**$ 20.00**

# TEST BANK FOR Classical Mechanics and Thermodynamics Chapters 1-9 By Sonntag, Borgnakke

- From Engineering, Fluid Mechanics

- A-Grades
- Rating : 0
- Grade :
**No Rating** - Questions : 0
- Solutions : 275
- Blog : 0
- Earned : $35.00

**The correspondence between the problem set in this fifth edition versus theproblem set in the 4'th edition text. Problems that are new are marked new andthose that are only slightly altered are marked as modified (mod).New Old New Old New Old1 4 mod 21 13 41E 33E mod2 new 22 14 42E 34E mod3 new 23 15 43E 35E4 7 mod 24 17 44E 36E5 2 mod 25 18 45E 37E6 new 26 new 46E 38E7 new 27 19 47E 39E8 new 28 20 48E 40E9 5 mod 29 21 49E 41E10 6 30 2211 8 mod 31 2312 new 32 2413 9 mod 33 new14 10 mod 34 25 mod15 11 35 26 mod16 new 36 27 mod17 new 37 2818 16 mod 38 2919 new 39E 31E mod20 12 40E 32E2-22.1 The “standard” acceleration (at sea level and 45° latitude) due to gravity is 9.80665m/s2. What is the force needed to hold a mass of 2 kg at rest in this gravitationalfield ? How much mass can a force of 1 N support ?Solution:ma = 0 = å F = F - mgF = mg = 2 ´ 9.80665 = 19.613 NF = mg => m = F/g = 1 / 9.80665 = 0.102 kg2.2 A model car rolls down an incline with a slope so the gravitational “pull” in thedirection of motion is one third of the standard gravitational force (see Problem2.1). If the car has a mass of 0.45 kg. Find the acceleration.Solution:ma = å F = mg / 3a = mg / 3m = g/3 = 9.80665 / 3 = 3.27 m/s22.3 A car drives at 60 km/h and is brought to a full stop with constant deceleration in 5seconds. If the total car and driver mass is 1075 kg. Find the necessary force.Solution:Acceleration is the time rate of change of velocity.ma = å F ; a = dV / dt = (60 ´ 1000) / (3600 ´ 5) = 3.33 m/s2Fnet = ma = 1075 ´ 3.333 = 3583 N2.4 A washing machine has 2 kg of clothes spinning at a rate that generates anacceleration of 24 m/s2. What is the force needed to hold the clothes?Solution:F = ma = 2 kg ´ 24 m/s2 = 48 N2.5 A 1200-kg car moving at 20 km/h is accelerated at a constant rate of 4 m/s2 up to aspeed of 75 km/h. What are the force and total time required?Solution:a = dV / dt => Dt = dV/a = [ ( 75 - 20 ) / 4 ] ´ ( 1000 / 3600 )Dt = 3.82 sec ; F = ma = 1200 ´ 4 = 4800 N2-32.6 A steel plate of 950 kg accelerates from rest with 3 m/s2 for a period of 10s. Whatforce is needed and what is the final velocity?Solution:Constant acceleration can be integrated to get velocity.a = dV / dt => ò dV = ò a dt => DV = a Dt = 3 ´ 10 = 30 m/sV = 30 m/s ; F = ma = 950 ´ 3 = 2850 N2.7 A 15 kg steel container has 1.75 kilomoles of liquid propane inside. A force of 2kN now accelerates this system. What is the acceleration?Solution:ma = å F Þ a = å F / mm = msteel + mpropane = 15 + (1.75 ´ 44.094) = 92.165 kga = 2000 / 92.165 = 21.7 m/s22.8 A rope hangs over a pulley with the two equally long ends down. On one end youattach a mass of 5 kg and on the other end you attach 10 kg. Assuming standardgravitation and no friction in the pulley what is the acceleration of the 10 kg masswhen released?Solution:Do the equation of motion for the mass m2 along thedownwards direction, in that case the mass m1 movesup (i.e. has -a for the acceleration)m2 a = m2 g - m1 g - m1a(m1 + m2 ) a = (m2 - m1 )gThis is net force in motion directiona = (10 - 5) g / (10 + 5) = g / 3 = 3.27 m/s2g122.9 A bucket of concrete of total mass 200 kg is raised by a crane with an accelerationof 2 m/s2 relative to the ground at a location where the local gravitationalacceleration is 9.5 m/s2. Find the required force.Solution:F = ma = Fup - mgFup = ma + mg = 200 ( 2 + 9.5 ) = 2300 N2-42.10 On the moon the gravitational acceleration is approximately one-sixth that on thesurface of the earth. A 5-kg mass is “weighed” with a beam balance on the surfaceon the moon. What is the expected reading? If this mass is weighed with a springscale that reads correctly for standard gravity on earth (see Problem 2.1), what isthe reading?Solution:Moon gravitation is: g = gearth/6m mmBeam Balance Reading is 5 kgThis is mass comparisonSpring Balance Reading is in kg unitslength µ F µ gReading will be56kgThis is force comparison2.11 One kilogram of diatomic oxygen (O2 molecular weight 32) is contained in a 500-L tank. Find the specific volume on both a mass and mole basis (v and v ).Solution:v = V/m = 0.5/1 = 0.5 m3/kgv = V/n =Vm/M= Mv = 32 ´ 0.5 = 16 m3/kmol2.12 A 5 m3 container is filled with 900 kg of granite (density 2400 kg/m3 ) and the restof the volume is air with density 1.15 kg/m3. Find the mass of air and the overall(average) specific volume.Solution:mair = r V = rair ( Vtot - mgranite / r )= 1.15 [ 5 - (900 / 2400) ] = 1.15 ´ 4.625 = 5.32 kgv = V / m = 5 / (900 + 5.32) = 0.00552 m3/kg2.13 A 15-kg steel gas tank holds 300 L of liquid gasoline, having a density of 800kg/m3. If the system is decelerated with 6 m/s2 what is the needed force?Solution:m = mtank + mgasoline = 15 + 0.3 ´ 800 = 255 kgF = ma = 255 ´ 6 = 1530 N2-52.14 A vertical hydraulic cylinder has a 125-mm diameter piston with hydraulic fluidinside the cylinder and an ambient pressure of 1 bar. Assuming standard gravity,find the piston mass that will create a pressure inside of 1500 kPa.Solution:Force balance: F = PA = F¯ = P0A + mpg ; P0 = 1 bar = 100 kPaA = (p/4) D2 = (p/4) ´ 0.1252 = 0.01227 m2mp = (P-P0)A/g = ( 1500 - 100 ) ´ 1000 ´ 0.01227 / 9.80665 = 1752 kg2.15 A barometer to measure absolute pressure shows a mercury column height of 725mm. The temperature is such that the density of the mercury is 13550 kg/m3. Findthe ambient pressure.Solution:Hg : Dl = 725 mm = 0.725 m; r = 13550 kg/m3P = r gDl = 13550 ´ 9.80665 ´ 0.725 ´ 10-3 = 96.34 kPa2.16 A cannon-ball of 5 kg acts as a piston in a cylinder of 0.15 m diameter. As the gunpowderis burned a pressure of 7 MPa is created in the gas behind the ball. What isthe acceleration of the ball if the cylinder (cannon) is pointing horizontally?Solution:The cannon ball has 101 kPa on the side facing the atmosphere.ma = F = P1 ´ A - P0 ´ Aa = (P1 - P0 ) ´ A / m = ( 7000 - 101 ) p [ ( 0.152 /4 )/5 ] = 24.38 m/s22.17 Repeat the previous problem for a cylinder (cannon) pointing 40 degrees up relativeto the horizontal direction.Solution:ma = F = ( P1 - P0 ) A - mg sin 400ma = ( 7000 - 101 ) ´ p ´ ( 0.152 / 4 ) - 5 ´ 9.80665 ´ 0.6428= 121.9 - 31.52 = 90.4 Na = 90.4 / 5 = 18.08 m/s22-62.18 A piston/cylinder with cross sectional area of 0.01 m2 has a piston mass of 100 kgresting on the stops, as shown in Fig. P2.18. With an outside atmospheric pressureof 100 kPa, what should the water pressure be to lift the piston?Solution:Force balance: F = F¯ = PA = mpg + P0AP = P0 + mpg/A = 100 kPa + (100 ´ 9.80665) / (0.01 ´ 1000)= 100 kPa + 98.07 = 198 kPa2.19 The hydraulic lift in an auto-repair shop has a cylinder diameter of 0.2 m. To whatpressure should the hydraulic fluid be pumped to lift 40 kg of piston/arms and 700kg of a car?Solution:F¯ = ma = mg = 740 ´ 9.80665 = 7256.9 NForce balance: F = ( P - P0 ) A = F¯ => P = P0 + F¯ / AA = p D2 (1 / 4) = 0.031416 m2P = 101 + 7256.9 / (0.031416 ´ 1000) = 332 kPa2.20 A differential pressure gauge mounted on a vessel shows 1.25 MPa and a localbarometer gives atmospheric pressure as 0.96 bar. Find the absolute pressure insidethe vessel.Solution:Pgauge = 1.25 MPa = 1250 kPa; P0 = 0.96 bar = 96 kPaP = Pgauge + P0 = 1250 + 96 = 1346 kPa2.21 The absolute pressure in a tank is 85 kPa and the local ambient absolute pressure is97 kPa. If a U-tube with mercury, density 13550 kg/m3, is attached to the tank tomeasure the vacuum, what column height difference would it show?Solution:DP = P0 - Ptank = rgDlDl = ( P0 - Ptank ) / rg = [(97 - 85 ) ´ 1000 ] / (13550 ´ 9.80665)= 0.090 m = 90 mm2-72.22 A 5-kg piston in a cylinder with diameter of 100 mm is loaded with a linear springand the outside atmospheric pressure of 100 kPa. The spring exerts no force on thepiston when it is at the bottom of the cylinder and for the state shown, the pressureis 400 kPa with volume 0.4 L. The valve is opened to let some air in, causing thepiston to rise 2 cm. Find the new pressure.Solution:A linear spring has a force linear proportional to displacement. F = k x, sothe equilibrium pressure then varies linearly with volume: P = a + bV, with anintersect a and a slope b = dP/dV. Look at the balancing pressure at zero volume (V-> 0) when there is no spring force F = PA = PoA + mpg and the initial state.These two points determine the straight line shown in the P-V diagram.Piston area = AP = (p/4) ´ 0.12 = 0.00785 m2400106.2210 0.4PV0.557P 2a = P0 +mpgAp= 100 +5 ´ 9.806650.00785= 106.2 kPa intersect for zero volume.V2 = 0.4 + 0.00785 ´ 20 = 0.557 LP2 = P1 +dPdV DV= 400 +(400-106.2)0.4 - 0(0.557 - 0.4)= 515.3 kPa2.23 A U-tube manometer filled with water, density 1000 kg/m3, shows a heightdifference of 25 cm. What is the gauge pressure? If the right branch is tilted tomake an angle of 30° with the horizontal, as shown in Fig. P2.23, what should thelength of the column in the tilted tube be relative to the U-tube?Solution:hH30°DP = F/A = mg/A = Vrg/A = hrg= 0.25 ´ 1000 ´ 9.807 = 2452.5 Pa= 2.45 kPah = H ´ sin 30°Þ H = h/sin 30° = 2h = 50 cm2-82.24 The difference in height between the columns of a manometer is 200 mm with afluid of density 900 kg/m3. What is the pressure difference? What is the heightdifference if the same pressure difference is measured using mercury, density13600 kg/ m3, as manometer fluid?Solution:DP = r1gh1 = 900 ´ 9.807 ´ 0.2 = 1765.26 Pa = 1.77 kPahhg = DP/ (rhg g) = (r1 gh1) / (rhg g) =90013600 ´0.2 = 0.0132 m= 13.2 mm2.25 Two reservoirs, A and B, open to the atmosphere, are connected with a mercurymanometer. Reservoir A is moved up/down so the two top surfaces are level at h3as shown in Fig. P2.25. Assuming that you know rA, rHg and measure theheights h1, h2 , and h3, find the density rB.Solution:Balance forces on each side:P0 + rAg(h3 - h2) + rHggh2 = P0 + rBg(h3 - h1) + rHggh1Þ rB = rAèç æø ÷ öh3 - h2h3 - h1+ rHgèç æø ÷ öh2 - h1h3 - h12.26 Two vertical cylindrical storage tanks are full of liquid water, density 1000 kg/m3,the top open to the atmoshere. One is 10 m tall, 2 m diameter, the other is 2.5 m tallwith diameter 4m. What is the total force from the bottom of each tank to the waterand what is the pressure at the bottom of each tank?Solution:VA = H ´ pD2 ´ (1 / 4) = 10 ´ p ´ 22 ´ ( 1 / 4) = 31.416 m3VB = H ´ pD2 ´ (1 / 4) = 2.5 ´ p ´ 42 ´ ( 1 / 4) = 31.416 m3Tanks have the same volume, so same mass of waterF = mg = r V g = 1000 ´ 31.416 ´ 9.80665 = 308086 NTanks have same net force up (holds same m in gravitation field)Pbot = P0 + r H gPbot,A = 101 + (1000 ´ 10 ´ 9.80665 / 1000) = 199 kPaPbot,B = 101 + (1000 ´ 2.5 ´ 9.80665 / 1000) = 125.5 kPa**

## [Solved] TEST BANK FOR Classical Mechanics and Thermodynamics Chapters 1-9 By Sonntag, Borgnakke

- This solution is not purchased yet.
- Submitted On 10 Feb, 2022 12:29:33

- A-Grades
- Rating : 0
- Grade :
**No Rating** - Questions : 0
- Solutions : 275
- Blog : 0
- Earned : $35.00