First Law For an Open System

3. First Law For an Open System

In open systems, mass crosses the system boundary (also known as the control surface cs which encloses a control volume cv). In addition to heat and work interactions with the en- vironment, interactions also occur through an exchange of constituent species between the system enclosure and its environment. Consequently the mass contained within the system may change. Examples of open systems include turbines which have a rigid boundary, thereby implying a fixed control volume (as in Figure 9 ) or automobile engine cylinders in which the cs deforms during the various strokes (as illustrated in Figure 10 ) We will initially restrict our analysis to situations for which boundary deformation occurs only in that part of the c.v. in which mass does not enter or exit the system (e.g., the portion H in Figure 10 ).

In general, the system properties are spatially nonuniform within the control volume, e.g., in the turbine illustrated in Figure 9 , T A ≠ T B ≠T C so that internal equilibrium for the

entire system mass cannot be assumed and hence a single property cannot be assigned for the whole control volume. However, the c.v. can be treated as though each elemental volume dV within it is internally in a state of quasiequilibrium, and constitutes a subsystem of the open composite system. The mass contained in any elemental volume (cf. Figure 9 ) is dV/(v(T,P)).

An open system energy conservation equation is equivalent to that for a closed system if the energy content of an appropriate fixed mass in the open system is temporally character- ized using the Lagrangian method of analysis. However, the problem becomes complicated if the matter contains multiple components. It is customary to employ an Eulerian approach that fixes the control volume, and analyzes the mass entering and leaving it. We now formulate the Eulerian mass and energy conservation equations, and illustrate their use by analyzing various flow problems. At the end of the chapter, we will also develop differential forms of these equations that are useful in problems involving fluid mechanics, heat transfer, and chemically reacting flows.

a. Conservation of Mass An elemental mass δm i awaits entry through the inlet port of an open system (such as

the automobile cylinder illustrated in Figure 10 ) at time t. The cs enclosing the open system is marked by the boundary FBHCDE. Another boundary AGFEDCHBA (called the control mass surface c.m.s) includes both the mass within the c.v. and the elemental mass δm i . During an

infinitesimal time period δt (does not necessarily denote an inexact differential), while the mass δm i enters the c.v, another elemental mass δm e exits it. Thus if mass in-flow rate is ˙ m i (say 0.2 kg/s) and if time period is δt (say 2 ms) then mass waiting outside the c.v. is δm i = m ˙ i δt (i.e 100 g). Thus every 2 ms, a slug of 100 g will enter our c.v. We will be concerned with mass and energy conservation equations within δt first. The piston moves simultaneously per-

forming deformation work δW d . As the mass δm i moves into the c.v., the boundary of the c.m.s moves from AG to BF and extends to LK, i.e., the c.m.s moves from AGFEDCHBA to

BFEDLKCHB in such a manner that it contains the same mass at both times t and t + δt. In summary, ˙ m

Quantity

At time t

At time t+ δt

mass in c.v

mc.v,t

m c.v,t+ δt

mass outside c.v

δm i

δm e

m c.v,t+ δt + δm e Since the mass enclosed within the c.m.s does not change during the time t,

mass within c.m.s

m c.v,t + δm i

m c.v,t + δm i =m c.v,t+ δt + δm e .

(30a)

Applying a Taylor series expansion (see Chapter 1) at time t+ δt to the RHS of Eq. (29),

m =m + (dm /dt) δt + (1/2!)(d 2 m / δτ c.v,t+ 2 δt c.v,t c.v c.v ) t ( δt) 2 + …, (30b) where (dm c.v, /dt) t denotes the time rate of change of mass in the c.v. at time t. Substituting Eq.

(30b) in Eq. (30a) δm

δt + (1/2!)(d 2 2 m 2 c.v,t + i =m c.v,t + (dm c.v /dt) t m c.v /dt ) t ( δt) +…+ δm e . Simplifying, and dividing throughout by δt,

(31) xvi. Nonsteady State

δm i / δt= (dm c.v /dt) t + (1/2!)(d 2 m c.v /dt 2 ) t ( δt) + … + δm e / δt.

In the limit δt → 0, the higher order terms in Eq. (31), namely d 2 m c.v /dt 2 and so on vanish. Therefore,

(32) i.e., the rate of mass accumulation within the c.v. equals the difference between the mass flow

dm c.v /dt = ˙ m i – ˙ m e ,

into it and that out of it. In Eq. (32) ˙ m i and ˙ m e , respectively, denote the mass flow rate cross- ing the system boundary at its inlet and its exit, and its LHS the rate of change of mass within

the c.v. The relation is derived for a c.v. containing a single inlet and exit. The mass in the control volume can be evaluated in terms of density and the volume. The density may be spatially nonuniform ( Figure 9 ). Considering an elemental volume dV, and evaluating the elemental mass as ρdV, an expression for m c.v. can be obtained in the form

m c.v. = ∫ cv ρdV , (33) Substituting in Eq. (32),

G δδδδm e , u i ,ke i ,pe i

δδδδm i ,

δδδδm i inside the cv

B δ W shaft

H δ W other

Figure 10 : Mass and energy conservation in an open system at: a: time t; and b: time t+ δt.

(34) xvii. Elemental Form

d( ∫ cv ρdV )/dt = ˙ m i – ˙ m e .

For a small infinitesimal time period t, Eq. (32) may be written in the form

(35) where dm i = ˙ m i d t denotes the elemental mass entering the c.v. during the time period dt, dm e

dm c.v. = dm i – dm e ,

= ˙ m e dt is the mass that exits during that time, and dm c.v. is the elemental mass that accumulates within the c.v. over the same period.

xviii. Steady State Steady state prevails when the system properties and characteristics are temporally

invariant. (Property gradients within the system may exist at steady state, e.g., the spatial non–uniformities in a turbine even though the local property values within the turbine are in- variant over time.) Therefore,

dm c.v, /dt = 0, (36) and Eq. (32) implies that ˙ m i = ˙ m e , since, at steady state, m c.v. = constant. The mass within the

c.v. is time independent in a steady flow open system. Although the steady flow open system exchanges mass with its environment, while the closed

δδδδm i

system does not, both systems contain

constant mass.

xix. Closed System Since mass cannot cross the

system boundary ˙ m i = ˙ m e = 0, and,

Area,A i

hence at steady state, once again Eq.

dx i

(34) applies so that m c.v. = constant. Equation (36) implies that the mass within c.v. is time independent even in

Work= P i A i dX i

a steady flow open system. Note that the steady flow open system ex- changes mass with its environment even though it has constant mass

Figure 11: Illustration of flow work. within c.v., while the closed system

does not allow mass to cross the boundaries.

b. Conservation of Energy The specific energy “e” of a mass of matter δm i entering the inlet port of an open

system during an arbitrary time interval δt is due to its kinetic, potential, and internal energies, ke i , pe i , and u i (e.g., as in the automobile engine illustrated in Figure 10 ). Applying the First Law to the c.m.s,

δQ c.m.s – δW c.m.s = dE c.m.s , (2 ) where δQ c.m.s and δW c.m.s refer to the heat and work transfer across the c.m.s boundary. The

δW c.m. includes electrical work δW elec , shaft work δW shaft , deformation work δW d (=PdV) etc..The energy accumulation within the c.m.s during δt is

dE c.m.s =E c.m.s,t+ δt –E c.m.s,t . (37)

The energy in the system at time t

E t =E c.v,t + δm i e i ,

where E c.v,t denotes the energy within the c.v, δm i e i the energy due to the mass m i , and e i is the specific energy of the inlet mass. The specific energy

e = u + ke + pe. (38) At time t = t + δt, the energy content of the c.m.s is

(39) and the subscript e refers to the exit conditions. Using Eqs. (2), (37), and (39),

E c.m.s,t+ δt =E c.v,t+ δt + δm e e e ,

(40) Expanding E c.v,t+ δt using a Taylor's series,

δQ c.m.s – δW c.m.s =E c.v,t+ δt + δm e e e –E c.v,t – δm i e i .

E =E + (dE /dt) δt +(1/2)(d 2 E 2 c.v,t+ 2 δt c.v,t c.v t c.v /dt ) t ( δt) + …, and using this result in Eq. (40), and simplifying

δQ c.m.s – δW

c.m.s = (dE c.v /dt) t δt +(1/2)(d E c.v /dt ) t ( δt) +…+ δm e e e – δm i e i . (41) In general, the work interaction through the c.m. boundary (analogous to closed sys-

tem work) can involve the shaft work δW shaft , the c.v. boundary deformation work δW d (e.g., at boundary H in Figure 10 ), flow work δW f (a kind of boundary work involving the deformation of c.m. boundary; for example, the boundary AG in Figure 11 is pushed in by applying inlet

pressure P i so that the mass δm i can enter the c.v. within dt, and due to exit pressure P e that pushes out the mass δm e within dt),and other work forms δW other (e.g., electrical work).

(42) xx. Flow Work

δW c.m.s = δW shaft + δW f + δW d + δW other .

The control mass boundary deforms at the inlet and exit due to the mass entering and leaving the c.v. Therefore,

δW f = δW f,i + δW f,e . (43) At the inlet, the boundary AG is pushed towards BF as illustrated in Figure 11 . Work is per-

formed on the control mass by pushing it through a distance dx i during time δt, i.e.,

δW f,i = –P i A i dx i .

The distance dx i =V i δt, where V i denotes the inlet velocity. Since P i A i have positive values, the negative sign is added in order to satisfy the sign convention for work input into the system.

The previous expression may be written as

δW f,i = –P i A i V i δt = –P i m ˙ i v i δt.

(44) Similarly, at the exit the work done by the system in pushing the mass dm e out is

δW f,e = –P e A e V e δt = –P e m ˙ e v e δt.

Therefore,

(45) Further, substituting Eqs. (42), and (45) in Eq. (41) and dividing throughout by δt δQ/δt – δW shaft / δt + (P i m ˙ i v i –P e m ˙ e v e )– δW d / δt – δW other / δt

δW f =P e m ˙ e v e δt – P i m ˙ i v i δt.

c.v = (dE /dt) t δt +(1/2)(d 2 E 2 c.v /dt ) t ( δt) +…+( δm e / δt)e e –( δm i / δt)e i . (46) xxi. Nonsteady State

In the limit δt → 0 in the context of Eq. (46), δm i shrinks to an infinitesimally small volume (see the remarks below), but the ratio δm e / δt is still finite, the boundary AG ap- proaches BF (cf. Figure 11 and Figure 10 ), and the boundary LK approaches DC. Since the

c.m.s is virtually identical with the cs, δQ c.m.s / δt = Q ˙ cv , δW shaft / δt= W ˙ shaft , δW d / δt= W ˙ d , δW other / δt= W ˙ other . Using the expression δW c.v. = δW shaft + δW d + δW other , and the definition w ˙ c.v. = w ˙ shaft + w ˙ d + w ˙ other , Eq. (46) assumes the form Eq. (46) assumes the form

(47) Simplifying this expression

Q ˙ cv – W ˙ cvt = (dE c.v /dt) + ˙ m e e e –˙ m i ei – P i m ˙ i v i +P e m ˙ e v e .

(48) where e T = h + ke + pe is called the methalpy or total enthalpy. The Pv term in the enthalpy is

Q ˙ cv – W ˙ cv = (dE c.v /dt) + ˙ m e e T,e –˙ m i e T,i ,

due to the work flow of matter into and out of the c.v. Equation (48) may be rewritten in the form

(49) the physical meaning of which is as follows: The energy accumulation rate = Energy added

(dE c.v. /dt) = Q ˙ cv – W ˙ cv + ˙ m i e T,i – ˙ m e e T,e ,

through the c.s. by heat transfer – Energy transfer through work interactions + Methalpy addi- tion by advection – Methalpy expulsion through advection. The term E c.v. must be evaluated for the entire open system in which property gradients may exist. For instance, in a steam turbine ( Figure 10 ) as steam is admitted in order to start it, the turbine c.v. is warmed up and E c.v. in- creases over time. However, the temperature and pressure near its inlet are higher than at the exit so that the specific energy varies within the c.v. We can evaluate E c.v .by using the relation

(50) For a nondeformable c.v., such as a turbine, w ˙ d = 0 so that w ˙ c.v. = w ˙ shaft + w ˙ other .

(dE c.v. /dt) = d/dt( ∫ cv ρedV )= Q ˙ cv – W ˙ cv + ˙ m i e T,i – ˙ m e e T,e .

The term w ˙ c.v. does not include the flow work which is already accounted for in the enthalpy term (i.e., h = u + Pv =internal energy + flow work). The elemental mass δm i in Figure 10 is the mass waiting outside control volume that

will subsequently enter the c.v. within the duration δt. As δt → 0, δm i and its volume → 0, and δm e and its volume → 0. In this case the c.m.s. → c.v., and the two bounda- ries merge.

The heat transfer across the c.v. boundary ˙ Q c.v. = δQ/δt ≠ 0 as δt → 0, although δQ c.m. → 0. The term E c.v. = (U + KE + PE) c.v .

The terms dE c.v /dt, ˙ Q , and W ˙ are expressed in similar units and the dot over symbols Q W and ˙,˙, m ˙ is used to indicate heat, work and mass transfer across the c.s., and time

differentials, e.g., dE c.v /dt, indicate accumulation of properties within the c.v.

Equations (49) and (50) can be applied to various cases such as steady (∂/∂t = 0), adiabatic ( ˙ Q c.v = 0), closed systems ( ˙ m i = 0, ˙ m e = 0), and heat exchange devices like boilers ( W ˙ c.v. = 0).

xxii. Elemental Form Upon multiplying Eq. (49) by δt we obtain

dE c.v. = Q ˙ cv dt– W ˙ cv dt+ ˙ m i e T,i dt– ˙ m e e T,e dt, = δQ c.v dt– δW c.v dt+ ˙ m i e T,i dt– ˙ m e e T,e dt.(51) In Eq. (51) dE c.v. denotes the energy accumulation, and δQ c.v. and δW c.v. the heat and work

transfer over a small time period dt. xxiii. Steady State

Open systems, e.g., turbines, compressors, and pumps, often operate at steady state,

i.e., when dE c.v. /dt = 0, dm c.v. /dt = 0, and ˙ m i = ˙ m e = 0. Hence,

(52) xxiv. Rate Form

Q ˙ cv – W ˙ cv + ˙ m i e T,i – ˙ m e e T,e = 0.

Consider the special case of a single inlet and exit with no boundary work. At steady state, properties within the c.v. do not vary over time, although spatial variations may exist. Therefore, Eq. (49) simplifies to the form

Q ˙ cv – W ˙ cv = m ˙ ∆e T , (53) where ∆e T =e T,i –e T,e . xxv. Unit Mass Basis

The unit mass–based equation may be obtained by dividing Eq. (53) by mass, i.e., ˙q cv – ˙ w cv = ∆e T .

(54a) where q c.v. = Q ˙ cv /˙ m ,and w c.v = W ˙ c.v / m ˙ . For an elemental section of a turbine Eq. (54a) can be

written as, δq c.v. – δw c.v. = de T .

If the KE and PE are neglected, δq c.v - δw c.v =dh.

(54b) In a Lagrangian reference frame, a unit mass enters a turbine (as illustrated in Figure 12 ) with

an inlet energy e T,i which decreases due to heat loss to the ambient (i.e., δq c.v. < 0) and work output ( δw c.v. > 0). At the same time the unit mass undergoes deformation due to changes in volume. If one travels with the mass, then

δq c.m. - δw c.m = du c.m. (54c) where

δw Pdv for a reversible process. Since the internal energy change du = dh - P dv - v dP, then the equation (54c) becomes

δw c.m is the work involved within c.m. But

δw c.m =

δq + v dP = dh, where the subscript c.m. has been omitted. Upon comparison with Eq. (54c) with Eq. (54b),

the reversible shaft work

δw c.v,rev = - vdP. xxvi. Elemental Form

Over a small time period δt during which a mass dm both enters and leaves a steady state open system, Eq. (52) yields

δQ c.v. – δW c.v. = dm( ∆e T ). (55) This expression may also be obtained by multiplying Eq. (53) by dm.

xxvii. Closed System Equation (36) is also an expression of closed system mass conservation. The energy

conservation expression of Eq. (50) can be applied to closed systems. Since ˙ m i = ˙ m e = 0, (dE c.v. /dt) = ˙ Q– W ˙ .

(56) The subscript c.v. has been omitted for the closed system. The elemental form of Eq. (56),

namely, δQ – δW = dE, is identical to Eq. (2). xxviii. Remarks

For a nondeformable c.v, such as a turbine, W d = 0 so that W c.v. =W shaft +W other . The term W c.v. does not include the flow work which is accounted for through the enthalpy term (h = u + Pv = internal energy + flow work). The elemental mass δm i in Figure 10

is the mass waiting outside control volume that will subsequently enter the c.v. within the duration δt. As δt → 0, δm i and its volume → 0, and δm e and its volume → 0.

However, δm i / δt ≠ 0, and δm e / δt ≠ 0. In this case the c.m.s. → c.s., and the two sur- faces merge. The heat transfer across the c.v. boundary Q ˙ cv = δQ/δt ≠ 0 as t → 0, although

δQc.m.s → 0. The control volume energy E c.v. = (U + KE + PE) c.v . The dot over symbols Q, W, and m is used to indicate heat, work and mass transfer across the cs and time differentials, e.g., dE c.v /dt, indicate accumulation. Equations (49) and (50) can be applied to various cases such as steady ( ∂/∂t = 0),

adiabatic ( Q ˙ cv = 0), closed systems ( ˙ m i = 0, ˙ m e = 0), and heat exchange devices such as boilers ( W ˙ cv = 0).

xxix. Steady State Steady Flow (SSSF) Steady flow need not necessarily result in steady state, e.g., during the mixing of a hot and cold fluid. Likewise, during intensive steady state, i.e., when the properties are temporally uni- form, a system may not experience steady flow, e.g., as a fluid is drained from a vessel.

j. Example 10 As liquid water flows steadily through an adiabatic valve the pressure decreases from P 1 = 51 bar to P 2 = 1 bar. If the inlet water temperature is 25ºC, what is the exit tem- perature? Assume that the specific volume of water is temperature independent and 3 –1 –1

equal to 0.001 m /kg, and that u = cT, where c = 4.184 kJ kg K . Neglect effects due to the kinetic and potential energies.

Solution Mass conservation implies that dm c.m. /dt = 0. Therefore, ˙ m i = ˙ m e = m ˙ . Furthermore, dE c.v /dt = 0, and Q ˙ cv = W ˙ cv = 0. Applying Eq. (50), m ˙ ∆e T = 0.

Since e T = h + ke + pe, this implies that h 2 =h 1 . A process during which the enthalpy is unchanged (i.e., h 2 =h 1 ) is called a throttling process. Furthermore, since v 2 =v 1 =

v, and u 2 +P 2 v 2 =u 1 +P 1 v 1 (as a consequence of h 2 =h 1 ),

Slug 5

Inlet

Slug 1

qw

cv

Slug 5

Exit

Slug 1

h e ,P e , TV

Figure 12: First Law applied to a steady state open system in a Lagrangian reference frame.

c (T 2 –T 1 ) = –v(P 2 –P 1 ), i.e.,

T –T = (0.001 × (51–1) bar × 100 kPa bar –1 ) × (4.184 kJ kg –1 K –1 2 1 ) = 1.2 K, and T 2 = (298 +1.2) = 299.2 K.

Remarks The temperature increases during the adiabatic throttling of liquids. We saw that when a fluid of fixed mass is compressed adiabatically in a closed system, the bound- ary or deformation work raises the internal energy and, hence, the temperature. If the fluid is incompressible, liquid deformation work is absent and u and T cannot change. On the other hand, if there is some flow inside the closed system, e.g., incompressible water around inside a piston–cylinder assembly, applying the relation δq - δw = du +

d(ke) = 0 - 0 = du + d(ke). If the water slows down due to friction effects at the walls, du increases while d (ke) decreases, i.e., the temperature changes for an incompressi- ble substance occur only due to friction. The internal energy and temperature of a compressible fluid changes during an adiabatic process due to Pdv work during com- pression due to frictional heating. We will now consider a throttling process. Water at the inlet must overcome an inlet pressure over a cross-sectional area A. Consider an elemental mass having dimen- sions of area A and distance L. This mass is pushed into the control volume with a pressure of P. Therefore, the work done by applying a force P A to move along a dis- tance L is equal to P A L. Mathematically, P A L = P V = P v m (which is the inlet flow work). Using the subscript 1 to denote conditions at the inlet, the total energy of

the mass after it enters the inlet (that includes the work required to push it) is u 1 +P 1 v 1 which will increase the c.v. internal energy u c.v.

(see also Example 14). In order to maintain steady state conditions, a similar mass must be pushed out of the exit by flow work equal to P 2 v 2 m. However, P 2 <P 1 , and the outlet flow work is lower than the inlet flow work. Consequently, energy starts accumulating within the control vol- ume heating the water. Therefore, the mass leaving the c.v. must be at a higher tem- (see also Example 14). In order to maintain steady state conditions, a similar mass must be pushed out of the exit by flow work equal to P 2 v 2 m. However, P 2 <P 1 , and the outlet flow work is lower than the inlet flow work. Consequently, energy starts accumulating within the control vol- ume heating the water. Therefore, the mass leaving the c.v. must be at a higher tem-

a sudden expansion process that is Figure 13: Steady flow through a capillary tube. also called a throttling process. In

that case v 2 >v 1 . k. Example 11

A fluid flows through a capillary tube with an inlet velocity V i and exit velocity V e . Apply the mass and energy conservation equations for steady flow and simplify the expression.( Figure 13 )

Solution Mass conservation implies that ˙ m i = ˙ m e = m ˙ . From energy conservation dE c.v /dt = 0,

and W ˙ cv = 0 so that Eq. (50) implies that Q ˙ cv – m ˙ ∆e T = 0, i.e., q c.v – ∆e T = 0.

(A) For an adiabatic system, Eq. (A) assumes the form

∆(h + ke) + ∆pe = 0, or (h + ke) + pe = Constant. (B) The sum (h + ke) is called the stagnation enthalpy and is commonly used in fluid dy-

namics analyses. If u is a function of temperature alone, e.g., as for an ideal gas or in- compressible liquid, ∆u = 0 and ∆h = ∆ (Pv). In this case,

(Pv + ke) + pe = Constant, or P/ ρ+V 2 /2 + gZ = Constant. (C) Remarks

Suppose eq. (C) is applied to an incompressible fluid between inlet and exit. Then

(D) Rewriting this relation, we have

u i +P i v + ke i +pe i =u e +P e v + ke e + pe e .

(E) The term e m = P/ ρ+V 2 /2 + gz is the mechanical portion of the energy. Rewriting Eq.

P i / ρ+V 2 i /2 + gz i –P e / ρ+V 2 e /2 + gz e =u i –u e .

(E),

e mi , − e me . = u i −. u e (F) From Eq. (E),

(G) where P =P+ ρV 2 m /2 + ρ gz denotes the mechanical pressure, or

P mi . − P me , = ( u i − u e ) ρ

(H) where, H m = P/ ρg + V 2 /2g + Z. The difference H m,i -H m,e is the mechanical head loss

H mi , − H me , = ( u i − u e )/ g

H m,L , i.e.,

H m,L =H m,i -H m,e = (u i -u e )/g. If there is no frictional loss for an incompressible fluid, u i =u e, and Eq. (E) yields

(I) Eq. (I) is the Bernoulli energy equation which is well known in the field of fluid me-

Pv + ke + pe or P/ ρ+V 2 /2 + gZ = e m = Constant.

chanics. Therefore,

e m,i =e m,e ;P m,i =P m,e ;H m,i =H m,e . Instead of a capillary tube with constant mass flow, consider natural gas flow through

a pipeline of variable cross section. The velocity distribution across the pipe may be spatially nonuniform, and the density also may vary axially as the flow pro- ceeds. If an imaginary capillary tube of very small cross section is inserted into the pipe, such that velocity across the capillary tube is spatially invariant at both inlet and outlet, the energy change for a non-adiabatic elemental mass flow through it is given by Eq. (B), and Eq. (D) if adiabatic. Such a tube is called a stream tube, and if we imagine ourselves to be situated on top of a unit mass travelling through the stream tube, the energy of the mass is governed by Eqs. (B), (D) or (I). An infinite number of stream tubes can

Figure 14: Boiler with multiple inlets and exits. theoretically be inserted across a cross section, and the total energy change can be calculated.

c. Multiple Inlets and Exits For the mass boiler illustrated in Figure 14 ,

dm c.v /dt = ∑ m ˙ i – ∑ m ˙ e , and

(57a)

(dE c.v. /dt) = Q ˙ cv – W ˙ cv + ∑ m ˙ i e T,i – ∑ m ˙ e e T,e ,

(57b)

d. Nonreacting Multicomponent System xxx. Mass Conservation

For multicomponent nonreacting systems having a single inlet and exit, Eq. (32) may

be written in terms of each component, namely, dm k,c.v /dt = ˙ m ki , – ˙ m ke , .

(58) The mass flow rate of a component is the product of the component molar flow rate multiplied

by its molecular weight, i.e., ˙ m k = N ˙ k M K . The mole balance equation for each species is written in the form

dN k /dt = N ˙ ki , – N ˙ ke , . (59) and the overall mass balance assumes the form

(60) The mass conservation equation will be extended to reacting systems in Chapter 11.

dm c.v /dt = d ∑ m k cv , /dt = – ∑ m ˙ e .

xxxi. Energy Conservation The energy conservation may be written in the molal form as

(dE c.v /dt) = ˙ Q c.v. - W ˙ c.v. + Σ N ˙ ki , ˆe k,T,i - Σ N ˙ ke . e ˆ kTe ,, .

The advection energy for mixtures is defined as ∑ m ˙ i e T,i = Σ N ˙ ki , ˆe k,T,i , and the advection enthalpy m ˙ h for a mixture mh ˙ = ∑ k Nh k ˆ k , )

where e k,T,i and ˆh k denote the energy and enthalpy of the k–the component in the mixture (cf. Chapter 1). If we assume that ˆ e k,T,i = e k,T,i and ˆh k = h k (which is the enthalpy of component k

in its pure form at the same temperature and pressure – discussed in greater detail in Chapter 8),

(dE c.v /dt) = ˙ Q c.v. - W ˙ c.v. + Σ N ˙ ki , e k,T,i - Σ N ˙ ke . e k,T,e .