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Unformatted text preview: PR 1111M 11 A pressure gauge on a cylinder of gas reads 1.05 MPa when the barometer reads 95.3
kPa. What is the pressure of the gas? SQHEMAILC: PM” = PG“, + Panama, = 1.05 x 103 kPa + 95.3 th1 = 1.1453 x 103 kPa <=<=¢=¢=¢=<= EMMA/LE
Km A block of ice at 10"C is placed in water at 25'C. E [ED ', What is the absolute temperature of the ice and of the water? What is the initial temperature difference between the ice and the water in degrees Celsius, Kelvin, '
Fahrenheit and Rankine? H AT °
W
BRQPERTIES:
LY I °
K = “C + 273.15 “F  32.0 = 'C (9l5)
R = “F + 459.67
Absolute Temperature: Ice: Tim = 10 + 273.15 = 263.2 K Water: Twaler = 25 + 273.15 = 298.2 K
AT = 25  (10) = 35°C ¢=<=<=¢=¢=<=
AT = 298.2  263.2 = 35 K ¢=c=<=c=c=c=
Tim = 10“C = 14°F = 473.7 R. Twme, = 25°C = 77°F = 536.7 R
AT = 77  (14) = 63°F <=¢=<=<=c=¢=
AT = 536.7  (473.7) = 63 R <=¢=¢=¢=¢=¢= .1.._._ PR BLE 24 K W Fl D°
Rework Problem 23 (e) after converting the density, pressure and temperature to
English units. The gas constant, R, is 55.16 fl'lbfllbm'R HE I 
AWE: Ideal Gas Law.
BBQEERIIES:
ANALYSIS;
P(VOI.) = MRT
Density is p = (V1311 ) Therefore P = pRT is the equation of state.
= L'
or p RT
1 (1617/1111 _ 5
P =1 1 MPa— 1.1 x 106 x 145. 03 x 1045: 159. 5 16111112 (psi)
T=20x9lS+32+460= 528R
R = 5.16 n Hlbf/lbm R: 661. 9 1n lbfllbm R 5
= L_ __§2t5__ 4 3 = 3 ¢=¢=¢=¢=¢=¢=
P R T 661 9 528 4. 565 x 10 lbm/in O78861bmfft EBQELEM 25 W A piston that has a mass of 2.5 kg is closely ﬁtted into a cylinder with a diameter of 0.080 m. The local acceleration of gravity is 9.80 111/32 and the local baromeuic
pressure is 0.100 MPa. A certain mass, M. is placed on top of the piston as shown in
Fig. P25. and the pressure gauge indicates 12.0 kPa. BEL; Calculate the value of the mass, M and the absolute pressure of the gas. SQHEMATIQ; W
ERQPERI IE5;
AEALXSLS;
Mp = 2.5 kg; dp = 0.08 m Pgage = 12.0 kPa Paun = 0.10 MPa
ZFveitical = 0 (Static Equilibrium) Ap = 113de!!! = 0.00503 1112 Due to absolute pressure; Fp = PamAP = (Pgage + Pam)Ap = (12.0 + PalmAP W
Due to the mass of the piston and M: Fm = Mg + Mpg = (M + 2.S)g Due to ambient pressure: Pam, Ap I'l.g!M+25! +P
M: —.l3g§§Okg Pabs=Pgage+Patm= 12.0+0. 10x 103: ll20kpa <=¢=¢=¢=<=¢= _ (M 2.5.9.307 _
=(12.0+Pmm}Ap, +1030 _(12.0)(o.00503) atm'Ap L 2 _ KMZEN & FIED; An electrical heating element is installed in an insulated tank containing water. When
the electrical current flows to the heater and the temperature of the heater exceeds that of
the water, is there any heat transfer to or from the system if the system is: (a) The water only?
(b) The tank (including the water, heating element, and the tank walls)? g R IE  ._ . _ A .
(a) If the water is the system then the heater element is part of the surroundings and there is a AT at the system boundary. Conseqmmly, heat is transferred. =‘=‘=‘=¢=¢=
(b) The heater element is now part of the system. There is no AT at the system boundary and: No heat is transferred. =====¢= R BLE 214 'II: _ W .. State whether the heat and work transfers are positive, negative or zero. The system to be
'  :3; . considered is given in italics.
it (a) A battery is discharged through a resistance wire in a light bulb.
' (b) The battery operates an electric motor which runs a table fan. The electric motor
has 100% efﬁciency (mechanical power output = electrical power in).
(c) The battery runs an electrical motor which runs a table fan. The efﬁciency of
the motor is 90%. Heat Transfer = 0 Work Transfer = The electrical work output of the battery, positive. =¢===== (b) Heat Transfer = 0
Work Transfer =  Electrical work input from battery + electrical work Output to fan. Work Transfer = 0 (since 100% efﬁcient) ¢=~==~=~=¢= (C) Heat Transfer = Negative (Since motorism efficient, it is likely that it is losing heat
to the surroundings as a result of friction)
Work Transfer = Negative (Electrical work input is greater than work output.) i=4=¢=<=<=<= E 1 W ._.. . a...
..._..——————————————— What are the signs of the work transfer associated with the following processes. The systems are identiﬁed by italics. Are the work transfers reversible? (a) A balloon that is initially ﬂat is inﬂated slowly. During this process. the fabric
of the balloon is stretched. (b) A balloon that is initially ﬂat is slowly inflated. The balloon fabric does not
distend during the process. (c) The balloon is slowly inﬂated by helium from a rigid tank containing
compressed helium. Take the helium in the balloon + the tank as the
system. (d) Helium in a balloon gradually leaks out into the atmosphere thereby deﬂating
the balloon. The pressure of helium in the balloon is greater than
atmospheric pressure as this takes place. (e) A balloon ﬁlled with helium at a pressure slightly higher than atmospheric is
pricked with a pin. The balloon explodes and the helium escapes into
the atmosphere. ’ SQHEMAJLQ:
W ERQPERTIES;
AW (3) The balloon is stretched. Work is done on the balloon.
Work Transfer is negaﬂm and mm, <=<=¢=<=c=4= (b) The balloon fabric is not stretched. No work is done on the balloon.
Work Transfer is zero (= 0) ¢=¢=¢=¢=
(c) The system boundary (which is the total mass of air in the tank and the balloon)
expands as a result of the inﬂation process.
Work Transfer = Ensiﬁm and mule. =¢=¢=¢=¢=¢=
(d) Air leaks out of the balloon. Since air is at a higher pressure inside the balloon. it
occupies less volume in the balloon than it does outside the balloon after leaking out
Thus, the system boundary expands as a result of the process.
The work done is m and Milli =4=¢=¢=4=¢=
(e) Again, the air expands as a result of the process. So. work is sitive.
But pricking the balloon causes the process to take place quickly 
therefore it is marble. ¢=¢=¢=4=¢=¢=
WE; In parts (0) and (d). the work is reversible mechanical work (PdV) but the processes are clearly
irreversible. The irreversibility is in the throttling process not in the work. i, W
I N N F D' I Consider heating a glass of water in a conventional microwave oven. If the water is the
system. what kind of energy transfer interaction occurs?  t SQHEMATIQ; 1,. i W
.I ; mums: :1 1 ANALYSIS;
1 . The energy interaction is work because heat transfer requires a temperature gradient in order to
‘ occur and there is no such temperature gradient between the food and the microwave oven. " I e=e=e=e=e=¢=
I
l i W In a standard oven. however. the energy interaction is heat transfer from the heating coils to the food.
in this case the work is done by the electmmagnetic radiation. W
K F ° ' A 20 kg. mass is placed on a horizontal frictionless surface. The mass is accelerated at a constant value of 5.0 m/82 by a horizontal force. Taking the mass as the system,
determine the work done in a 3 seconds interval of time. SQHEMAIIQ
ASSUMPTIQNS; PRQPERTIES; answers; _
F = Ma = (20)(5.0) = 100 N w = F(Dtstance) Distance = (Average velocity)'(time) = 3
Initial velocity: V; = 0.0; Final velocity: Vf = at
Therefore: 5 = 1/2 at2 ; s = 1/2 (5.0)(3.0)2 = 22.50 m W = Fs = (100)«(22.5) = 2250.0 I ¢=¢=<=¢=<=¢=
It should be noted that all the work goes into kinetic energy thus:
w = Mv2 [2 The ﬁnal velocity is V = at = (5.0)(3.0) = 15.0 m/s W = (20.0)(15)2 [2 = 2250.0 J ¢=¢=¢=¢=¢=¢= ...
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 Fall '07
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