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Unformatted text preview: _,_____...._._—..————~— Problem 13 Taking the :11 gas as “system", we write thefirst iaw for the process (134(2), Q1—2  W10. = U2  U1 which means 0~P1A<V2—y1>=%cV<T2T1)+%Elcv<T2T1)+%cv<T2—T1> 01‘ Pi PIA P113 P10 '\ m m In
mRIZ ”g RTI g“ RT1 ? RTI
— P1A( ‘ — “' = 111% (T2 — T1) Noting that P; 5151A, the above statement can be written as T_2= 1 +1+(P1A/P1B)+(P1A/P1C].3
T1 1+R/cv 3(1+R/ev) c:v
P P
_ﬂ+£(1+i\ AA)
Cp 30p PIB P1c 14 Problem 1.4 The process is one of heating at constant Volume. Let Inf and mg represent the
instantaneous liquid and vapor inventories in the system, 1111: + mg = m, (constant) Fulthennore, the constant—volume constraint reads m f+maV =V, (constant)
fV a g. I The ﬁrst law of thermodynamics requires on a per unit time basis that or, since W = O,
.. _ d _
Q — E (mef + mgug]
dmf duf drng dug
‘ =UfF+me+ug—d{'+mng‘ (3)
‘ A A
% g dug a:
dP dt dP (it The time derivatives dﬂ'lf/dt and ding/dt follow from solving the system of two equations
a ‘ 5;
dt (1) and dt (2) _ The solution is dmf A dmg A 
_ = _ —— — = —— 4 .
t dt ‘ vfg and ‘ dt vfg ( )
Where ' de dvg de dP dVg d? ='_\. _._ _.=s _.s_ __
A $3.01: mgdtt mfdP dt “9% dP dt Combining (3) and (4) we obtain after a fen! manipulations
' dp Q/rn 15 W Problem 1.5 After the seal is broken the atmosPheric air rushes into the glass tube and the
pressure inside the tube becomes atmospheric. If we wait long enough, the temperature inside the
tube becomes equal to the atmospheric temperature also (this scenario is described in Example 1.2
in the text). In that example we learned that en route to thermal equilibrium the bottle air rejects
heat to the ambient. This means that innnediately after we break the seal the air that occupies the
glass tube is warmer than the ambient. If we dip the open end of the tube into the pool of water
immediately after breaking the seal, the water will rise into the tube as the air mass that is trapped insde the glass tube cools down (and Shrinks) to atmOSpheric temperature._'
7 , I , ,
Problem 1.6 (21) Applying the ﬁrst law to the water container as an Open system, we have > d' . . ,
a“; (mu) = (mmin '” (mmout (1)
where
. V I.nin I V ‘—___ Iinout
m = T : constant ——r —> Mass conservation dictates
i (111) ~— m , m — 0
dt " in out ‘
hence min = rhout = m. The firstlaw (1) reads finally % %lt‘: IIn(hin’”hout) _ For an incompressible fluid we also have du = ch and db 2 ch + deP In the present case Pin = Point, therefore hin ‘ hout 3 Cam “ Tout) = C(TZ ‘ T) Equation (1) becomes y—wc~(¥=mc(T2—T) which, integrated from O to t, means 16 ‘ (b) The mass of hot water that raises the contianer water temperature from 10°C to 20°C is W
Problem 1.8 Selected for analysis is the system that contains the two masses (m1, m2). In the
initial state (a) the velocities of the two masses are different (V1, V2), while in the ﬁnal state (b)
mutual friction brings the velocities to the same level (V00). Since there are no forces between the
system and its environment, the total momentum of the ensemble is conserved, m1V1+m2V2=(m1+rng)Vm (1) The initial and final kinetic energy inventories of the ensemble are KEa1 = %— mlvl’z + 712—1112 V22 (2)
KEb = %(m1 + 1112) V062 (3) The evolution of the total kinetic energy during the process (a) — (b) is described by the ”eficiency"
ratio KEb <4) Eliminating V00 between Eqs. (1) and (3), the efﬁciency can be expressed in terms of the initial
mass and velocity ratios ”12/ ml and V2 /' V1, < 1 _ <5) It can be shown analytically that n is less than 1 as soon as V2 is different than V1, for any value of
the ratio tn2 / m1. Two limits of Eq. (5) are worth noting, 1 V2 '
_2 1
m1 +1
1 V2
n = [— e co) (7)
m2 'With the special case n = 1 when VI = V2 for any rug/ml. Equations (5) — (7) show that the order of magnitude of n is 1 when 1312/ m1 is a number of order 1. 1—9 In conclusion, the kinetic energy of the system decreases from state (a) to state (b).
According to the first law of thermodynamics, this decrease is balanced by the other energy interactons and energy changes of the system,
Qa—b ' Wab : Ub _ Ua + KEb " KE'a (8) Where me = 0. If the process is adiabatic, QM) : 0, then the KB decrease is balanced by an
increase in U, 'UbUar—KEaKEb ' ' (9) If the system boundary is diathermal, and (a) and (b) are states of thermal equilibrium with the
ambient temperature reservoir (To), then Qa—b=Ub_Ua+KEb“KEa (10) If m1 and 1112 are two incompressible substances then U = U(T), and at thermal equilibrium (To)
the energy change Uh — Ua is zero, QaszEb'K'Ea<0 (11) M 1—10 ...
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 Spring '08
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