This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 153 The absolute pressure in water at a speciﬁed depth is given. The local atmospheric pressure and the
absolute pressure at the same depth in a different liquid are to be determined. Assumptions The liquid and water are incompressible. Properties The speciﬁc gravity of the ﬂuid is given to be SG = 0.85. We take the density of water to be 1000 kg/m3. Then density of the liquid is obtained by multiplying its speciﬁc gravity by the density of
water, v [3:3pr =(0.85)(1000kg/m3)=850 k m3
H10 Analysis (a) Knowing the absolute pressure, the atmospheric
pressure can be determined from Patm =P—pgh
3 2 lkPa
=(145kPa)—(1000 kg/m )(9.81m/s )(5m) ‘52
/
.1961) kPa . lOOONm (b) The absolute pressure at a depth of 5 m in the other liquid is P = Pam +pgh
= (96 OkPa) +(850 kg/m3)(9.8l m/sz)(5 )[ “(Pa 2]
:13?“ kPa 1000 N/m Discussion Note that at a given depth, the pressure in the lighter ﬂuid is lower, as expected. l60E A pressure gage connected to a tank reads 50 psi. The absolute pressure in the tank is to be
determined. Properties The density of mercury is given to be p = 848.4 lbm/ﬁs. Analysis The atmospheric (or barometric) pressure can be expressed as Pm = pgh 4 lbm/ﬂ3 32 2 ﬁ/ 2 )(29 1/12 a) “bf lﬁz Pa“ 50 W
= 848. . s . ——
( X 32.21bmﬂ/sz 144 m2
= 14.29 psia Then the absolute pressure in the tank is P
\ abs = Pgagc +Patm = 50+14'29 = 64'3 pSia 172 The air pressure in a duct is measured by a mercurylmanontileéer.
For a given mercurylevel difference between the two co umns, absolute pressure in the duct is to be determined. T
3
Properties The density of mercury giventotbe p. = 13_,600 kg/m . Analysis (a) The pressure in the duct is above atmospheric pressure AIR 5
since the ﬂuid column on the duct side is at a lower level. (b) The absolute pressure in the duct is determined ﬁ'om P = Patm +pgh 2 1 N l kPa
=(100 kPa)+(13,600 kg/m3)(981m/S X00451“) #1“ W52 —100‘0'N,m2
—106 kPa ' l40C The pressure relative to the atmospheric pressure is called the gage pressure, and the pressure
relative to an absolute vacuum is called absolute pressure. A l25C A process during which the temperature remains constant is called isothermal; a process during which the pressure remains constant is called isobaric; and a process during which the volume remains
constant is called isochoric. ' l23C For a system to be in thermodynamic equilibrium, the temperature has to be the same throughout but
the pressure does not. HOWCver, there should be no unbalanced pressure forces present. The increasing
pressure with depth in a ﬂuid, for example, should be balanced by increasing weight. 134E A temperature is given in °C. It is to be expressed in °F, K, and R.,
Analysis Using the conversion relations between the various temperature scales,
T(K] = T(°C) + 273 = 18°C + 273 = 291 K
T(°F] = 1.8T(°C) + 32 = (1.8)(18) + 32 = 64.4°F
RR] = JI°F) + 460 = 64.4 + 460 = 524.4 R l37E The temperature of oil given in °F unit is to be converted to °C unit.
Analysis Using the conversion relation between the temperature scales, T(°C): T( F)—32 =150—32 1.8 1.8 = 65.6°C 138E The temperature of air given in °C unit is to be converted to °F unit. Analysis Using the conversion relation between the temperature scales, T(°F) = 1.8T(°C) +32 = (1.8)(150) +32 = 302°F 1—2C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus
the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of energy principle.
J 18E The weight of a man on earth is given. His weight on the moon is to be determined. Analysis Applying Newton's second law to the weight force gives 2
W 1801bf [32.1741bmﬁ/s]=180.41bm W= ———>m=——=
mg g 32.10%2 llbf l have the same mass on the moon. Then, his weight on the moon will be “bf J: 30.71131
32.174 lbmﬁ/sz Mass is invariant and the man wil W = mg = (180.41bm)(5.47 n/s2)[ mass and weight of the air in the room are to be 19 The interior dimensions of a room are given. The determined.
nsity of air is constant throughout the room. Assumptions The de
1.16 kg/ms. Properties The density of air is given to be p = Analysis The mass of the air in the room is m = pV =(1.16 kg/m3)(6x6><8 m3)=334.1 kg Thus, IN
= =334.1k 9.31m/s2 . =3277N
W mg ( g)( {lam/5,] 112 The acceleration of an aircraft is given in g’s. The net upward force acting on a man in the aircraft is to be determined.
Analysis From the Newton‘s second law, the force applied is IN 2 =5297 N
1kgm/s F = ma = m(6 g) = (90 kg)(6><9.81rn/sz)[ "V 115 Gravitational. acceleration g and thus the weight of bodies decreases with increasing elevation. The
percent reductlon 1n the weight of an airplane cruising at 13,000 m is to be determined. Properties The gravitational acceleration g is iven to be 9.807 m/s2 at l 2
altitude “13,000 m. g I sea evel and 9.767 m/s at an Analysis Weight is proportional to the gravitational acceleration g, and thus the
percent reduct1on 1n weight is equivalent to the percent reduction in the gravitational acceleration, which is determined from ‘
%Reduction in weight = %Reduction in g = Agx 100 = 100 = 0 41% . g 9.807 ' ‘
Tiltierfore, the airplane and the people in it will weight 0.41% less at 13,000 m
a 1tu e. Discussion Note that the weight loss at cruising altitudes is negligible. ...
View
Full
Document
This note was uploaded on 04/07/2008 for the course ME 104 taught by Professor Gomatam during the Spring '08 term at Lehigh University .
 Spring '08
 Gomatam

Click to edit the document details