Thermo 5th Chap01P085 - Solucionario Capítulo 1 Termodinámica Cengel 6ta edición problemas 85- PDF

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Solucionario Capítulo 1 Termodinámica Cengel 6ta edición problemas 85-125...

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Review Problems 1-85 A hydraulic lift is used to lift a weight. The diameter of the piston on which the weight to be placed is to be determined. Assumptions 1 The cylinders of the lift are vertical. 2 There are no leaks. 3 Atmospheric pressure act on both sides, and thus it can be disregarded. Analysis Noting that pressure is force per unit area, the pressure on the smaller piston is determined from F m1 g P1 = 1 = A1 πD1 2 / 4 =

F1 25 kg

Weight 2500 kg

F2

10 cm

D2

 (25 kg)(9.81 m/s 2 )  1 kN    2 2 π (0.10 m) / 4  1000 kg ⋅ m/s 

= 31.23 kN/m 2 = 31.23 kPa From Pascal’s principle, the pressure on the greater piston is equal to that in the smaller piston. Then, the needed diameter is determined from  (2500 kg)(9.81 m/s 2 )  1 kN  → D 2 = 1.0 m 2 2 2   A2 πD 2 / 4 πD 2 / 4  1000 kg ⋅ m/s  Discussion Note that large weights can be raised by little effort in hydraulic lift by making use of Pascal’s principle. P1 = P2 =

F2

=

m2 g

 → 31.23 kN/m 2 =

1-86 A vertical piston-cylinder device contains a gas. Some weights are to be placed on the piston to increase the gas pressure. The local atmospheric pressure and the mass of the weights that will double the pressure of the gas are to be determined. Assumptions Friction between the piston and the cylinder is negligible. Analysis The gas pressure in the piston-cylinder device initially depends on the local atmospheric pressure and the weight of the piston. Balancing the vertical forces yield

WEIGTHS

GAS

 (5 kg)(9.81 m/s 2 )  1 kN   = 95.66 kN/m 2 = 95.7 kPa 2 2   A π (0.12 m )/4 1000 kg ⋅ m/s  The force balance when the weights are placed is used to determine the mass of the weights ( mpiston + mweights ) g P = Patm + A (5 kg + m weights )(9.81 m/s 2 )   1 kN   200 kPa = 95.66 kPa + → m weights = 115.3 kg  2 2  ⋅ π (0.12 m )/4 1000 kg m/s   A large mass is needed to double the pressure. Patm = P −

m piston g

= 100 kPa −

1-35

1-87 An airplane is flying over a city. The local atmospheric pressure in that city is to be determined. Assumptions The gravitational acceleration does not change with altitude. Properties The densities of air and mercury are given to be 1.15 kg/m3 and 13,600 kg/m3. Analysis The local atmospheric pressure is determined from Patm = Pplane + ρ gh  1 kN =58 kPa +(1.15 kg/m 3 )(9.81 m/s 2 )(3000 m)  1000 kg ⋅ m/s 2  The atmospheric pressure may be expressed in mmHg as h Hg =

Patm

ρg

=

  = 91.84 kN/m 2 = 91.8 kPa  

 1000 Pa  1000 mm   = 688 mmHg   (13,600 kg/m )(9.81 m/s )  1 kPa  1 m  91.8 kPa 3

2

1-88 The gravitational acceleration changes with altitude. Accounting for this variation, the weights of a body at different locations are to be determined. Analysis The weight of an 80-kg man at various locations is obtained by substituting the altitude z (values in m) into the relation

 1N − W = mg = (80 kg)(9.807 − 3.32 × 10 6 z m/s 2 ) 1kg ⋅ m/s 2  Sea level: Denver: Mt. Ev.:

   

(z = 0 m): W = 80×(9.807-3.32x10-6×0) = 80×9.807 = 784.6 N (z = 1610 m): W = 80×(9.807-3.32x10-6×1610) = 80×9.802 = 784.2 N (z = 8848 m): W = 80×(9.807-3.32x10-6×8848) = 80×9.778 = 782.2 N

1-89 A man is considering buying a 12-oz steak for $3.15, or a 320-g steak for $2.80. The steak that is a better buy is to be determined. Assumptions The steaks are of identical quality. Analysis To make a comparison possible, we need to express the cost of each steak on a common basis. Let us choose 1 kg as the basis for comparison. Using proper conversion factors, the unit cost of each steak is determined to be

12 ounce steak:

 $3.15   16 oz   1 lbm   = $9.26/kg Unit Cost =       12 oz  1 lbm   0.45359 kg 

320 gram steak:  $2.80   1000 g     = $8.75/kg Unit Cost =   320 g   1 kg  Therefore, the steak at the international market is a better buy.

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1-90 The thrust developed by the jet engine of a Boeing 777 is given to be 85,000 pounds. This thrust is to be expressed in N and kgf. Analysis Noting that 1 lbf = 4.448 N and 1 kgf = 9.81 N, the thrust developed can be expressed in two other units as 4.448 N  5 Thrust = (85,000 lbf ) Thrust in N:  = 3.78 ×10 N  1 lbf 

Thrust in kgf:

 1 kgf  = 3.85 ×10 4 kgf Thrust = (37.8 ×105 N )    9.81 N 

1-91E The efficiency of a refrigerator increases by 3% per °C rise in the minimum temperature. This increase is to be expressed per °F, K, and R rise in the minimum temperature. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the increase in efficiency is (a) 3% for each K rise in temperature, and (b), (c) 3/1.8 = 1.67% for each R or °F rise in temperature.

1-92E The boiling temperature of water decreases by 3°C for each 1000 m rise in altitude. This decrease in temperature is to be expressed in °F, K, and R. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the decrease in the boiling temperature is (a) 3 K for each 1000 m rise in altitude, and (b), (c) 3×1.8 = 5.4°°F = 5.4 R for each 1000 m rise in altitude.

1-93E The average body temperature of a person rises by about 2°C during strenuous exercise. This increase in temperature is to be expressed in °F, K, and R. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the rise in the body temperature during strenuous exercise is (a) 2 K (b) 2×1.8 = 3.6°°F (c) 2×1.8 = 3.6 R

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1-94E Hyperthermia of 5°C is considered fatal. This fatal level temperature change of body temperature is to be expressed in °F, K, and R. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the fatal level of hypothermia is (a) 5 K (b) 5×1.8 = 9°°F (c) 5×1.8 = 9 R

1-95E A house is losing heat at a rate of 4500 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. The rate of heat loss is to be expressed per °F, K, and R of temperature difference between the indoor and the outdoor temperatures. Analysis The magnitudes of 1 K and 1°C are identical, so are the magnitudes of 1 R and 1°F. Also, a change of 1 K or 1°C in temperature corresponds to a change of 1.8 R or 1.8°F. Therefore, the rate of heat loss from the house is (a) 4500 kJ/h per K difference in temperature, and (b), (c) 4500/1.8 = 2500 kJ/h per R or °F rise in temperature.

1-96 The average temperature of the atmosphere is expressed as Tatm = 288.15 – 6.5z where z is altitude in km. The temperature outside an airplane cruising at 12,000 m is to be determined. Analysis Using the relation given, the average temperature of the atmosphere at an altitude of 12,000 m is determined to be Tatm = 288.15 - 6.5z = 288.15 - 6.5×12 = 210.15 K = - 63°°C Discussion This is the “average” temperature. The actual temperature at different times can be different.

1-97 A new “Smith” absolute temperature scale is proposed, and a value of 1000 S is assigned to the boiling point of water. The ice point on this scale, and its relation to the Kelvin scale are to be determined. Analysis All linear absolute temperature scales read zero at absolute zero pressure, and are constant multiples of each other. For example, T(R) = 1.8 T(K). That is, K S multiplying a temperature value in K by 1.8 will give the same temperature in R. 1000 373.15 The proposed temperature scale is an acceptable absolute temperature scale since it differs from the other absolute temperature scales by a constant only. The boiling temperature of water in the Kelvin and the Smith scales are 315.15 K and 1000 K, respectively. Therefore, these two temperature scales are related to each other by 1000 T (S ) = T ( K ) = 2.6799 T(K ) 373.15 The ice point of water on the Smith scale is T(S)ice = 2.6799 T(K)ice = 2.6799×273.15 = 732.0 S 0

1-38

1-98E An expression for the equivalent wind chill temperature is given in English units. It is to be converted to SI units. Analysis The required conversion relations are 1 mph = 1.609 km/h and T(°F) = 1.8T(°C) + 32. The first thought that comes to mind is to replace T(°F) in the equation by its equivalent 1.8T(°C) + 32, and V in mph by 1.609 km/h, which is the “regular” way of converting units. However, the equation we have is not a regular dimensionally homogeneous equation, and thus the regular rules do not apply. The V in the equation is a constant whose value is equal to the numerical value of the velocity in mph. Therefore, if V is given in km/h, we should divide it by 1.609 to convert it to the desired unit of mph. That is, . − [914 . − Tambient (° F)][0.475 − 0.0203(V / 1609 . ) + 0.304 V / 1609 . ] Tequiv (° F) = 914

or Tequiv (° F) = 914 . − [914 . − Tambient (° F)][0.475 − 0.0126V + 0.240 V ]

where V is in km/h. Now the problem reduces to converting a temperature in °F to a temperature in °C, using the proper convection relation: 1.8Tequiv (° C ) + 32 = 914 . − [914 . − (18 . Tambient (° C) + 32)][ 0. 475 − 0. 0126V + 0. 240 V ]

which simplifies to Tequiv (° C) = 33.0 − (33.0 − Tambient )(0.475 − 0.0126V + 0.240 V )

where the ambient air temperature is in °C.

1-39

1-99E EES Problem 1-98E is reconsidered. The equivalent wind-chill temperatures in °F as a function of wind velocity in the range of 4 mph to 100 mph for the ambient temperatures of 20, 40, and 60°F are to be plotted, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Obtain V and T_ambient from the Diagram Window" {T_ambient=10 V=20} V_use=max(V,4) T_equiv=91.4-(91.4-T_ambient)*(0.475 - 0.0203*V_use + 0.304*sqrt(V_use)) "The parametric table was used to generate the plot, Fill in values for T_ambient and V (use Alter Values under Tables menu) then use F3 to solve table. Plot the first 10 rows and then overlay the second ten, and so on. Place the text on the plot using Add Text under the Plot menu." Tambient [F] -25 -20 -15 -10 -5 0 5 10 15 20 -25 -20 -15 -10 -5 0 5 10 15 20 -25 -20 -15 -10 -5 0 5 10 15 20 -25 -20 -15 -10 -5 0 5 10 15 20

V [mph] 10 10 10 10 10 10 10 10 10 10 20 20 20 20 20 20 20 20 20 20 30 30 30 30 30 30 30 30 30 30 40 40 40 40 40 40 40 40 40 40

20

W ind Chill Temperature

10 0 -10

W ind speed =10 mph

-20

T W in dChill

Tequiv [F] -52 -46 -40 -34 -27 -21 -15 -9 -3 3 -75 -68 -61 -53 -46 -39 -32 -25 -18 -11 -87 -79 -72 -64 -56 -49 -41 -33 -26 -18 -93 -85 -77 -69 -61 -54 -46 -38 -30 -22

-30

20 mph

-40 -50

30 mph -60 -70

40 mph

-80 -30

-20

-10

0

10

20

T ambient

60 50

Tamb = 60F

40 30

F[ iv u q e

20

T amb = 40F 10

T

0 -10

T amb = 20F -20 0

20

40

60

V [mph]

80

100

1-40

1-100 One section of the duct of an air-conditioning system is laid underwater. The upward force the water will exert on the duct is to be determined. Assumptions 1 The diameter given is the outer diameter of the duct (or, the thickness of the duct material is negligible). 2 The weight of the duct and the air in is negligible. Properties The density of air is given to be ρ = 1.30 kg/m3. We take the density of water to be 1000 kg/m3. D =15 cm Analysis Noting that the weight of the duct and the air in it is L = 20 m negligible, the net upward force acting on the duct is the buoyancy FB force exerted by water. The volume of the underground section of the duct is V = AL = ( πD2 / 4) L = [ π (0.15 m)2 /4](20 m) = 0.353 m 3

Then the buoyancy force becomes   1 kN  = 3.46 kN FB = ρgV = (1000 kg/m 3 )(9.81 m/s 2 )(0.353 m 3 )  1000 kg ⋅ m/s 2    Discussion The upward force exerted by water on the duct is 3.46 kN, which is equivalent to the weight of a mass of 353 kg. Therefore, this force must be treated seriously.

1-101 A helium balloon tied to the ground carries 2 people. The acceleration of the balloon when it is first released is to be determined. Assumptions The weight of the cage and the ropes of the balloon is negligible. Properties The density of air is given to be ρ = 1.16 kg/m3. The density of helium gas is 1/7th of this. Analysis The buoyancy force acting on the balloon is

V balloon = 4π r 3 /3 = 4 π(5 m)3 /3 = 523.6 m3 FB = ρ air gV balloon

  1N  = 5958 N = (1.16 kg/m 3 )(9.81m/s 2 )(523.6 m 3 ) 2  1 kg ⋅ m/s    The total mass is

  1.16 kg/m 3 (523.6 m 3 ) = 86.8 kg m He = ρ HeV =    7 m total = m He + m people = 86.8 + 2 × 70 = 226.8 kg

The total weight is  1N W = m total g = (226.8 kg)(9.81 m/s 2 )   1 kg⋅ m/s2  Thus the net force acting on the balloon is Fnet = FB − W = 5958 − 2225 = 3733 N

  = 2225 N  

Then the acceleration becomes a=

Fnet m total

=

3733 N  1kg ⋅ m/s 2 1N 226.8 kg 

  = 16.5 m/s 2  

m = 140 kg

1-41

1-102 EES Problem 1-101 is reconsidered. The effect of the number of people carried in the balloon on acceleration is to be investigated. Acceleration is to be plotted against the number of people, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Given Data:" rho_air=1.16"[kg/m^3]" "density of air" g=9.807"[m/s^2]" d_balloon=10"[m]" m_1person=70"[kg]" {NoPeople = 2} "Data suppied in Parametric Table" "Calculated values:" rho_He=rho_air/7"[kg/m^3]" "density of helium" r_balloon=d_balloon/2"[m]" V_balloon=4*pi*r_balloon^3/3"[m^3]" m_people=NoPeople*m_1person"[kg]" m_He=rho_He*V_balloon"[kg]" m_total=m_He+m_people"[kg]" "The total weight of balloon and people is:" W_total=m_total*g"[N]" "The buoyancy force acting on the balloon, F_b, is equal to the weight of the air displaced by the balloon." F_b=rho_air*V_balloon*g"[N]" "From the free body diagram of the balloon, the balancing vertical forces must equal the product of the total mass and the vertical acceleration:" F_b- W_total=m_total*a_up

NoPeople 1 2 3 4 5 6 7 8 9 10

30 25 20

a up [m/s^2]

Aup [m/s2] 28.19 16.46 10.26 6.434 3.831 1.947 0.5204 -0.5973 -1.497 -2.236

15 10 5 0 -5 1

2

3

4

5

6

NoPeople

7

8

9

10

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1-103 A balloon is filled with helium gas. The maximum amount of load the balloon can carry is to be determined. Assumptions The weight of the cage and the ropes of the balloon is negligible. Properties The density of air is given to be ρ = 1.16 kg/m3. The density of helium gas is 1/7th of this. Analysis In the limiting case, the net force acting on the balloon will be zero. That is, the buoyancy force and the weight will balance each other: W = mg = FB m total =

Helium balloon

5958 N FB = = 607.3 kg g 9.81 m/s 2

Thus, m people = m total − m He = 607.3 − 86.8 = 520.5 kg

m

1-104E The pressure in a steam boiler is given in kgf/cm2. It is to be expressed in psi, kPa, atm, and bars. Analysis We note that 1 atm = 1.03323 kgf/cm2, 1 atm = 14.696 psi, 1 atm = 101.325 kPa, and 1 atm = 1.01325 bar (inner cover page of text). Then the desired conversions become:

In atm:

  1 atm  = 89.04 atm P = (92 kgf/cm 2 )   1.03323 kgf/cm 2   

In psi:

 1 atm P = (92 kgf/cm 2 )   1.03323 kgf/cm2 

  14.696 psi    = 1309 psi   1 atm   

In kPa:

 1 atm P = (92 kgf/cm 2 )   1.03323 kgf/cm2 

  101.325 kPa    = 9022 kPa   1 atm   

In bars:

 1 atm P = (92 kgf/cm 2 )   1.03323 kgf/cm2 

  1.01325 bar    = 90.22 bar   1 atm   

Discussion Note that the units atm, kgf/cm2, and bar are almost identical to each other.

1-43

1-105 A barometer is used to measure the altitude of a plane relative to the ground. The barometric readings at the ground and in the plane are given. The altitude of the plane is to be determined. Assumptions The variation of air density with altitude is negligible. Properties The densities of air and mercury are given to be ρ = 1.20 kg/m3 and ρ = 13,600 kg/m3. Analysis Atmospheric pressures at the location of the plane and the ground level are Pplane = ( ρ g h) plane  1 kPa   1N   = (13,600 kg/m 3)(9.81 m/s 2 )(0.690 m)  1 kg ⋅ m/s 2  1000 N/m 2      = 92.06 kPa Pground = ( ρ g h) ground

 1N = (13,600 kg/m 3)(9.81 m/s 2)(0.753 m)  1 kg ⋅ m/s 2  = 100.46 kPa

 1 kPa   1000 N/m 2 

   

Taking an air column between the airplane and the ground and writing a force balance per unit base area, we obtain Wair / A = Pground −Pplane

h

0 Sea level

( ρ g h)air = Pground − Pplane  1N (1.20 kg/m 3 )(9.81 m/s 2 )( h)  1 kg ⋅ m/s 2  It yields h = 714 m which is also the altitude of the airplane.

 1 kPa   (100.46 − 92.06) kPa   1000 N/m 2  =  

1-106 A 10-m high cylindrical container is filled with equal volumes of water and oil. The pressure difference between the top and the bottom of the container is to be determined. Properties The density of water is given to be ρ = 1000 kg/m3. The specific gravity of oil is given to be 0.85. Analysis The density of the oil is obtained by multiplying its specific gravity by the density of water, 3

ρ = SG × ρ H2 O = (0.85)(1000 kg/m ) = 850 kg/m

3

The pressure difference between the top and the bottom of the cylinder is the sum of the pressure differences across the two fluids, ∆Ptotal = ∆Poil + ∆Pwater = ( ρgh) oil + ( ρgh) water

Oil SG = 0.85 h = 10 m Water

 1 kPa   = (850 kg/m 3)(9.81 m/s 2)(5 m) + (1000 kg/m 3 )(9.81 m/s 2 )(5 m)  1000 N/m 2    = 90.7 kPa

[

]

1-44

1-107 The pressure of a gas contained in a vertical piston-cylinder device is measured to be 250 kPa. The mass of the piston is to be determined. Assumptions There is no friction between the piston and the cylinder. Patm Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield W = PA − Patm A mg = ( P − Patm ) A

 1000 kg/m⋅ s2 (m )(9.81 m/s2 ) = (250 − 100 kPa)(30 × 10−4 m2 )  1kPa 

It yields

   

P

W = mg

m = 45.9 kg

1-108 The gage pressure in a pressure cooker is maintained constant at 100 kPa by a petcock. The mass of the petcock is to be determined. Assumptions There is no blockage of the pressure release valve. Patm Analysis Atmospheric pressure is acting on all surfaces of the petcock, which balances itself out. Therefore, it can be disregarded in calculations if we use the gage pressure as the cooker pressure. A force balance on the petcock (ΣFy = ...