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Solutions Manual for Fundamentals of Thermodynamics 11th Edition by Borgnakke ISBN: 9781394212910

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Solution Manual Chapter 1

August 2025

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

CONTENT CHAPTER 1 SUBSECTION Concept Problems Properties, Units and Force Specific Volume Pressure Manometers and Barometers Energy and Temperature Review problems

PROB NO. 1 - 18 19 - 25 26 - 28 29 - 42 43 - 53 54 - 58 59 - 62

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

In-Text Concept Questions

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.a Make a control volume around the turbine in the steam power plant in Fig. 1.2 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.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

1.b Take a control volume around your kitchen refrigerator and indicate where the components shown in Figure 1.3 are located and show all flows of energy transfers. 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.

The compressor sits at the bottom.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.c Why do people float high in the water when swimming in the Dead Sea as compared with swimming in a fresh water lake?

As the dead sea is very salty its density is higher than fresh water density. The buoyancy effect gives a force up that equals the weight of the displaced water. Since salt water density is higher the displaced volume is smaller for the same force. F = mH2O salt g – mH2O fresh g = (V) H2O salt g – (V)H2O fresh g

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.d Density of liquid water is  = 1008 – T/2 [kg/m3] with T in °C. If the temperature increases, what happens to the density and specific volume? Solution: The density is seen to decrease as the temperature increases.  = – T/2 Since the specific volume is the inverse of the density v = 1/ it will increase.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

1.e A car tire gauge indicates 195 kPa; what is the air pressure inside? The pressure you read on the gauge is a gauge pressure, P, so the absolute pressure is found as P = Po + P = 101 + 195 = 296 kPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.f

Can I always neglect P in the fluid above location A in figure 1.13? What does that depend on? If the fluid density above A is low relative to the manometer fluid then you neglect the pressure variation above position A, say the fluid is a gas like air and the manometer fluid is like liquid water. However, if the fluid above A has a density of the same order of magnitude as the manometer fluid then the pressure variation with elevation is as large as in the manometer fluid and it must be accounted for.

1.g A U tube manometer has the left branch connected to a box with a pressure of 110 kPa and the right branch open. Which side has a higher column of fluid?

Solution: Box

Since the left branch fluid surface feels 110 kPa and the right branch surface is at 100 kPa you must go further down to match the 110 kPa. The right branch has a higher column of fluid.

Po H

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

1.h If the right side pipe section in Fig. 1.13 is V shaped like a funnel does that change the pressure at location B?

The shape does not affect the pressure only depth from surface at Po matters.

Fluid Po H

Comment: the slanted surface has a component of the pressure (normal to the surface) that points upwards. cb

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.i If the cylinder pressure in Ex. 1.3 does not give Fnet = 0 what happens? If:

Fnet = ma  0



a  0

The piston will accelerate up if Pcyl > 250 kPa given F = 932.9 N or down up if Pcyl < 250 kPa given F = 932.9 N which changes the cylinder volume. Thus by controlling the pressure you can move the piston, which is the basis for the hydraulic cylinder used in a bulldozer, a backhoe or front loader.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

Concept Problems

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.1

Separate the list P, F, V, v, , T, a, m, L, t, and V into intensive, extensive, and nonproperties. Solution: Intensive properties are independent upon mass: P, v, , 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.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.2 A tray of liquid water is placed in a freezer where it cools from 20℃ to -5℃. Show the energy flow(s) and storage and explain what changes.

Inside the freezer box, the walls are very cold as they are the outside of the evaporator, or the air is cooled and a small fan moves the air around to redistribute the cold air to all the items stored in the freezer box. The fluid in the evaporator absorbs the energy and the fluid flows over to the compressor on its way around the cycle, see Fig. 1.3. As the water is cooled it eventually reaches the freezing point and ice starts to form. After a significant amount of energy is removed from the water it is turned completely into ice (at 0℃) and then cooled a little more to -5℃. The water has a negative energy storage and the energy is moved by the refrigerant fluid out of the evaporator into the compressor and then finally out of the condenser into the outside room air.

©C. Borgnakke

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.3 The overall density of fibers, rock wool insulation, foams and cotton is fairly low. Why is that? Solution: All these materials consist of some solid substance and mainly air or other gas. The volume of fibers (clothes) and rockwool that is a solid substance is low relative to the total volume that includes air. The overall density is m msolid + mair =V= V solid + Vair where most of the mass is the solid and most of the volume is air. If you talk about the density of the solid only, it is high.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.4 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.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.5 Water in nature exists in different phases such as solid, liquid and vapor (gas). Indicate the relative magnitude of density and specific volume for the three phases. Solution: Values are indicated in Figure 1.8 as density for common substances. More accurate values are found in Tables A.3, A.4 and A.5 Water as solid (ice) has density of around 900 kg/m3 Water as liquid has density of around 1000 kg/m3 Water as vapor has density of around 1 kg/m3 (sensitive to P and T)

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

1.6 What is the approximate mass of 1 L of gasoline? Of helium in a balloon at To, Po? Solution: Gasoline is a liquid slightly lighter than liquid water so its density is smaller than 1000 kg/m3. 1 L is 0.001 m3 which is a common volume used for food items. A more accurate density is listed in Table A.3 as 750 kg/m3 so the mass becomes m = ρ V = 750 kg/m3 × 0.001 m3 = 0.75 kg The helium is a gas highly sensitive to P and T, so its density is listed at the standard conditions (100 kPa, 25°C) in Table A.5 as ρ = 0.1615 kg/m3, m = ρ V = 0.1615 kg/m3 × 0.001 m3 = 1.615 × 10-4 kg

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.7 You must mail 100 cm3 of copper (Cu), how heavy a package is that? Solution: The density of copper is 8300 kg/m3 from Table A.3. Therefore the mass is m = V = 8300 kg/m3  100 cm3 = 8300 kg × 100 (cm/m)3 = 0.83 kg = 830 g

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.8 A heavy refrigerator has four height-adjustable feet. What feature of the feet will ensure that they do not make dents in the floor? Answer: The area that is in contact with the floor supports the total mass in the gravitational field. F = PA = mg so for a given mass the smaller the area is the larger the pressure becomes.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.9 A swimming pool has an evenly distributed pressure at the bottom. Consider a stiff steel plate lying on the ground. Is the pressure below it just as evenly distributed? Solution: The pressure is force per unit area from Eq. 1.3: P = F/A = mg/A The steel plate 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 actual plate area. Thus the local pressure at the contact locations is much larger than the average indicated 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.

Steel plate Ground

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.10 If something floats in water, what does it say about its density? Solution: The density must be less than the density of the water.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

1.11 Two divers swim at 20 m depth. One of them swims right in under a supertanker; the other stays away from the tanker. Who feels a greater pressure? Solution: Each one feels the local pressure which is the static pressure only a function of depth. Pocean= P0 + P = P0 + gH So they feel exactly the same pressure.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.12 An operating room has a positive gage pressure, whereas an engine test cell has a vacuum; why is that? Solution: For the operating room any air leak should be out so there will not be a chance that any microbes or other matter could come in from the outside and contaminate the patient. For the engine test cell you do not want any leak of fuel, oil fumes or exhaust gasses to go out. They are exhausted by a fan to the outside where they will be highly diluted. If in a manufacturing situation you exhaust large amounts of fumes (like in a paint operation or chemical process) the fumes must be burned in an incinerator (thermal oxidicer) before exhausted to the outside.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

1.13 Mention some reasons to have hydraulic cylinders and some reasons to have pneumatic cylinders for motion activation. Hydraulic cylinders are used when you want to have a large pressure in a cylinder. The hydraulic fluid is a liquid typically one that does not corrode the cylinder/piston walls. Used for cylinders in bulldozers, backhoes, tractors and equipment where you transmit large forces by a shaft over relatively short distances. Pneumatic cylinders uses a gas typically air to activate a piston. Used for lower pressure applications like motion of a door, flap or contact in control equipment. It is typical for a manufacturing facility to have pressurized air distributed to all working stations or places. It is economical and easier to deal with than the high pressure liquid except you must make sure the air is dried so water vapor cannot condense. Since air is highly compressible it is ill suited to the high pressure applications.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.14 A water skier does not sink too far down in the water if the speed is high enough. What makes that situation different from our static pressure calculations? The water pressure right under the ski is not a static pressure but a static plus dynamic pressure that pushes the water away from the ski. The faster you go, the smaller the amount of water is displaced, but at a higher velocity which requires a greater force as the water must accelerated significantly to be moved away.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.15 What is the lowest temperature in degrees Celsuis? In degrees Kelvin? Solution: The lowest temperature is absolute zero which is at zero degrees Kelvin at which point the temperature in Celsius is negative TK = 0 K = 273.15°C

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.16 How cold can it be on Earth and in empty space? Solution: The coldest place on earth is the South Pole Winter average:

T = – 60°C

Summer average:

T = – 28°C

Empty space About 3 K caused by background radiation. It obviously depends on where you are.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.17 A thermometer that indicates the temperature with a liquid column has a bulb with a larger volume of liquid, why is that? Solution: The expansion of the liquid volume with temperature is rather small so by having a larger volume expand with all the volume increase showing in the very small diameter column of fluid greatly increases the signal that can be read.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.18 How can you illustrate the binding energy between the three atoms in water as they sit in a tri-atomic water molecule. Hint: imagine what must happen to create three separate atoms. Answer: If you want to separate the atoms you must pull them apart. Since they are bound together with strong forces (like non-linear springs) you apply a force over a distance which is work (energy in transfer) to the system and you could end up with two hydrogen atoms and one oxygen atom far apart so they no longer have strong forces between them. If you do not do anything else the atoms will sooner or later recombine and release all the energy you put in and the energy will come out as radiation or given to other molecules by collision interactions.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

Properties, Units, and Force

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.19 One kilopond (1 kp) is the weight of 1 kg in the standard gravitational field. How many Newtons (N) is that?

F = ma = mg 1 kp = 1 kg  9.807 m/s2 = 9.807 N

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

1.20 A fuel container has 1/3 kmol of ethanol and 1/4 kmol of methanol. What is the total mass of fuel?

Table A.5: Methanol = 46.069 ; Mmethanol = 32.042 1

methanol = neth. Meth. = 3 × 46.069 = 15.356 kg 1

mmethanol = nmeth. Mmeth. = 4 × 32.042 = 8.0105 kg mtot = meth. + mmeth. = 15.356 + 8.0105 = 23.337 kg

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.21 A steel cylinder of mass 4 kg contains 4 L of dry sand at 25℃ at 100 kPa. Find the total mass and volume of the system. List two extensive and three intensive properties of the sand. Solution: Density of steel in Table A.3: Volume of steel:

 = 7820 kg/m3 4 kg V = m/ = = 0.000 512 m3 7820 kg/m3

Density of sand in Table A.3:  = 1500 kg/m3 Mass of sand:

m = V = 1500 kg/m3  0.004 m3 = 6 kg

Total mass:

m = msteel + msand = 4 + 6 = 10 kg

Total volume:

V = Vsteel + Vsand = (0.000 512 + 0.004) m3 = 0.004 512 m3 = 4.51 L

Extensive properties: m, V Intensive properties:  (or v = 1/), T, P

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

1.22 As a clothes washer is nearly done it spins 5 kg of wet clothes with 20 m/s at a radius of 0.3 m. What is the force on the rotating drum? Hint: acceleration in rotation is V2/r. Solution: Fcentrifugal = m V2/r = 5 kg × (20 m/s)2 / 0.3 m = 6667 kg m/s2 = 6667 N

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.23 A short freight train moves at 60 km/h with a total mass of 55 tonne and starts to decelerate at a constant rate of 0.5 m/ s2 to a full stop. What are the force and total time required?

Solution: dV V a = dt = => t (60  0) km/h  1000 m/km V t = a = = 33.3 sec 3600 s/h  0.5 m/s2 F = ma = 55 000 kg  0.5 m/s2 = 27 500 N Comment: 55 tonne it is a light load for a freight train.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.24 At a building site the work elevator has a mass of 550 kg and it has supplies with a total mass of 400 kg. How much force should the cable pull up with to have an acceleration of 1.5 m/s2 in the upwards direction? Solution:

The total mass moves upwards with an acceleration plus the gravitations acts with a force pointing down.

ma =  F = F – mg F = ma + mg = m(a + g) = (550 + 400) kg  (1.5 + 9.81) m/s2 = 10 745 N

m g

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.25 If a worker, 85 kg, gets on the elevator in the previous problem how much weight does this person feel when the elevator starts moving? Solution: The equation of motion is ma =  F = F – mg so the force from the floor becomes F = ma + mg = m(a + g) = 85 kg  (1.5 + 9.81) m/s2 = 961.35 N = x kg  9.81 m/s2 Solve for x x = 961.35 N/ 9.81 m/s2 = 98 kg The person then feels like having a mass of 98 kg instead of 85 kg. The weight is really force so to compare to standard mass we should use kp. So in this example the person is experiencing a force of 98 kp instead of the normal 85 kp.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

Specific Volume

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.26 A power plant that separates carbon-dioxide from the exhaust gases compresses it to a density of 110 kg/m3 and stores it in an un-minable coal seam with a porous volume of 100 000 m3. Find the mass they can store. Solution: m =  V = 110 kg/m3  100 000 m3 = 11  10 6 kg Comment: Just to put this in perspective a power plant that generates 2000 MW by burning coal would make about 20 million tons of carbon-dioxide a year. That is 2000 times the above mass so it is nearly impossible to store all the carbon-dioxide being produced.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.27 A cement mixer should deliver 2 m3 mix of granite stone, dry sand and water in a mass ratio of 2 to 1 to 1 (cement included with the sand). Use properties from tables A.3 and A.4 and find the mass of each component. Solution: Specific volume and density are ratios of total mass and total volume. mTOT = mstone + msand + mliq = mscale (2 + 1 + 1) VTOT = Vstone + Vsand + Vliq = (m/)stone+ (m/)sand + (m/)liq = mscale [(2/)stone+ (1/)sand + (1/)liq] 2 1 1 = mscale [ 2750 + 1500 + 997 ]

1 m3 = m × 0.002397 scale kg kg/m3

m3 mscale = 2 m3/( 0.002397 kg ) = 834.4 kg mstone = 2 mscale = 2 × 834.4 kg = 1669 kg msand = mscale = 834.4 kg mliq = mscale = 834.4 kg

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.28 A container is filled with 1100 kg of granite (density 2400 kg/m3 ) together with 5 kg air, with density 1.15 kg/m3. Find the total volume and the overall (average) specific volume. Solution: Vtot = Vair + Vgranite = mair/air + mgranite/granite = 5 kg/1.15 kg/m3 + 1100 kg/2400 kg/m3 = 4.348 m3 + 0.4583 m3 = 4.806 m3 V 4.806 m3 v = m = 1100 + 5 kg = 0.00435 m3/kg Comment: Because the air and the granite are not mixed or evenly distributed in the container the overall specific volume or density does not have much meaning.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

Pressure

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.29 A 5000-kg elephant has a cross sectional area of 0.02 m2 on each foot. Assuming an even distribution, what is the pressure under its feet? Force balance:

ma = 0 = PA – mg P = mg/A = 5000 kg  9.81 m/s2 /(4  0.02 m2) = 613 125 Pa = 613 kPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.30 A valve in a cylinder has a cross sectional area of 11 cm2 with a pressure of 735 kPa inside the cylinder and 99 kPa outside. How large a force is needed to open the valve? Fnet = PinA – PoutA = (735 – 99) kPa  11 cm2

Pcyl

= 6996 kPa cm2 = 6996 

kN  10-4 m2 2 m

= 700 N

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.31 The piston cylinder in Fig. P1.30 has a diameter of 10 cm, inside pressure 1235 kPa. What is the force holding the massless piston up as the piston lower side has P0 besides the force? Solution: Force acting on the mass by the inside pressure F = P A Force balance:

F = F + P0 A = F => F = (P - P0) A A =  D2 (1 / 4) = 0.007854 m2 F = (1235 – 101) kPa × 0.007854 m2 = 8.91 kN

Pcyl cb

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.32 A hydraulic lift has a piston diameter of 10 cm. What should the hydraulic fluid pressure be in order to lift a mass of 1250 kg? Solution: With the piston at rest the static force balance is F = P A = F = mg A =  r2 =  D2/4 =  × 0.12 m2/4 = 0.007854 m2 PA = mg



P = mg/A

1250 × 9.81 kg m/s2 N P = 0.007854 = 1561 306 2 = 1561 kPa 2 m m

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.33 A hydraulic cylinder has a 125-mm diameter piston with an ambient pressure of 1 bar. Assuming standard gravity, find the total mass the piston can lift if the inside hydraulic pressure is 2000 kPa. Solution: Force balance:

F = PA = F = P0A + m g; P0 = 1 bar = 100 kPa = 100 kN/ m2

= 100  1000 kg m/s2m2 = 100  1000 kg/ms2 A = (/4) D2 = (/4)  0.1252 m2 = 0.01227 m2 A 0.01227 m2 m = (P P0) g = ( 2000  100 ) kPa  1000 Pa/kPa  9.80665 m/s2 = 2377 kg

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.34 A 75-kg human footprint is 0.05 m2 when the human is wearing boots. Suppose you want to walk on snow that can at most support an extra 3 kPa; what should the total snowshoe area be? Force balance:

ma = 0 = ΔP A – mg mg 75 kg  9.81 m/s2 A = ΔP = = 0.245 m2 3 kPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.35 A piston/cylinder with diameter of 0.11 m has a piston mass of 60 kg plus a force of 700 N resting on the stops, as shown in the figure. With an outside atmospheric pressure of 101 kPa, what should the water pressure be to lift the piston? Solution: The force acting down on the piston comes from gravitation and the outside atmospheric pressure acting over the top surface. Force balance:

F = F = PA = P0A + mpg + F

F cb

Now solve for P divide by 1000 to convert to kPa for 2nd + 3rd terms

Water

A = D2/4 =  × 0.112 m2/4 = 0.0095 m2 mpg P = P0 + A + F/A = 101 kPa +

60 kg  9.80665 m/s2 700 N + 2 2 0.0095 m  1000 Pa/kPa 0.0095 m  1000 Pa/kPa

= 101 kPa + 61.94 kPa + 73.68 kPa = 236.6 kPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

1.36 A 2.5 m tall steel cylinder has a cross sectional area of 1.5 m2. At the bottom with a height of 0.5 m is liquid glycol on top of which is a 1.5 m high layer of engine oil shown in Fig. P1.36. The oil surface is exposed to atmospheric air at 101 kPa. What is the pressure at the bottom and at the interphase between the glycol and the oil? Solution: The pressure in the fluid goes up with the depth as P = Ptop + P = Ptop + gh

Air

1.5 m

Oil

0.5 m

Glycol

and since we have two fluid layers we get P = Ptop + [(h)oil + (h)glycol] g The densities from Table A.4 are: glycol = 1120 kg/m3;

oil = 885 kg/m3

2 2 9.807 m/s Pbot = 101 kPa + [885  1.5 + 1120  0.5] kg/m  1000 Pa/kPa

= 119.5 kPa 9.807 m/s2 Pint = 101 kPa + 885  1.5 kg/m2  1000 Pa/kPa = 114 kPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.37 An underwater buoy is anchored at the seabed with a cable, and it contains a total mass of 250 kg in a 0.4 m3 volume. What is the cable force? Solution: We need to do a force balance on the system at rest and the combined pressure over the buoy surface is the buoyancy (lift) equal to the “weight” of the displaced water volume The buoyancy force is shown in Eq. 1.8 where Flift comes from the pressure distribution around the object. Fbuoyancy = Flift = mH2Og = H2OVg ma = 0 = mH2Og – mg – F = Fbuoyancy – Fgravitation – Fcable = H2OVg – mg – F

F = (H2OV – m) g = (997 kg/m3 × 0.4 m3 – 250 kg) × 9.81 m/s2 = 1460 N

Fig. P1.37

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.38 If an oil rig mass is 20 000 tonnes and the net pull of the cables should be 5000 kN from the seabed, how much water displacement is needed? Solution: We need to do a force balance on the system at rest and the combined pressure over the submerged surface is the buoyancy (lift) equal to the “weight” of the displaced water volume ma = 0 = mH2Og – Fcable – mrig g mH2O = (Fcable + mrig g)/g = Fcable/g + mrig = 5000 × 1000 N/9.81 m/s2 + 20 000 × 1000 kg = 509.7 × 1000 kg + 20 000 × 1000 kg = 20.51 × 106 kg V = mH2O/H2O = 20.51 × 106 kg / 997 kg/m3 = 20 600 m3

Fig. P1.38-cb

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.39 You go parasailing getting up 55 m in the air and later dive 10 m down in the ocean. Assume the atmospheric pressure is 1025 mbar at the beach, density of the salt water is 1000 kg/m3 and the air density is 1.18 kg/m3. What is the pressure each place? Solution: P = gh, Units from A.1:

1 mbar = 100 Pa (1 bar = 100 kPa).

Ppara = P0  P = 1025  100 Pa  1.18 kg/m3  9.81 m/s2  55 m = 1.0186  105 Pa = 101.9 kPa Pocean= P0 + P = 1025  100 Pa + 1000 kg/m3  9.81 m/s2  10 m = 2.006  105 Pa = 201 kPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.40 A steel tank of cross sectional area 2 m2 and 13 m tall weighs 10 000 kg and it is open at the top, as shown in Fig. P1.40. We want to float it in the ocean so it sticks 10 m straight down by pouring concrete into the bottom of it. How much concrete should I put in? Solution: The force up on the tank is from the water pressure at the bottom times its area. The force down is the gravitation times mass and the atmospheric pressure. F = PA = (oceangh + P0)A F = (mtank + mconcrete)g + P0A

Air Ocean 10 m Concrete Fig P1.40-cb

The force balance becomes F = F = (oceangh + P0)A = (mtank + mconcrete)g + P0A Solve for the mass of concrete mconcrete = (oceanhA - mtank) = 997 kg/ m3  10 m  2 m2 – 10 000 kg = 9 940 kg Notice: The first term is the mass of the displaced ocean water. The force up is the weight (mg) of this mass called buoyancy which balances with gravitation and the force from P0 cancel.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.41 A piston, mp = 5 kg, is fitted in a cylinder, A = 15 cm2, that contains a gas. The setup is in a centrifuge that creates an acceleration of 25 m/s2 in the direction of piston motion towards the gas. Assuming standard atmospheric pressure outside the cylinder, find the gas pressure. Solution: Force balance: mpg P = P0 + A = 101.325 kPa +

F = F = P0A + mpg = PA

5  25 1000  0.0015

gas

kPa kg m/s2 Pa m2

= 184.7 kPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

1.42 A container ship is 240 m long and 32 m wide. Assume the shape is like a rectangular box. How much mass does the ship carry as load if it is 10 m down in the water and the mass of the ship itself is 30 000 tonnes? Balance forces F = F = mdispl H2O g = (mship + mcargo) g The displacement volume of water has a mass that equals the total mass of the ship and its cargo. Vdispl = H A = 10 m × 240 m × 32 m = 76 800 m3 mdispl H2O = ocean Vdispl = 1000 kg/ m3 × 76 800 m3 = 76.8 × 106 kg The cargo mass then is mcargo = mdispl H2O  mship = 76.8 × 106 kg 30 000 t = (76.8 30) × 106 kg = 46.8 × 106 kg = 46 800 tonnes

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

Manometers and Barometers

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.43 A probe is lowered 16 m into a lake. Find the absolute pressure there? Solution: The pressure difference for a column is from Eq.1.2 and the density of water is from Table A.4. P = gH = 997 kg/m3  9.81 m/s2  16 m = 156 489 Pa = 156.489 kPa Plake = P0 + P = 101.325 kPa + 156.489 kPa = 257.8 kPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

1.44 A hang glider is about 22 m2 for the sail with 45 kg mass plus the person, 75 kg. How large a gage pressure under the sail is needed for level flight? Suppose you accelerate down with 2 m/s2 how large must the gage pressure then be? Solution: Force acting on the mass by the gravitational field at level flight F = ma = mg = F = ( P  P0 ) A = ΔP A ΔP = F/A = mg/A = (75 + 45) kg  9.80665 m/s2 / 22 m2 = 53.5 Pa = 0.054 kPa Flying downwards F = m(g  a) = ΔP A ΔP = m(g  a)/A = (75 + 45) kg  (9.80665 – 2) m/s2 / 22 m2 = 42.6 Pa = 0.043 kPa Recall 1 Pa = 1 N/m. The kPa matches with F in kN. This assumes an even pressure distribution under the sail. The small ΔP is due to the large sail.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.45 A vacuum pump keeps a gage pressure of 3.1 cm water column underneath a basement floor to suck out radon gas. What force does that put under a floor of area 20 m2. Solution: ΔP = F/A

F = ΔP A =  L g A

F = 997 kg/m3  0.031 m  9.80665 m/s2  20 m2 = 6062 kg m/s2 = 6062 N Since it is a vacuum it is a force down on the floor.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.46 A barometer to measure absolute pressure shows a mercury column height of 745 mm. The temperature is such that the density of the mercury is 13 550 kg/m3. Find the ambient pressure. Solution: Hg : L = 745 mm = 0.745 m;  = 13 550 kg/m3 The external pressure P balances the column of height L so from Fig. 1.14 P =  L g = 13 550 kg/m3  9.80665 m/s2  0.745 m  10-3 kPa/Pa = 99 kPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.47 A differential pressure gauge mounted on a vessel shows 1.25 MPa and a local barometer gives atmospheric pressure as 0.96 bar. Find the absolute pressure inside the vessel. Solution: Def. of gauge pressure:

P = Pgauge + P0

Convert all pressures to units of kPa. Pgauge = 1.25 MPa = 1250 kPa; P0 = 0.96 bar = 96 kPa P = Pgauge + P0 = 1250 + 96 = 1346 kPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.48 A barometer measures 760 mmHg at street level and 745 mmHg on top of a building. How tall is the building if we assume air density of 1.15 kg/m3? Solution: P = gH H = P/g =

760 – 745 mmHg 133.32 Pa = 177 m 1.15  9.807 kg/m2s2 mmHg

Conversion mmHg to Pa in Table A.1

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.49 An exploration submarine should be able to go 2000 m down in the ocean. If the ocean density is 1025 kg/m3 what is the maximum pressure on the submarine hull? Solution: Assume we have atmospheric pressure inside the submarine then the pressure difference to the outside water is P = Lg = (1025 kg/m3  2000 m  9.807 m/s2) / (1000 Pa/kPa) = 20 104 kPa = 20.1 MPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.50 The absolute pressure in a tank is 245 kPa and the local ambient absolute pressure is 101 kPa. If a U-tube with mercury, density 13 550 kg/m3, is attached to the tank to measure the gage pressure, what column height difference would it show? Solution: P = Ptank  P0 = g H H = (Ptank  P0)/g = [(245  101)  1000] Pa/(13550 kg/m3  9.81 m/s2) = 1.08 m

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.51 An absolute pressure gauge attached to a steel cylinder shows 135 kPa. We want to attach a manometer using liquid water a day that Patm = 101 kPa. How high a fluid level difference must we plan for? Solution: Since the manometer shows a pressure difference we have P = Pcyl - Patm =  L g L = P / g =

(135 – 101) kPa 1000 Pa -3 kPa 2 997 kg m  9.807 m/s

= 3.48 m

Comment: You would not use a manometer to measure this, as it is impractical with such a height. A pressure transducer will be used with an electronic readout.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.52 A pipe flowing light oil has a manometer attached as shown in Fig. P1.52. What is the absolute pressure in the pipe flow? Solution: Table A.3:

oil = 910 kg/m3;

water = 997 kg/m3

Pbot = P0 + water g Htot = P0 + 997 kg/m3  9.807 m/s2  0.8 m = Po + 7822 Pa Ppipe = Pbot – water g H1 – oil g H2 = Pbot – 997 kg/m3  9.807 m/s2  0.1 m

– 910 kg/m3  9.807 m/s2  0.2 m = Pbot – 977.7 Pa – 1784.9 Pa Ppipe = Po + (7822 – 977.7 – 1784.9) Pa = Po + 5059.4 Pa = 101.325 + 5.06 = 106.4 kPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.53 A piece of experimental apparatus is located where g = 9.5 m/s2 and the temperature is 5°C. An air flow inside the apparatus is determined by measuring the pressure drop across an orifice with a mercury manometer (density 13 580 kg/m3) showing a height difference of 200 mm. What is the pressure drop in kPa? Solution: P = gh ;

Hg = 13 580 kg/m3

P = 13 580 kg/m3  9.5 m/s2  0.2 m = 25802 Pa = 25.80 kPa

Air

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

Energy and Temperature

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.54 A 0.25 m3 piece of softwood is lifted up to the top shelf in a storage bin that is 4 m above the ground floor. How much increase in potential energy does the wood get?

The potential energy is from Eq.1.5 pe = gz and the mass becomes m =  V = 510 kg/m3  0.25 m3 = 127.5 kg PE = mgH = 127.5 kg  9.81 m/s2  4 m = 5003 J = 5.0 kJ

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.55 A racecar of mass 1275 kg travels with a velocity of 20 km/h. Find the kinetic energy. How high should it be lifted in the standard gravitational field to have a potential energy that equals the kinetic energy? Solution: Standard kinetic energy of the mass is 20  10002 KE = ½ m V2 = ½  1275 kg   3600  m2/s2   = ½  1275  5.5556 Nm = 3541.7 J = 3.54 kJ Standard potential energy is POT = mgh 3541.7 Nm h = ½ m V2 / mg = = 0.28 m 1275 kg  9.807 m/s2

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.56 What is a temperature of –15°C in degrees Kelvin? Solution:

The offset from Celsius to Kelvin is 273.15 K, so we get TK = TC + 273.15 = 5 + 273.15 = 258.15 K

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.57 A mercury thermometer measures temperature by measuring the volume expansion of a fixed mass of liquid mercury due to a change in density as Hg = [13595  2.5 T ] kg/ m3

T in Celsius

Find the relative change (%) in volume for a change in temperature from 10°C to 25°C. From 10°C to 25°C At 10°C : Hg = 13595 – 2.5  10 = 13570 kg/m3 At 25°C : Hg = 13595 – 2.5  25 = 13532.5 kg/m3 The volume from the mass and density is: Relative Change = =

V = m/

V25 – V10 (m/25) – (m/10) = V10 m/10 10 13570 – 1 = 13532.5 – 1 = 0.0028 or +0.3% 25

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.58 Density of liquid water is  = 1008 – T/2 [kg/m3] with T in °C. If the temperature increases 10°C how much deeper does a 1 m layer of water become? Solution: The density change for a change in temperature of 10oC becomes  = – T/2 = –5 kg/m3 from an ambient density of  = 1008 – T/2 = 1008 – 25/2 = 995.5 kg/m3 Assume the area is the same and the mass is the same m = V = AH, then we have m = 0 = V + V  V = – V/ and the change in the height is V HV –H – (–5) H = A = V = = 1 m  995.5 = 0.005 m  barely measurable.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1

Review Problems

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.59

Repeat problem 1.53 if the flow inside the apparatus is liquid water,   1000 kg/m3, instead of air. Find the pressure difference between the two holes flush with the bottom of the channel. You cannot neglect the two unequal water columns. Solution: P1 . H

Balance forces in the manometer: P2 ·

(H - h2) - (H - h1) = hHg = h1 - h2 h2

P1A + H Oh1gA + Hg(H - h1)gA 2

= P2A + H Oh2gA + Hg(H - h2)gA 2

 P1 - P2 = H O(h2 - h1)g + Hg(h1 - h2)g 2

P1 – P2 = HghHgg – H OhHgg 2

= (13 580  0.2  9.5 – 1000  0.2  9.5 ) kg/m3  m  m/s2 = (25 802 – 1900) Pa = 23 902 Pa = 23.9 kPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.60 A dam retains a lake 6 m deep as shown in Fig. P1.60. 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 = gh = 997 kg/m3  9.807 m/s2 6 m = 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 kPa  5 m  6 m = 880 kN

Fleft

Fright

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.61 In the city water tower, water is pumped up to a level 25 m above ground in a pressurized tank with air at 150 kPa over the water surface. This is illustrated in Fig. P1.61. Assuming the water density is 1000 kg/m3 and standard gravity, find the pressure required to pump more water in at ground level. Solution: P =  L g = 1000 kg/m3  25 m  9.807 m/s2 = 245 175 Pa = 245.2 kPa Pbottom = Ptop + P = 150 + 245.2 = 395 kPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke, Solutions Chapter 1 1.62 The main waterline into a tall building has a pressure of 600 kPa at 5 m elevation below ground level. How much extra pressure does a pump need to add to ensure a water line pressure of 200 kPa at the top floor 150 m above ground? Solution: The pump exit pressure must balance the top pressure plus the column P. The pump inlet pressure provides part of the absolute pressure. Pafter pump = Ptop + P P = gh = 997 kg/m3  9.807 m/s2  (150 + 5) m = 1 515 525 Pa = 1516 kPa Pafter pump = 200 + 1516 = 1716 kPa Ppump = 1716 – 600 = 1116 kPa

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Solution Manual Chapter 1 English units

August 2025

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag

Concept Problems

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag

1.63E A mass of 2 lbm has acceleration of 15 ft/s2, what is the needed force in lbf? Solution: Newtons 2nd law:

F = ma

F = ma = 2 lbm  15 ft/s2 = 30 lbm ft/s2 30 = 32.174 lbf = 0.932 lbf

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.64E How much mass is in 1 gallon of gasoline? If helium in a balloon at atmospheric P and T? Solution: A volume of 1 gal equals 231 in3, see Table A.1. From Table F.3 the density is 46.8 lbm/ft3, so we get m = V = 46.8 lbm/ft3  1  (231/123 ) ft3 = 6.256 lbm For the helium we see Table F.4 that density is 10.08  10-3 lbm/ft3 so we get m = V = 10.08  10-3 lbm/ft3  1  (231/123 ) ft3 = 0.00135 lbm

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag

1.65E Can you easily carry a one gallon bar of solid gold? Solution: The density of solid gold is about 1205 lbm/ft3 from Table F.2, we could also have read Figure 1.7 and converted the units. V = 1 gal = 231 in3 = 231  12-3 ft3 = 0.13368 ft3 Therefore the mass in one gallon is m = V = 1205 lbm/ft3  0.13368 ft3 = 161 lbm and some people can just about carry that in the standard gravitational field.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.66E What is the temperature of –15F in degrees Rankine? Solution: The offset from Fahrenheit to Rankine is 459.67 R, so we get TR = TF + 459.67 R = –15 + 459.67 = 444.67 R

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.67E What is the lowest temperature in degrees Fahrenheit you can have? Rankine? Solution: The lowest temperature is absolute zero which is at zero degrees Rankine at which point the temperature in Fahrenheit is negative TR = 0 R = 459.67 F

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.68E What is the relative magnitude of degree Rankine to degree Kelvin Look in Table A.1 p. A-6: 1 K = 1 oC = 1.8 R = 1.8 F 5 1 R = 9 K = 0.5556 K

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag Force, Energy, Density

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag

1.69E The Rover Explorer has a mass of 410 lbm, how much does this weigh on the Moon (g = gstd/6) and on Mars where g = 12.3 ft/s2 Solution: Density is mass per unit volume Moon: F = mg = 410 lbm  (32.174 / 6) ft/s2 = 68.33 lbf Mars: F = mg = 410 lbm  12.3 ft/s2 = 5043 lbm ft/s2 = 156.7 lbf The 410 lbm on earth will give a weight of 410 lbf

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.70E A short freight train moves at 45 mi/h with a total mass of 55 ton and starts to decelerate at a constant 2 ft/s2 to a full stop. What are the force and the time required? Solution: dV V a = dt =  t t =

V t = a

45 mi/h  1609.34 m/mi  3.28084 ft/m = 33 sec 3600 s/h  2 ft/s2

F = ma = 55  2000 lbm  2 ft/s2 / (32.174 lbm ft /lbf-s2 ) = 6838 lbf

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.71E A building site elevator brings material with mass 900 lbm and a 800 lbm cage up with an acceleration of 3 ft/s2. What is the needed force in the cable? Solution: The total mass moves upwards with an acceleration plus the gravitations acts with a force pointing down. ma =  F = F – mg

F = ma + mg = m(a + g) = (900 + 800) lbm  (3 + 32.174) ft/s2 = 59 796 lbm ft/s2 = 1859 lbf

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.72E A car of mass 4000 lbm travels with a velocity of 60 mi/h. Find the kinetic energy. How high should the car be lifted in the standard gravitational field to have a potential energy that equals the kinetic energy? Solution 60 mi/h  1609.3 m/mi 2 ] 3600 s/h  0.3048 m/ft = 15 487 153 lbm ft2/s2 = 618.6 Btu

KIN = 0.5 m V2 = 0.5  4000 lbm  [

POT = mgH => H = POT/mg =

15 487 153 lbm ft 2 /s 2 = 120.3 ft 4000 lbm  32.174 ft/s2

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.73E A powerplant that separates carbon-dioxide from the exhaust gases compresses it to a density of 8 lbm/ft3 and stores it in an un-minable coal seam with a porous volume of 3 500 000 ft3. Find the mass they can store.

Solution: m =  V = 8 lbm/ft3  3 500 000 ft3 = 2.8  10 7 lbm Just to put this in perspective a power plant that generates 2000 MW by burning coal would make about 44 billion pounds of carbon-dioxide a year. That is 1500 times the above mass so it is nearly impossible to store all the carbon-dioxide being produced.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.74E The piston cylinder in Fig. P1.30 has a diameter of 4 in., inside pressure of 100 psia. What force must hold the massless piston up as the piston lower side has Po besides the force? Fnet = PinA – PoutA = (100 – 14.7) psia  π22 in2

Pcyl

= 1071.9 psi  in2 = 1072 lbf

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.75E A laboratory room keeps a vacuum of 1 in. of water due to the exhaust fan. What is the net force on a door of size 6 ft by 3 ft? Solution: The net force on the door is the difference between the forces on the two sides as the pressure times the area F = Poutside A – Pinside A = P  A = 1 in H2O  6 ft  3 ft = 1  0.036126 lbf/in2  18 ft2  144 in2/ft2 = 93.64 lbf Table A.1: 1 in. H2O is 0.036 126 lbf/in2, lbf/in2 is also often listed as psi. Pabs = Po - P P = 1 in H2O

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.76E A person, 175 lbm, wants to hoover on a 4 lbm skateboard of size 2 ft by 0.8 ft. How large a gauge pressure under the board is needed? Solution: Force acting on the mass by the gravitational field F = ma = mg = F = ( P - P0 ) A = ΔP A ΔP = F/A = mg/A = (175 + 4) lbm  32.174 ft/s2 / (2 ft  0.8 ft) = 179 lbf/ 1.6 ft2 = (111.9 / 144) lbf/in2 = 0.777 psi Recall 1 lbf = 32.174 lbm ft/s2.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag

1.77E If an oil rig mass is 20 000 ton and the net pull of the cables should be 1.1 ×106 lbf from the seabed, how much water displacement is needed? Solution: We need to do a force balance on the system at rest and the combined pressure over the submerged surface is the buoyancy (lift) equal to the “weight” of the displaced water volume ma = 0 = mH2Og – F – mrigg = H2OVg – F – mrigg V = (F + mrigg)/H2Og = (1.1 ×106 + 20 000  2000) lbf / (32.174 ft/s2  62.2 lbm/ft3 ) = 20 537 ft3 Recall 1 lbf = 32.174 lbm ft/s2, 1 ton = 2000 lbm

77E Fig. P1.37-cb

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.78E A container ship is 790 ft long and 100 ft wide. Assume the shape is like a rectangular box. How much mass does the ship carry as load if it is 30 ft down in the water and the mass of the ship itself is 30 000 tons? Balance forces F = F = mdispl H2O g = (mship + mcargo) g The displacement volume of water has a mass that equals the total mass of the ship and its cargo. Vdispl = H A = 30 ft × 790 ft × 100 ft = 2 370 000 ft3 mdispl H2O = ocean Vdispl = 62.2 lbm/ ft3 × 2 370 000 ft3 = 147.4 × 106 lbm The cargo mass then is mcargo = mdispl H2O  mship = 147.4 × 106 lbm30 000 × 2000 lbm = (147.4 0) × 106 lbm = 87.4 × 106 lbm = 43 700 tons

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag

1.79E

A manometer shows a pressure difference of 3.5 in. of liquid mercury. Find P in psi. Solution: Hg: L = 3.5 in; Pressure:

 = 848 lbm/ft3 from Table F.3

1 psi = 1 lbf/ in2

The pressure difference P balances the column of height L so from Eq.2.2 P =  g L = 848 lbm/ft3  32.174 ft/s2  (3.5/12) ft = 247.3 lbf/ft2 = (247.3 / 144) lbf/in2 = 1.72 psi

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag

1.80E What pressure difference does a 300-ft column of atmospheric air show? Solution: Air: L = 300 ft;  = 0.073 lbm/ft3 from Table F.4 The pressure difference P balances the column of height L so from Eq.1.2 P =  g L = 0.073 lbm/ft3  32.1741 fts-2  300 ft = 21.9 lbf/ft2 = ( 21.9 / 144) lbf/in2 = 0.152 psi

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.81E A piston/cylinder with cross-sectional area of 0.1 ft2 has a piston mass of 100 lbm and a force of 180 lbf resting on the stops, as shown in Fig. P1.35. With an outside atmospheric pressure of 1 atm, what should the water pressure be to lift the piston? Solution: The force acting down on the piston comes from gravitation and the outside atmospheric pressure acting over the top surface. Force balance:

F = F = PA = mpg + P0A + F cb

Now solve for P (multiply by 144 to convert from ft2 to in2)

Water

mpg P = P0 + A + F/A = 14.696 psia +

100  32.174 psi + 180 lbf /(144 × 0.1) in2 0.1  144  32.174

= 14.696 psia + 6.944 psi + 12.5 psi = 34.14 lbf/in2

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.82E The main waterline into a tall building has a pressure of 90 psia at 16 ft elevation below ground level. How much extra pressure does a pump need to add to ensure a waterline pressure of 30 psia at the top floor 450 ft above ground? Solution: The pump exit pressure must balance the top pressure plus the column P. The pump inlet pressure provides part of the absolute pressure. Pafter pump = Ptop + P P = gh = 62.2 lbm/ft3  32.174 ft/s2  (450 + 16) ft 

1 lbf s2 32.174 lbm ft

= 28 985 lbf/ft2 = 201.3 lbf/in2 Pafter pump = 30 + 201.3 = 231.3 psia Ppump = 231.3 – 90 = 141.3 psi

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.83E As a clothes washer is nealr done it spins 10 lbm of wet clothes with 7 ft/s at a radius of 1 ft. What is the force on the rotating drum? Hint: acceleration in rotation is V2/r Solution: Fcentrifugal = m V2/r = 10 lbm × (7 ft/s)2 / 1 ft = 490 lbm ft/s2 = 15.23 lbf

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag Temperature

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual

Borgnakke and Sonntag 1.84E The human comfort zone is between 18 and 24°C. What is that range in Fahrenheit? T = 18°C = 32 + 1.8  18 = 64.4 F T = 24°C = 32 + 1.8  24 = 75.2 F So the range is like 64 F to 75 F.

Fundamentals of Thermodynamics 11e Borgnakke Solutions Manual