[Solutions Manual] Fundamentals.of.Thermodynamics.[Sonntag-Borgnakke-Van.Wylen]...

(Parte 1 de 13)

CHAPTER 2

The correspondence between the problem set in this fifth edition versus the problem set in the 4'th edition text. Problems that are new are marked new and those that are only slightly altered are marked as modified (mod).

2.1The “standard” acceleration (at sea level and 45° latitude) due to gravity is 9.80665 m/s2. What is the force needed to hold a mass of 2 kg at rest in this gravitational field ? How much mass can a force of 1 N support ?

Solution: ma = 0 = å F = F - mg

F = mg=>

m = F/g = 1 / 9.80665 = 0.102 kg

F = mg = 2 · 9.80665 = 19.613 N

2.2A model car rolls down an incline with a slope so the gravitational “pull” in the direction of motion is one third of the standard gravitational force (see Problem 2.1). If the car has a mass of 0.45 kg. Find the acceleration.

Solution:

ma = å F = mg / 3 a = mg / 3m = g/3 = 9.80665 / 3 = 3.27 m/s2

2.3A car drives at 60 km/h and is brought to a full stop with constant deceleration in 5 seconds. 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) = 3.3 m/s2

Fnet = ma = 1075 · 3.3 = 3583 N

2.4A washing machine has 2 kg of clothes spinning at a rate that generates an acceleration of 24 m/s2. What is the force needed to hold the clothes?

Solution:

F = ma = 2 kg · 24 m/s2 = 48 N

2.5A 1200-kg car moving at 20 km/h is accelerated at a constant rate of 4 m/s2 up to a speed 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 N

2.6A steel plate of 950 kg accelerates from rest with 3 m/s2 for a period of 10s. What force is needed and what is the final velocity?

Solution:

Constant acceleration can be integrated to get velocity.

a dt	=> DV = a Dt = 3 · 10 = 30 m/s
V = 30 m/s ;	F = ma = 950 · 3 = 2850 N

a = dV / dt => ò dV = ò

2.7A 15 kg steel container has 1.75 kilomoles of liquid propane inside. A force of 2 kN now accelerates this system. What is the acceleration?

Solution:

ma = å F Þ a = å F / m m = msteel + mpropane = 15 + (1.75 · 4.094) = 92.165 kg a = 2000 / 92.165 = 21.7 m/s2

2.8A rope hangs over a pulley with the two equally long ends down. On one end you attach a mass of 5 kg and on the other end you attach 10 kg. Assuming standard gravitation and no friction in the pulley what is the acceleration of the 10 kg mass when released?

Solution:

Do the equation of motion for the mass m2 along the downwards direction, in that case the mass m1 moves up (i.e. has -a for the acceleration)

	m2 a = m2 g - m1 g - m1a
	(m1 + m2 ) a = (m2 - m1 )g

This is net force in motion direction

a = (10 - 5) g / (10 + 5) = g / 3 = 3.27 m/s2

2.9A bucket of concrete of total mass 200 kg is raised by a crane with an acceleration of 2 m/s2 relative to the ground at a location where the local gravitational acceleration is 9.5 m/s2. Find the required force.

Solution:

F = ma = Fup - mg Fup = ma + mg = 200 ( 2 + 9.5 ) = 2300 N

2.10On the moon the gravitational acceleration is approximately one-sixth that on the surface of the earth. A 5-kg mass is “weighed” with a beam balance on the surface on the moon. What is the expected reading? If this mass is weighed with a spring scale that reads correctly for standard gravity on earth (see Problem 2.1), what is the reading?

Moon gravitation is: g = gearth/6

Solution: m m m m

Beam Balance Reading is 5 kg This is mass comparison Spring Balance Reading is in kg units length µ F µ g

Reading will be

6 kg

This is force comparison

Solution: v = V/m = 0.5/1 = 0.5 m3/kg v = V/n = V m/M = Mv = 32 · 0.5 = 16 m3/kmol

2.12A 5 m3 container is filled with 900 kg of granite (density 2400 kg/m3 ) and the rest of the volume is air with density 1.15 kg/m3. Find the mass of air and the overall (average) specific volume.

Solution:

= 1.15 [ 5 - (900 / 2400) ] = 1.15 · 4.625 = 5.32 kg

mair = r V = rair ( Vtot - mgranite / r ) v = V / m = 5 / (900 + 5.32) = 0.00552 m3/kg

2.13A 15-kg steel gas tank holds 300 L of liquid gasoline, having a density of 800 kg/m3. If the system is decelerated with 6 m/s2 what is the needed force?

Solution:

m = mtank + mgasoline = 15 + 0.3 · 800 = 255 kg F = ma = 255 · 6 = 1530 N

2.14A vertical hydraulic cylinder has a 125-m diameter piston with hydraulic fluid inside 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 = Ffl = P0A + mpg ;

P0 = 1 bar = 100 kPa

A = (p/4) D2 = (p/4) · 0.1252 = 0.01227 m2 mp = (P-P0)A/g = ( 1500 - 100 ) · 1000 · 0.01227 / 9.80665 = 1752 kg

2.15A barometer to measure absolute pressure shows a mercury column height of 725 m. The temperature is such that the density of the mercury is 13550 kg/m3. Find the ambient pressure.

Solution:

Hg : Dl = 725 m = 0.725 m;

r = 13550 kg/m3

P = r gDl = 13550 · 9.80665 · 0.725 · 10-3 = 96.34 kPa

(Parte 1 de 13)