly requires selecting the INLET and OUTLET locations. To start,
lets pick location (1) as the inlet and (4) as outlet. Both locations are just outside the fluid services.
Having done so, we can decide on the following:
(1) both pressures are atmospheric (ie, zero gauge readings), therefore ∆p = 0.
(2) The surface area at (1) is much much larger than the surface area of the pipe exit at (2), therefore
u1 = 0.
(3) ∆z = 40 ft
(4) Ƒ = (2.5)(252 ft2/s2) = 1562.5 ft2/s2
Substituting into the energy equation, gives:
0+
25ଶ
݂‫ ݐ‬ଶ
− 0ଶ
݂‫ݐ‬
݂‫ ݐ‬ଶ
‫ݏ‬ଶ
+ ൬32.2 ଶ ൰ (40 ݂‫ )ݐ‬+ 1562.5 ଶ + ‫ = ݓ‬0
2
‫ݏ‬
‫ݏ‬
Hence the work (w) per unit mass of the flowing fluid is,
݂‫ ݐ‬ଶ
3,163 ଶ
݂‫ ݐ‬ଶ
‫ = ݏ‬−98.3 ݈ܾ௙ ௙௧
‫ = ݓ‬−3,163 ଶ = −
݈ܾ௠ ௙௧
‫ݏ‬
݈ܾ௠
32.2
݈ܾ௙ ‫ ݏ‬ଶ
The minus sign indicates that the work was supplied to the system in the form of ‘shaft work’ (pump).
The power required to drive the pump motor is:
ܲ = ‫= ݉ݓ‬
ቀ48.2
݈ܾ௙ ௙௧
݈ܾ௠
ቁ ൬98.3
൰
‫ݏ‬
݈ܾ௠
= 6.42 ܹ݇
݈ܾ௙ ݂‫ݐ‬
737.6
‫ܹ݇ ݏ‬
But the pump/motor is only 75% efficient, so the total power required is:
ܲ (‫= )݈ܽݐ݋ݐ‬
ܲ 6.42 ܹ݇
=
= 8.56 ܹ݇
ߟ
0.75
The pressure at the pump inlet is similarly obtained by applying the energy equation between locations
(1) and (2): Again, the velocity at location (1) is negligible, and the pressure is atmospheric (ie., zero
gauge). Solving for P2 yields,
5
CHE 204: TRANSPORT PHENOMENA I
ܲଶ = ߩ ቈ݃(‫ݖ‬ଵ − ‫ݖ‬ଶ ) −
TERM 102
ALSHAMI
‫ݑ‬ଶ
39.3
25ଶ
቉=
ቈ(32.2)(4.5) −
቉ == −1.42 ‫ = ݃݅ݏ݌‬13.28 ‫ܽ݅ݏ݌‬
(32.2)(144)
2
2
Since pressure remains above the fluid vapor pressure (ie., 10.3 psia), it will remain liquid as it enters
the pump. If pressure was to be below the fluid vapor pressure, it will vaporize and the pump will be
damaged due to cavitations.
P3 is found exactly as in finding P2, by applying the energy equation between inlet and outlet (ie,
locations 2 and 3). Here velocity does not change, and viscous losses (dissipation) are neglected:
P3 = 25.2 psig
6