Unformatted text preview: Chapter 2 Practice Problems
(P2.1) Energy Balance: …………………………………………………………Eqn. 2.27
. d u2
gz in = ∑ H +
m − ∑ H +
m out + Q + W EC + W S dt 2 g C g C inlets 2gC gC 2g C gC outlets &
closed system, no mass flow ⇒ m in = m out = 0 ,
valve stuck preventing any steam from going out ⇒ m out = 0 , no change in volume
. occurs (no expansion or contraction) ⇒ W EC = 0 , no work has been added
. or removed, and there is no pump nor turbine ⇒ W S = 0 .
Ignore K.E. and P.E.
⇒ the final energy balance becomes.
(mU ) = Q
dt Before the valve is stuck, ⇒ there is steam escaping and therefore,
there is an enthalpy for the vapor leaving, no expansion / contraction
work, no shaft work, no kinetic or potential energy occurred. &
⇒ energy balance will be .
(mU ) = −m out H out + Q , note m = −m out
dt (P2.3) (Solution for gas furnace). Assume the house is air-tight. Furnace has been heating the
home steadily, ⇒ the system (furnace) is a steady-state open system, ⇒
gz = 0 ∴ m in = m out ,
+ dt 2 g C g C Furnace is fixed size, no expansion/contraction work occurs. Consider just the hot side of
the heat exchanger.
and air . &
∴ m out H out − min H in = Q Furnace &
House Combustion gases
(Chap 14 will consider reactions such as
combustion, but the E-balance doesn’t need to be
Note: if the small shaft work from blower is included, then the boundary includes the
house air in and out of the furnace, and the boundary includes both sides of the heat To accompany Introductory Chemical Engineering Thermodynamics
© J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/31/2001) 1 of 8 Chapter 2 Practice Problems
exchanger. For this boundary, all heat transfer is within the furnace, and there is no Q
term, but there are four mass flow terms (fuel/air in, combustion gas out, house air in,
house air return) and a small (negligible compared to other terms) Ws term.
(P2.4) (Solution for gas furnace.) Assume house is air-tight, m in = m out = 0 .
System: the house and all contents, consider just the house side of heat exchanger ⇒
closed, unsteady-state system.
⇒ there is heat gain and heat lost in the house.
No expansion / contraction work nor kinetic, potential energy change.
The internal energy of the system could change with respect to time depending on heat
gain and heat loss, or could be zero if they are balanced. If the furnace is properly sized,
there will be a net gain in temperature of the house! Fuel
and air Furnace &
Q loss &
House Combustion gases .
(mU ) = Q gain + Q loss , where Q is negative.
loss Note: if blower is included within system, then you also should be clear whether the
boundary includes the entire heat exchanger (in which case the Qgain term is replaced with
enthalpy flow terms), or just the house side of the heat exchanger. In either event, the Ws
term will be small relative to other terms since there is so much heat coming from
(P2.6) System: the bulb and its contents.
(∀) = 0 …. ∀ is left side of equation.
There is no work acting on the bulb and no mass flow in or out.
⇒ ∆U = 0
System is closed steady state system, ⇒ Photons &
Q loss &
Consider photons as heat transfer by radiation Q photons and heat loss from bulb, Q bulb ,
Q net = 0 To accompany Introductory Chemical Engineering Thermodynamics
© J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/31/2001) 2 of 8 Chapter 2 Practice Problems
(P2.7) System: the sunbather at 12:00 noon (open system).
No work has been done, no kinetic or potential energies.
No mass flow in, but perspiration loss, mbody = − m out Q photons &
m out Unsteady state, still getting hot.
Consider photons to be heat transfer by radiation.
⇒ by using the equation 2.64 (text book) and drop out all the zero values,
⇒ energy balance will be: d
( mbodyU body ) = mbody H V + Q
dt (steam tables could be used to estimate the enthalpy of the evaporating perspiration if the
T is known or estimated).
(P2.8) System: the balloon and its contents.
Unsteady state system, open system.
No heat gain or loss.
No shaft work.
Mass changing with respect to time, as it gets flat, m = − m out
Volume of the balloon is changing the work of expansion / contraction occurs.
Kinetic energy effected by changing the velocity of the balloon.
No potential energy change.
2 mu balloon u 2 dm &
⇒ energy balance is:
+ W EC
mu + = +H + dt 2g C 2 g C dt (P2.9) (a) Potential energy (P.E),
⇒ ∆PΕ = m*g*∆h, ( 1J = 1Nm )
where m = mass,
g = gravity,
h = height or distance that the body moved by.
⇒ ∆P.E = mg∆h / g c , ⇒ ∆h =
mg / g c 1kg * 9.8066 N / kg ⇒ h = 101.97m This is why potential energy is often ignored when thermal changes are present– the
height change must be very large to be significant.
(b) Kinetic energy = potential energy
1 1 2 g c K .E 2 2 * 1(kgm / Ns 2 ) * 1000 Nm 2 ⇒ K .E =
mu , ⇒ u = = 2gC
m 1/ 2 ∴ u = 44.72m / s This is why kinetic energy is often ignored when thermal changes are present– the
velocity change must be very large to be significant. To accompany Introductory Chemical Engineering Thermodynamics
© J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/31/2001) 3 of 8 Chapter 2 Practice Problems
(P2.10)Closed system, unsteady-state, 1 block, 2 water, 3 tank.
Energy balance ∆U=0
m1* ∆U1+ m2* ∆U2 + m3* ∆U2 = 0
(∆U = T2 ∫ T1 1 CvdT ), ∆U = Cv∆T Q=0 2 WEC = 0 3 (T – 400)*200*0.380 + 4.184*4000*(T-300)+500*0.380*(T-300) = 0
Solve for T, ⇒ T = 300.447K
∆UBLOCK = 200*0.38*(400-300.447) = -7566J
∆UWATER = 4000*4.184*(300.447-300) = 7480.9J (P2.11) Unsteady-state, closed system, Energy balance with block at 50m deep in the water.
m1* ∆U1+ m2* ∆U2 + m3* ∆U2 +m4(kg)*∆z(m)*g/gc(N/kg) = 0 (1J=1kgm2/s2) 200*.38*(T-400)-0.2*50*9.81+4000*4.184*(T-300)+500*.38*(T-300) = 0 ⇒
T = 300.453 ⇒ ∆UBLOCK = -7566 (P2.12) a. Unsteady-state closed system WEC = 0(rigid ) ∆U = Q ∆Ugas = nCv∆T=5mole*200*5cal/mol-K =5000cal.;
with gas AND vessel
∆U = n gas Cv gas ∆T + mvessel Cv vessel ∆T =
5000 + mvessel Cv∆T = 5000 + 80(0.125)200 = 7000 cal
b. ∆U = Q + W EC ⇒ ∆U − W EC = Q ⇒ ∆ H = Q = mCp∆T
5mole*200*7cal/mol-K = 7000cal (P2.13) Unsteady state, closed system.
Assume Ideal Gas. ⇒ Cp = Cv + R, and Cp = 7R/2 (diatomic)
⇒7R/2 – R = Cv;
Q = ∆U + W ⇒ Cv = 5R/2 = 5*8.314/2 = 20.785J/g-k
( no work ) ⇒ Q = ∆U = 20.785* (400 – 300)
⇒ Q = 2079J/mol
To accompany Introductory Chemical Engineering Thermodynamics
© J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/31/2001) 4 of 8 Chapter 2 Practice Problems (P2.14) Steady state, open system.
660oF = 348.88oC T(oC) = (T(oF) – 32)*5/9 &
Through the valve ⇒ 0 = m in H in − m out H out ⇒ H in = H out
From steam tables (back of the book).
Interpolate for 348.88oC to find Hin
⇒ 350 − 348.88
2563.64 − H
350 − 345
2563.64 − 2594.9 ⇒ Hin = 2570.64 J/g Outlet at 1 atm. Use sat T table to find P=1.014 MPa (close enough to 1 atm). Hout is less
than HsatV= 2676 J/g, so the outlet is two-phase.
∆Hvap = 2256 J/g and find HsatL = 419.2 J/g
H = q∆H vap + H SatL
⇒ 2570.64 = q*(2256) + 419.2
⇒ q = 0.954
(P2.16) Steady state, open system.
⇒ 0 = m in H in − m out H out + Q
a) P1 = 15 MPa , T1 = 600oC
P2 = 10MPa , T2 = 700oC ⇒ H1 = 3583.1 kJ/kg (Steam Table)
⇒ H2 = 3870 kJ/kg (Steam Table) ∆H = Heat = Q = H2 – H1 = -287 kJ/kg
b) for H ig take lowest P at 600oC and 0.01MPa
H = 3706 J/g ( at 600 C and 0.1MPa) H in − H ig
(3583.1J / g − 3706 J / g ) * 18 g / mole
= - 0.305
8.314 J / mole − K * (600 + 273.15) K
NOTE: (answer typed wrong in the book printings 1-3) To accompany Introductory Chemical Engineering Thermodynamics
© J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/31/2001) 5 of 8 Chapter 2 Practice Problems
(P2.17) There are several ways to look at this problem, and some choices of system boundries are
more difficult than others. One method is presented.
(a). System: piston, donote with subscript p
0.1m 2 area closed system,
no Q (neglect)
no flow (rigid body)
no change of internal energy (piston doesn’t change T) 0.1m3
Pi = 10bar g
∆Z + m p
= W S,p
surface forces cause movement; treat this as Ws. mp Now we need to determine Ws. The work done by the gas (subscript g) and the
atmosphere (subscript a) will be equal to the work done on the piston. The gas and
atmosphere are closed systems that change size.
W EC , g = − ∫ Pg d V g ..............................................................................(2)
Note : dV g = − d V a
W EC , a = − ∫ Pa d V a = ∫ Pa d V g ............................................................(3)
Combining work interaction at boundary
W s , p = −W EC , g − W EC ,a Which results in the working equation.
∆Z + m p
= W S , p = ∫ Pg d V g − ∫ Pa d V g ...........................(4)
2gC W S , p = n g ∫ ( RT / V )dV − ∫ Pa d V
WS,p = (ngRT1* ln V2
- 0.1*(0.25 - 0.1)) = (P1V1* ln
0.1 WS,p = (1.0*0.1*ln2.5-.015)*1E6cm3/m3 = 76629J
From E. Bal.
mu2/2gc = WS,p - mg∆z/gc = 76629J - 700*9.81*(2.5m-1m) = 66328.6J....(6)
u = 13.76m/s To accompany Introductory Chemical Engineering Thermodynamics
© J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/31/2001) 6 of 8 Chapter 2 Practice Problems
Free flight, system: Piston, closed system, constant T. (b) Equation (1) applies where WS,p = 0 (no surface forces acting on piston, because we will
ignore air resistance). The kinetic energy will go to zero at the top of flight. The initial
kinetic energy from part (a).
∆z + m p
= 700(9.81)∆z − 66328.6 J = 0.....................................(7)
∆ z = 9 .6 m
(not counting 2.5 meters of tube).
(d) PgVg = constant, since isothermal, P2 = P1V1/V2 = 4 bar
= 5.88, ⇒ V 2 = 0.588
plugging into (5)
WS,p = (1.0*0.1*ln5.88-0.1*(0.588-0.1))*1E6 cm3/m3
WS,P = 128350 J
mu2/2gc = 128350 – 700*9.81(5.88 –1) = 94840 J
plugging into (7)
⇒ 700 * 9.81 * ∆z = 94840 J ,
⇒ ∆z = 13.8m, free, flight (e) the piston will accelerate when the upword force (Fup) > down ward force (Fdown). The
piston will decelerate when Fup< Fdown. The maximum exit velocity will be obtained if the
piston leaves the cylinder at the condition Fup = Fdown.
Fup = P2 A
Fdown = Pa A +
P2 A = Pa A + mg
gC P2 can be related to V2 since gas is isothermal, PVg = constant,
Substituting (9) into (8) , and inserting the values. P2 = (1E 6 Pa)(V1 V2 )(0.1m 2 ) = (0.1E 6 Pa)(0.1m 2 ) + 700 * 9.81
To accompany Introductory Chemical Engineering Thermodynamics
© J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/31/2001) 7 of 8 Chapter 2 Practice Problems
= 5.93, ⇒ V 2 = 0.593m 3
Answer will be almost the same as (d)
Plugging into (5)
WS,p = (1.0*0.1*ln5.93-0.1*(0.593-0.1))*1E6 cm3/m3
WS,p = 128700 J
mu2/2gc = 128700 – 700*9.81(5.93 –1) = 94846 J
sub. Into (7)
⇒ 700 * 9.81 * ∆z = 94846 J ,
⇒ ∆z = 13.8m, free, flight
V1 Initial and final conditions , P1 = 50 psig = 64.7 psia = 0.4462MPa.
ball: 3in*2.54cm/in=7.62 cm dia
cylinder volume: V2=100cm*3.14*(7.62/2)2 = 4558.1cm3
ball mass: 0.125lbm*.454kg/lbm=.057kg ;
exit velocity: 40mi/hr *5280ft/mi*0.3048m/ft*1hr/3600sec = 17.88 m/s
see E. Bal. From (P2.17), except potential energy is not included since piston is
horizontal, ball replaces the piston of (P2.17).
adapting (4), (5) from (P.2.17)
WS,ball = 0.057/2*(17.88)2 = 9.11 J
m∆u2/2gc =WS,ball = 0.4462*V1*ln(V2/V1) - 0.1*(4558.1-V1) = 9.11J
Using solver⇒ V1 = 387cm3 not counting Vball. To accompany Introductory Chemical Engineering Thermodynamics
© J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/31/2001) 8 of 8 ...
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