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Exercícios resolvidos do livro Van Wylen, G. e Sonntag, R.E. Fundamentos da Termodinâmica Clássica - Ed. Blücher
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Sixth Edition
SONNTAG • BORGNAKKE • VAN WYLEN
Correspondence table Concept-Study Guide Problems 1- Properties and Units 23- Force and Energy 27- Specific Volume 38- Pressure 44- Manometers and Barometers 58- Temperature 77- Review Problems 81-
th
The correspondence between the problem set in this sixth edition versus the problem set in the 5'th edition text. Problems that are new are marked new and those that are only slightly altered are marked as modified (mod).
Study guide problems 2.1-2.22 and 2.23-2.26 are all new problems.
New 5 th^ Ed. New 5 th^ Ed. New 5 th^ Ed. 27 1 47 new 67 24 28 new 48 16 68 new 29 2 49 17 69 new 30 new 50 new 70 23 31 3 51 new 71 new 32 new 52 19 72 30 33 5 53 new 73 32 34 6 54 34 74 33 35 7 55 29 75 new 36 9 56 new 76 37 37 10 57 28 mod 77 27 38 12 58 new 78 new 39 new 59 20 79 38 40 new 60 26 80 new 41 new 61 new 81 31 42 11 62 21 82 new 43 13 63 new 83 22 44 new 64 new 84 35 45 18 65 15 85 36 46 14 66 new 86 new
English Unit Problems New 5 th^ Ed. SI New 5 th^ Ed. SI 87 new - 97 43E 43 88 new 11 98 new 50 89 new 12 99 new 53 90 new 19 100 45E 70 91 new 20 101 46E 45 92 new 24 102 new 82 93 39E 33 103 48E 55 94 40E - 104 new 80 95 new 47 105 47E 77 96 42E 42 Design and Open ended problems 106-116 are from 5th^ edition problems 2.50-
Make a control volume around the whole power plant in Figure 1.2 and with the help of Fig. 1.1 list what flows of mass and energy are in or out and any storage of energy. Make sure you know what is inside and what is outside your chosen C.V.
Solution:
Smoke stack
Boiler building
Coal conveyor system
Dock
Turbine house
Storage gypsum Coal storage flue gas
cb
Underground
power cable
Welectrical
Hot water
District heating
m
Coal
m
m
Flue gas
Storage for later
Gypsum, fly ash, slag
transport out: Cold return m
m
Combustion air
Make a control volume that includes the steam flow around in the main turbine loop in the nuclear propulsion system in Fig.1.3. Identify mass flows (hot or cold) and energy transfers that enter or leave the C.V.
Solution:
Welectrical
1
2
WT
1
3
Electric power gen.
(^5 )
6 7 Cooling by seawater
Condensate to steam gen. cold
Hot steam from generator
cb
The electrical power also leaves the C.V. to be used for lights, instruments and to charge the batteries.
An electric dip heater is put into a cup of water and heats it from 20 o^ C to 80 o^ C. Show the energy flow(s) and storage and explain what changes.
Solution:
Electric power is converted in the heater element (an electric resistor) so it becomes hot and gives energy by heat transfer to the water. The water heats up and thus stores energy and as it is warmer than the cup material it heats the cup which also stores some energy. The cup being warmer than the air gives a smaller amount of energy (a rate) to the air as a heat loss.
Welectric
Q (^) loss
Separate the list P, F, V, v, ρ, T, a, m, L, t and V into intensive, extensive and non- properties.
Solution:
Intensive properties are independent upon mass: P, v, ρ, T Extensive properties scales with mass: V, m Non-properties : F, a, L, t, V
Comment: You could claim that acceleration a and velocity V are physical properties for the dynamic motion of the mass, but not thermal properties.
Water in nature exist in different phases like solid, liquid and vapor (gas). Indicate the relative magnitude of density and specific volume for the three phases.
Solution:
Values are indicated in Figure 2.7 as density for common substances. More accurate values are found in Tables A.3, A.4 and A.
Water as solid (ice) has density of around 900 kg/m^3 Water as liquid has density of around 1000 kg/m^3 Water as vapor has density of around 1 kg/m^3 (sensitive to P and T)
Is density a unique measure of mass distribution in a volume? Does it vary? If so, on what kind of scale (distance)?
Solution:
Density is an average of mass per unit volume and we sense if it is not evenly distributed by holding a mass that is more heavy in one side than the other. Through the volume of the same substance (say air in a room) density varies only little from one location to another on scales of meter, cm or mm. If the volume you look at has different substances (air and the furniture in the room) then it can change abruptly as you look at a small volume of air next to a volume of hardwood.
Finally if we look at very small scales on the order of the size of atoms the density can vary infinitely, since the mass (electrons, neutrons and positrons) occupy very little volume relative to all the empty space between them.
How much mass is there approximately in 1 L of mercury (Hg)? Atmospheric air?
Solution:
A volume of 1 L equals 0.001 m^3 , see Table A.1. From Figure 2.7 the density is in the range of 10 000 kg/m^3 so we get
m = ρV = 10 000 kg/m^3 × 0.001 m^3 = 10 kg A more accurate value from Table A.4 is ρ = 13 580 kg/m^3.
For the air we see in Figure 2.7 that density is about 1 kg/m^3 so we get
m = ρV = 1 kg/m^3 × 0.001 m^3 = 0.001 kg
A more accurate value from Table A.5 is ρ = 1.17 kg/m^3 at 100 kPa, 25 o^ C.
Can you carry 1 m^3 of liquid water?
Solution:
The density of liquid water is about 1000 kg/m^3 from Figure 2.7, see also Table A.3. Therefore the mass in one cubic meter is
m = ρV = 1000 kg/m^3 × 1 m^3 = 1000 kg
and we can not carry that in the standard gravitational field.
A manometer shows a pressure difference of 1 m of liquid mercury. Find ∆P in kPa.
Solution:
Hg : L = 1 m; ρ = 13 580 kg/m^3 from Table A.4 (or read Fig 2.7) The pressure difference ∆P balances the column of height L so from Eq.2.
∆P = ρ g L = 13 580 kg/m^3 × 9.80665 m/s 2 × 1.0 m × 10 -3 kPa/Pa = 133.2 kPa
What pressure difference does a 10 m column of atmospheric air show?
Solution: The pressure difference for a column is from Eq.2. ∆P = ρgH So we need density of air from Fig.2.7, ρ = 1.2 kg/m^3 ∆P = 1.2 kg/m^3 × 9.81 ms -2^ × 10 m = 117.7 Pa = 0.12 kPa
The pressure at the bottom of a swimming pool is evenly distributed. Suppose we look at a cast iron plate of 7272 kg lying on the ground with an area of 100 m^2. What is the average pressure below that? Is it just as evenly distributed?
Solution: The pressure is force per unit area from page 25: P = F/A = mg/A = 7272 kg × (9.81 m/s 2 ) / 100 m^2 = 713.4 Pa
The iron plate being cast can be reasonable plane and flat, but it is stiff and rigid. However, the ground is usually uneven so the contact between the plate and the ground is made over an area much smaller than the 100 m^2. Thus the local pressure at the contact locations is much larger than the quoted value above.
The pressure at the bottom of the swimming pool is very even due to the ability of the fluid (water) to have full contact with the bottom by deforming itself. This is the main difference between a fluid behavior and a solid behavior.
Iron plate Ground
A tornado rips off a 100 m^2 roof with a mass of 1000 kg. What is the minimum vacuum pressure needed to do that if we neglect the anchoring forces?
Solution:
The net force on the roof is the difference between the forces on the two sides as the pressure times the area
F = Pinside A – PoutsideA = ∆P A That force must overcome the gravitation mg, so the balance is ∆P A = mg
∆P = mg/A = (1000 kg × 9.807 m/s 2 )/100 m^2 = 98 Pa = 0.098 kPa
Remember that kPa is kN/m^2.
What is a temperature of –5 o^ C in degrees Kelvin?
Solution:
The offset from Celsius to Kelvin is 273.15 K, so we get
TK = TC + 273.15 = -5 + 273. = 268.15 K
