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SOLUTION MANUAL CHAPTER 2
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Borgnakke and Sonntag
CONTENT SUBSECTION Concept Problems Properties and Units Force and Energy Specific Volume Pressure Manometers and Barometers Temperature Review problems
PROB NO. 1-18 19-22 23-34 35-40 41-56 57-77 78-83 84-89
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Borgnakke and Sonntag
In-Text Concept Questions
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Borgnakke and Sonntag
2.a Make a control volume around the turbine in the steam power plant in Fig. 1.1 and list the flows of mass and energy that are there. Solution: We see hot high pressure steam flowing in at state 1 from the steam drum through a flow control (not shown). The steam leaves at a lower pressure to the condenser (heat exchanger) at state 2. A rotating shaft gives a rate of energy (power) to the electric generator set.
1
W
T
2
2.b Take a control volume around your kitchen refrigerator and indicate where the components shown in Figure 1.6 are located and show all flows of energy transfer. Solution:
The valve and the cold line, the evaporator, is inside close to the inside wall and usually a small blower distributes cold air from the freezer box to the refrigerator room.
Q leak
.
The black grille in the back or at the bottom is the condenser that gives heat to the room air.
Q
.
W
The compressor sits at the bottom.
cb
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Borgnakke and Sonntag
2.38
One kilogram of diatomic oxygen (O2 molecular weight 32) is contained in a 500L tank. Find the specific volume on both a mass and mole basis (v and v ). Solution: From the definition of the specific volume V 0.5 v = m = 1 = 0.5 m3/kg V V 3 v = = n m/M = M v = 32 × 0.5 = 16 m /kmol
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Borgnakke and Sonntag
2.85
A dam retains a lake 6 m deep. To construct a gate in the dam we need to know the net horizontal force on a 5 m wide and 6 m tall port section that then replaces a 5 m section of the dam. Find the net horizontal force from the water on one side and air on the other side of the port. Solution: Pbot = P0 + ∆P
∆P = ρgh = 997× 9.807× 6 = 58 665 Pa = 58.66 kPa
Neglect ∆P in air Fnet = Fright – Fleft = Pavg A - P0A
Pavg = P0 + 0.5 ∆P
Since a linear pressure variation with depth.
Fnet = (P0 + 0.5 ∆P)A - P0A = 0.5 ∆P A = 0.5 × 58.66 × 5 × 6 = 880 kN
Fleft
Frig h t
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96
Borgnakke and Sonntag
3.83
A cylindrical gas tank 1 m long, inside diameter of 20 cm, is evacuated and then filled with carbon dioxide gas at 20°C. To what pressure should it be charged if there should be 1.2 kg of carbon dioxide? Solution: Assume CO2 is an ideal gas, table A.5: R = 0.1889 kJ/kg K π Vcyl = A × L = 4(0.2)2 × 1 = 0.031416 m3
P V = mRT ⇒P=
=>
mRT P= V
1.2 kg × 0.1889 kJ/kg Κ × (273.15 + 20) K = 2115 kPa 0.031416 m3
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Borgnakke and Sonntag
4.34
A cylinder fitted with a frictionless piston contains 5 kg of superheated refrigerant R-134a vapor at 1000 kPa, 140°C. The setup is cooled at constant pressure until the R-134a reaches a quality of 25%. Calculate the work done in the process. Solution: Constant pressure process boundary work. State properties from Table B.5.2 State 1: v = 0.03150 m3/kg ,
State 2: v = 0.000871 + 0.25 × 0.01956 = 0.00576 m3/kg Interpolated to be at 1000 kPa, numbers at 1017 kPa could have
∫ P dV = P (V2-V1) = mP (v2-v1)
been used in which case: 1W2 =
v = 0.00566 m3/kg
= 5 × 1000 (0.00576 - 0.03150) = -128.7 kJ
P C.P.
T
C.P. P = 1000 kPa
2 1000
1
140
1
39
T
2 cb
v
v
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Borgnakke and Sonntag
4.129
Two springs with same spring constant are installed in a massless piston/cylinder with the outside air at 100 kPa. If the piston is at the bottom, both springs are relaxed and the second spring comes in contact with the piston at V = 2 m3. The cylinder (Fig. P4.129) contains ammonia initially at −2°C, x = 0.13, V = 1 m3, which is then heated until the pressure finally reaches 1200 kPa. At what pressure will the piston touch the second spring? Find the final temperature and the total work done by the ammonia. Solution : P
State 1: P = 399.7 kPa Table B.2.1 v = 0.00156 + 0.13×0.3106 = 0.0419
3
At bottom state 0: 0 m3, 100 kPa
2 1
0
2W3
1W2
P0
V
cb
0
1
State 2: V = 2 m3 and on line 0-1-2 Final state 3: 1200 kPa, on line segment 2.
V3
2
Slope of line 0-1-2: ∆P/ ∆V = (P1 - P0)/∆V = (399.7-100)/1 = 299.7 kPa/ m3 P2 = P1 + (V2 - V1)∆P/∆V = 399.7 + (2-1)×299.7 = 699.4 kPa
State 3: Last line segment has twice the slope. P3 = P2 + (V3 - V2)2∆P/∆V ⇒ V3 = V2+ (P3 - P2)/(2∆P/∆V) V3 = 2 + (1200-699.4)/599.4 = 2.835 m3
v3 = v1V3/V1 = 0.0419×2.835/1 = 0.1188 ⇒ T = 51°C 1
1
1W3 = 1W2 + 2W3 = 2 (P1 + P2)(V2 - V1) + 2 (P3 + P2)(V3 - V2)
= 549.6 + 793.0 = 1342.6 kJ
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Borgnakke and Sonntag
5.87
A 1 kg steel pot contains 1 kg liquid water both at 15oC. It is now put on the stove where it is heated to the boiling point of the water. Neglect any air being heated and find the total amount of energy needed. Solution: Energy Eq.:
U2 − U1= 1Q2 − 1W2
: The steel does not change volume and the change for the liquid is minimal, so 1W2 ≅ 0.
State 2:
T2 = Tsat (1atm) = 100oC
Tbl B.1.1 : u1 = 62.98 kJ/kg,
u2 = 418.91 kJ/kg
Tbl A.3 : Cst = 0.46 kJ/kg K Solve for the heat transfer from the energy equation 1Q2 = U2 − U1 = mst (u2 − u1)st + mH2O (u2 − u1)H2O = mstCst (T2 – T1) + mH2O (u2 − u1)H2O
1Q2 = 1 kg × 0.46 kg K ×(100 – 15) K + 1 kg ×(418.91 – 62.98) kJ/kg
kJ
= 39.1 + 355.93 = 395 kJ
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Borgnakke and Sonntag
6.i
If you compress air the temperature goes up, why? When the hot air, high P flows in long pipes it eventually cools to ambient T. How does that change the flow? As the air is compressed, volume decreases so work is done on a mass element, its energy and hence temperature goes up. If it flows at nearly constant P and cools its density increases (v decreases) so it slows down . for same mass flow rate ( m = ρAV ) and flow area.
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Borgnakke and Sonntag
8.3
CV A is the mass inside a piston/cylinder, CV B is that plus part of the wall out to a source of 1Q2 at Ts. Write the entropy equation for the two control volumes assuming no change of state of the piston mass or walls.
Po mp mA
Ts
Fig. P8.3 The general entropy equation for a control mass is Eq.8.37 2 dQ S2 – S1 = ⌠ T + 1S2 gen ⌡1 The left hand side is storage so that depends of what is inside the C.V. and the integral is summing the dQ/T that crosses the control volume surface while the process proceeds from 1 to 2. 2 dQ C.V. A: mA (s2 – s1) = ⌠ T + 1S2 gen CV A ⌡ A 1 2 ⌠ dQ
mA (s2 – s1) = T + 1S2 gen CV B ⌡1 s In the first equation the temperature is that of mass mA which possibly changes from 1 to 2 whereas in the second equation it is the reservoir temperature Ts. The two entropy generation terms are also different the second one includes the first one plus any s generated in the walls that separate the mass mA from the reservoir and there is a Q over a finite temperature difference. When the storage effect in the walls are neglected the left hand sides of the two equations are equal. C.V. B:
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Borgnakke and Sonntag
8.98
Argon in a light bulb is at 90 kPa and 20oC when it is turned on and electric input now heats it to 60oC. Find the entropy increase of the argon gas. Solution: C.V. Argon gas. Neglect any heat transfer. Energy Eq.5.11: m(u2 - u1) = 1W2 electrical in Entropy Eq.8.37: s2 - s1 = ∫ dq/T + 1s2 gen = 1s2 gen Process: v = constant and ideal gas =>
P2/ P1 = T2/T1
Evaluate changes in s from Eq.8.16 or 8.17 s2 - s1 = Cp ln (T2/T1) – R ln (P2/ P1) = Cp ln (T2/T1) – R ln (T2/ T1) = Cv ln(T2/T1)
Eq.8.16 Eq.8.17
60 + 273 = 0.312 kJ/kg-K × ln [ 20 + 273 ] = 0.04 kJ/kg K
cb
Since there was no heat transfer but work input all the change in s is generated by the process (irreversible conversion of W to internal energy)
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Borgnakke and Sonntag
9.89 A condenser in a power plant receives 5 kg/s steam at 15 kPa, quality 90% and rejects the heat to cooling water with an average temperature of 17°C. Find the power given to the cooling water in this constant pressure process and the total rate of enropy generation when condenser exit is saturated liquid. Solution: C.V. Condenser. Steady state with no shaft work term. Energy Eq.6.12:
. . . m hi + Q = mhe
Entropy Eq.9.8:
. . . . m si + Q/T + Sgen = m se
Properties are from Table B.1.2
hi = 225.91 + 0.9 × 2373.14 = 2361.74 kJ/kg ,
he= 225.91 kJ/kg
si = 0.7548 + 0.9 × 7.2536 = 7.283 kJ/kg K, se = 0.7548 kJ/kg K . . . Qout = –Q = m (hi – he) = 5(2361.74 – 225.91) = 10679 kW . . . Sgen = m (se – si) + Qout/T = 5(0.7548 – 7.283) + 10679/(273 + 17) = –32.641 + 36.824 = 4.183 kW/K Often the cooling media flows inside a long pipe carrying the energy away. cb
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Borgnakke and Sonntag
Closed Feedwater Heaters
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Borgnakke and Sonntag
14.64
Solve the previous Problem with Table B.3 values and find the compressibility of the carbon dioxide at that state. B.3: v = 0.05236 m3/kg, u = 327.27 kJ/kg,
A5: R = 0.1889 kJ/kg-K
Pv 1000 × 0.05236 Z = RT = 0.1889 × 293.15 = 0.9455
close to ideal gas
To get u* let us look at the lowest pressure 400 kPa, 20oC: v = 0.13551 m3/kg and u = 331.57 kJ/kg. Z = Pv/RT = 400 × 0.13551/(0.1889 × 293.15) = 0.97883 It is not very close to ideal gas but this is the lowest P in the printed table. u − u* = 327.27 – 331.57 = −4.3 kJ/kg
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Borgnakke and Sonntag
16.26
Find K for: CO2 ⇔ CO + 1/2O2 at 3000 K using A.11 The elementary reaction in A.11 is : 2CO2 ⇔ 2CO + O2 so the wanted reaction is (1/2) times that so 1/2
K = KA.11 = exp(-2.217) = 0.108935 = 0.33 or ln K = 0.5 ln KA.11 = 0.5 (−2.217) = −1.1085 K = exp(−1.1085) = 0.33
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