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,,"0_ Thermodynamics THERl\10DYNAMICS THERMOD YNAI~lICS PROPERTIES Thermodynamics - 1 (Math-ME Ed Oct. 1998) What is the pressure 8,000 ft (2000 m) below the surface of the ocean? Neglect the compressibility factor, in Sl units. Ao 21 Mpa C. 21 Mpa B. 20 I Mpa D. 22 Mpa SOLUTION: Sea water p = '>'1 h ~ ~ o ~ = r : 2 : ~ Using typical SG of sea water equal to 1 m p = (1 x 98 J )(2000) p = 20,208 Kpa P c 20 Mpa 1 P=wh Thermodynamics - 2 (Math-ME Ed Oct. 1998) What is the temperature at which water freezes using the Kelvin scale'! A. 373 C. 278 B. 273 D. 406 SOLUTION: Freezing temperature of water is o-c. "K = °C + 273 "K = 0 + 273 "K = 273 Thermodynamics - 3 (Math-ME Ed Oct. 1998) The SI unit of temperature is: A. of C. BTU B. oK D. oR 411.', B
iii ::' Thermodynamics Thermodynamics - 4 (Math-ME Bd Oct. 1998) The pressure reading of 35 psi in kpa is: A. 342 kpa C. 273. ~~ B. 724 D. 42730 fi'] ; i:~' SOL UIION' ~ Pg 35 (101325 14) Pg 241 kpag Pabs = Pg + Palm Pabs 24l" + 10132" Pabs 342 kpaa Thermodynamics - 5 (Math-ME Bd Apr. 1999) 1 torr is equivalent to pressure _ A. 1 atm C 14. C 2 mm Hg D. 1/760 atm An5. D Thermodynamics - 6 (Math-ME Bd Apr. 1999) What is the standard temperature in the US? A. Fahrenheit C Celsius B. Rankine D. Kelvin Ans. A Thermodynamics - 7 (Math-ME Bd Apr. 1999) Given steam pressure of 900 Ib/fe, temperature of 300°F, specific volume of 5 fellb. If the specific enthalpy is 9500 ft-lb/lb, what is the internal energy per Ib of the system? A. 4400 C. 3600 B. 3900 D. 4280 ~ Thermodynamics .J St lllJTlON h = u + Pv 9500 -r- u + 900(58) II ~c 4280/t-lh/lh Thermodynamics - 8 (Math-ME Bd Oct. 1997) The barometer reads 29 inches (737 mm) of mercury. What is the absolute pressure if a vacuum gage reads 9 psi (66 kpa) in 51? A. 3202 kpa C 31 kpa B. 3304 kpa. D. 31 kpa SOLUTION: 101. Patrn (29)(---) 29 Paun 98 kpa r., Pgage + Pann pabs -66 ~ 98 PaLs 32 kpa Thermodynamics - 9 (Math-ME Bd Oct. 1997) A fluid with a vapor pressure of 0 Pa and a specific gravity of 12 is used in a barometer. If the fluids column height is 1 rn, what is the atmospheric pressure? A. 150 kpa C 144 Kpa B. 115 Kpa D. 117 Kpa SOLUTION Pressure = (specific weight)(height) Pressure rr; (12 x 9)(1 m) Pressure c= 117 Kpa Thermodynamics - 10 (Math-ME Bd Apr. 1997) What is the atmospheric pressure on a planet if the pressure is 100 kpa and the gage pressure is 10 kpa? A. 10 kpa C. so kpa B. 100 kpa D. 90 kpa
6 Thermodynamics Thermodynamics 7 A. 25 A 102 KN/m] C. 150 KN/rn J en B. 26 D. 28 B. 132/m J D. 82 KN/m J SOLUTION: 6°C = °C2 - °C I ,:'.°C =40- 6°C = 15 t, =40°C 6°F 9 6°C 5 6°F 9 15 5 6°F 27°F Thermodynamics - 16 Wafer enters the heater with 28°C and leaves at 75°C. What is the temperature change in OF? A. 7480 C. 84. B 38 D. 57. SOLUT10N: 6°C = °C 2 - °C l 6°C = 75 - 28 6°C ~c 47°C
####### ~
i\OF/6°C = 9/5 HEATER 6°F/47 = 9/ 6°F = 84 of Thermodynamics - 17 (Math-ME Bd Apr. 1996) The specific gravity of mercury relative to water is 13. What is the specific weight of mercury? The specific weight of water is 62 lb per cubic foot. SOLUTION: Specific weight of mercury (Specific gravity) (Density of H 20) Specific weight of mercury (1355)(9) Specific weight of mercury 132 KlV/m J Thermodynamics - 18 (Math-ME Bd Apr. 1996) An iron block weighs 5 N and has a volume of 200 em]. What is the density of the block? A. 988 kg/m ' C. 2550 kg/rrr' B. 1255 kg/rn' D. 800 kg/rn ' SOLUTION: Mass = 5/9 Mass = 0 kg Volume = 200 cm ' (1/1003) Volume = 2 x 10. 4 m' Mass w Volume 0. w 2xl0- 4 w 2548 kg/m' Thermodynamics - 19 The suction pressure of a pump reads 540 mm Hg vaccum. What is the absolute pressure in Kpa? A. 40 C. 60 B. 3 D. 29 SOLUTION:
rr®] J)Pa b, = Pg + Paull Pabv -540 mrn Hg + 760 rnrn Hg 1\" 220 mm Hg 1'.11" 220 X (101/760) 1',,1," 29 Kpa p.= -540 mm Hg
8 Thermodynamics 9
Thermodynamics - 20
A boiler installed where the atmospheric pressure is 755 mm Hg. Has
a pressure 12 kg/ern", What is the absolute pressure in Mpa?
A. 1 C. 1.
B. 1 D. 1,
Patm=755 mm Hg
SOLUTION:
P'b, p ~ -'- Parm
Pabs ~ 12 x (101325/1) + 755 x (101/760)
r., 127769 Kpa
r., 1 Mpa
BOILER
Thermodynamics - 21
A storage tank contains oil with specific gravity of 0 and depth of 20
m. What is the hydrostatic pressure at the bottom of the tank in
kg/ern"?
A. 1 C. 60
R. 2 D. 3.
SOLUTION.
P - w x h
P - (0 x 981 KN/m J ) ( 20 m)
P - 172 Kpa ~G" 0 'd ,'om
P J72 x (1.033/101)
P - 1 kg/em'
Thermodynamics - 22 (Math-ME Bd Oct. 1997)
A batch of concrete consisted of 200 Ibs fine aggregate, 350 Ibs coarse
aggregate, 94 Ibs, cement, and 5 gallons water. The specific gravity of
the sand and gravel may be taken as 2 and that of the cement as
3. What was the weight of concrete in place per cubic foot?
A. 1721b C. 1621b
B. 236 Ib D. 153 Ib
Th crmodynamics
SOltiTlON:
Weight of water ~ !S.'7481)(62A)
Weight of water 4L7 lb
::200 + 350
Volume of sand and gravel
2(62)
Volume of sand and gravel cc 3 ft'
94
Volume of cement
3 O(62.)
Volume of cement -; 0 flO
Total weight 2 41.'7 -i- 200 .r. 350 ,. 9..+
Total weight - 685 lbs
Total volume 5!7! - 33::26 ' CdS
Total volurne- 4 jf
Weight of concrete per CLl. it. concrete 68".'7,4.
Weight of concrete/cu. n. concrete c-. 153 lb m
Thermodynamics - 23 (Math-ME Bd Oct. 19(5)
A batch of concrete consisted of 200 Ibs fine aggregate, 350 lhs (031 'it
aggregate, 94 Ills, cement, and 5 gallons water. The specific gr:n ity cf
the sand and gnwel may be taken as 2 and that of 011' m<:,- :b
3 O. Hew m uch by weight of cement is req uired to produce one cubic
yard'.'
A. 765 C. 675
13. 657 D. 567
SOLLTIC:
Volume of water ~'7 .48l
Volume of water C 0668 ft
]no+3:<J
Volume of sand and gravtl
':' t,5( 62)
Volume of sand ;mj zravel 3 .,:' 6 nJ
94
Volume of cement
.1. i () 62 A)
12 I nerrnouynumus 1,) Thermodynamics SOLUTION' A. 50 rnrn C. 30 mm B. 25 mm D. 35 mm Q= volume flow of water flowing Q = A x vel SO}"-,UTI()r~: v ~ rn/w "' t l +- V 10,000 kgll,OOO kg/rrr' , I
V = 10 m 3 ..-- ;i --. t+r, :r;
V = 4/3 it [3 I I
10 = 4/3 it [3 OIV
i r = 1336 m r = 1336 mm 1 2722/2-1336 1 = 24 mm Thermodynamics - 28 A cylindrical t~WK is filled with water at the rate of 5 gal/min. The height of water ill the tank after minutes is 20 ft. What is the diameter of the tank'? A 30 ft C. 20 ft B. 25 ft D SOLUTION: After J5 minutes, V c, 5000(15) ~, 75,000 gal V = 75,000x(l fe/7ASI gal) V = 10,025 ft3 20. V c. n'4 DC h 10,025 = n:/4 D 2 (20) D = 25ft. ~-_Q --+ Thermodynamics - 29 Water is flowing through a 1 foot diameter pipe at the rate of W It/sec, What is the volume flow of water flowing? A. 7 ft3/sec C. 7 m3/sec B. 7 ft3/sec D. 0 mzsec Q = n/4 D 2 x vel 1ft i Q ~ n/4(1)2(10) Q = 7/sec EI%1§ J
~ 'lS
Thermodynamics - 30 A certain fluid is flowing in a 0 m x 0 m channel at the rate of 3 m/sec and has a specific volume of 0 m3/kg. Dctermine the mass of water flowing in kg/sec. A 380 kg/sec C. 375 kg/sec i3 390 kg/sec D/sec
P-O
a SOLUTION: ~3m/sr=c_ O Q = A x vel Q (05 X 0)(3) Q OA5 m ' /sec Q = fiX v OA5 = rn (0) m = 375 kg/sec
J. IICI ".uuy"u,"'U.... ..., UWS OF THERMODYNAMICS Thermodynamics - 31 (Math-ME Bd. Apr. 1998) One useful equation used is the change of enthalpy of compressible liquid with constant specific heat is: hs ub2 - h,u b l = c(T, ub1 - T b l ) + V(P - P l ) where: T. ubu = temperature at state n p. u bn = pressure at state n v =specific volume of liquid Water with enthalpy with C,ubp = 4 KJIkg_oK and v = 1 x 10 to the _3rd power cu has the following states: State I: Tsu b l = 19°C P l = 1 x 10 to the 5th power Pa State II: T, ub2 = 30°C P b2 = 0 Mpa What is the change in enthalpy from state I to state II? A. 46 Kpa/kg C. 46 KJlkg B. 56 KJlkg D. 46 KNlkg SOLUTION: h2 - h. Cp(Tz-T 1) + v(P2-P t ) h z - hi 4(30 - 19) + 0(113 - 101) hz - h, 46KJ/kg Thermodynamics - 32 What is the potential energy of a 500 kg body if it is dropped to a height of 100 m? A. 490 KJ C. 560 KJ B. 765 KJ D. 645 KJ SOLUTION: Potential Energy = m x z Potential Energy = 500 x 100 Potential Energy = 50,000 kg x 0 KNlkg Potential Energy = 490 KJ ............ .. '~,.a ...... v Thermodynamics - 33 The flow energy of 124 IiImin of a fluid passing a boundary to a system is 2 KJ/sec. What is the pressure at this point. A. 100 Kpa C. 1,000 Kpa B. 140 psi D. 871 Kpa SOLUTION: W = Pressure x Volume 2 KJ/sec x 60sec/min = P(0 m3/min) P = 967 Kpa x 14/101. P = 140 psi Thermodynamics - 34 (Math-ME Bd Apr. 1996) Steam at 1000 Ib/fr pressure and 3000 R has a specific volume of 6. fe/lb and a specific enthalpy of 9800 ft-Ib/lb. Find the internal energy per pound mass of steam. A. 5400 C. 6400 B. 3300 D. 2500 SOLUTION: h=u+Pv 9800 = u + 1000(6) u = 3300 ft-lbfllbm Thermodynamics - 35 Air and fuel enter a furnace used for home heating. The air has an enthalpy of 302 KJ/kg and the fuel has an enthalpy of 43,207 KJ/kg. The gases leaving the furnace have an enthalpy of 616 KJ/kg. There
19VI Thermodynamics Thermodynamics (80)- (v ? i] ierrnodynamics 39 (M Bd. Oct 1986) 800 T = 750 -+- 2(9)(427) 2(9)(427) earn enters a turbine stage with a enthalpy of 3628 KJ/kg at 10 m/sec h,=800Kcal/kg ---<>----, d leaves the same stage 'with an enthalpy of 2846 KJ/kg and a Y 2 652 m/sec V,=80m/s
locity of 124 m/sec. Calculate the work done by steam. ---.,
A. 77676 KJ/kg C. 567 KJ/kg B 873 KJ/kg D. 923 KJ/kg TURBiNE ~ W SOLUTION: 0=0 V, Ih, =750kcal/kg For ITl ~c 1 kg (basis) By heat balance: ~ mh.! KE[ = mh2+KE 2 + Q -rW w Thermodynamics - 41 (Power-MEEd Apr. 1998)
I
A volume of 450 cc of air is measured at a pressure of 740 rnm Hg
W m(11:. - h 2 ) + --ill (v:. 2 - v 2 2 ) - Q------""----1' l absolute and a temperature of 20°C. What is the volume in cc at 760
,
t,
2g h,=2846KJlkg w mm Hg absolute and O°C?
I
f,
w 1 (3628 - 2846, -t 1 [(70)2 -(24)2)(0) A. 516 C. 620.
2(9) t;, B. 408 D. 375.
v,
£,
"~.
776 Kllkg SOLUTION:
1 ~ CD II @
I, V,= 450c:c: V,
~! PlY' P P, = 470 P,=
I 2Y]
t, =20°C t, =O°C
er ruodynamics - 40 (Math-ME Bd Apr. 1998) ~Ii T I T 2
r
am with all enthalpy of 800 Kcal/kg enters a nozzle at a velocity of 740( 450) (760)(Y
rn/sec, Find the velocity of the steam at the exit of the nozzle if its 2 )
halpy is reduced to 750 Kcal/kg, assuming the nozzle is hortzontal (20+273) (0 + 273)
! disregarding heat losses. Take g(9,81) m/sec and J = 427 kg-
(cal Y 2 = 408 cc
A. 56124 m/s C 52 m/s
B 142 m/s D. 652 mls
Thermodynamics - 42 (Power-ME Bd \pr. 1998)
SOLUTION:
Assuming compression is according to the law PV = constant.
V I 2 V 22 Calculate the initial volume of gas at a pressure of 2 bar which will
hi-+- --- = h 2 + -=-- occupy a volume of 6 cubic meters when it is compressed to a pressure
2gJ LgJ of 42 bar.
A. 126 m ' C. 130 m'
B. 120 m' D. 136 mY
20 Thermodynamics Th ermadynumics .!l SOLUTION Thermodynamics - 45 (Math-ME Bd Apr. 1998) PI V; P2 V 2 CD Q) The mass of air in the room 3 m x 5 m x 20 m is known to be 350 kg. 2 (VI) = 42 (6) P, = ·2bar P,=42 bar Find its density. V,=6m' V,=? . A 1617kg/m J C. 1/m J VI = 126 m 3 B.1/m' D I/m ' SOLUTION Thermodynamics - 43 (Power-ME Bd Apr. 1998) V ~ 3(5)(20) V - 300m' How much heat , KJ must be transferred to 20 kg of air to increase the
temperature from 20 degrees C to 280 degrees C if the pressure is m ,meso" [J
maintained constant. Density 5m
v 20
A 2500 C. 5200
B. 2050 D. 5500 t
350
r Density -
SOLUTION: 1,=20°C HEATER 280°C=t, \ 300
-------.
m=20kg f! Density 1 kg/m'
Q = m cp (t2 - t.)
Q = 20 (1) (280 - 20) r
Q = 5200 KJ Q ~
I
Thermodynamics - 46 (Math-ME Bd Apr. 1998)
I A transportation company specializes in the shipment of pressurized
Thermodynamics - 44 (Math-ME Bd Apr. 1998)
gaseous materials. An order is received for 100 liters of a particular
If air is at pressure, p, of 3200 Iblft\ and at a temperature, T, of 800 oR, I gas at STP (32°F and 1 atm). What minimum volume tank is
f. necessary to transport the gas at 80°F and a maximum pressure of 8
what is the specific volume, v?
A. I 4 fe/lb C. 11 ftJlIb a trn?
B. 13 ft 3 /1b D. 9 fellb A 16 liters C. 10 liters
B 14 liters D. 12 liters
SOLUTION
SOLUTION
PV = mRT CD @
V PI VI T\ V 2
P, = 1atm P,=8atm
v 1'[ 1', t, = 32°F t, = 80°F
m V,= 1001i V 2 =?
RT
-
v 1(100) (8)(V2 )
P - --- ---
(53)(800) (32 t 4(0) (80 + 460)
v c
3200
v = 13/f/lb V.' /4 liters
4 Thermodvnamics Thermodvnamics 25
'hermodynamics - 51
,II automobile tire is inflated to 30 psig pressure at 50°F. After being
riven, the temperature rise to i5°F. Determine the final gage pressure
ssuming the volume remains constant.
A. 32. J 9 psig c. 0 psig
B 55 psig D. 38 psig
SOLUTION:
P 2 PI T 2 / T 1
P 2 (75 + 460)
(30+147) (50+460)
P= = 46 89 psia
Peo 46 - 14
P 2 cc32 psig
______________________ T
.crmodynamics - 52
flO 111} of atmospheric air at zero degree centigrade temperature are
.ompressed to a volume of 1 m) at 100°C, what will be the pressure of
ur in Kpa?
A. J 500 C. 2,
B. J,384 D. JOOO
SOLUTION:
----_. PlY] P2Y 2
T 1 T 2
P, = 101 Kpa (atmospheric air)
(101)(\0) (p} )()
(0 + 273) (\ 00 + 272
P2 = 1384 Kpa
Thermodynamics - 53
If 8 Ihs of a substance receives 240 Btu of heat at constant volume and
undergo a temperature change of 150°F. What is the average specific
heat of the substance during the process?
A 0 Btu/lb-oF C G50 Btu/lb-oF
B. 025 Btu/Ib-oF D 0 Btu/lb· OF
SOLUTION
Q ~.' m c, VH) m = Bibs
aF= 150
240 = 8(c,)( 150)
c, = 0 Btu/lb- OF 0=240 Btu
Thermodynamics - 54 (ME Bd, Oct. 1995)
A certain gas at 101 Kpa and 16 cC whose volume is 2 mJ are
compressed into a storage vessel of 0] m' capacity, Before admission,
the storage vessel contained the gas at a pressure and temperature of
137.,8 Kpa and 24°C; after admission the pressure hasmcreased to
l1il Kpa. What should be the final temperature of the gas in the
vessel in Kelvin'!
A. 298 C. [
B. 319 [) 4200
SOLUTION
Solving for the mass of gas which is to be compressed:
PV -r- mRT
JOJ 32 ~( 283 ) = m ,R( J6 + 2 7J )
m , ~ o 9922/R
Solving fur the mass of gas Initially contained in the
vessel:
PV mRT
137(031) = m2R(24 + 273)
m=- 01433/R
Solving for the final temperature:
Thermodynamics Thermodynamics 27
p]v, ~ rnJ RT 3
m, = rnlT rn2 Thermodynamics - 57
m, . 0/R+01438/R
m, cc 1 6/R
What is the specific volume of an air at 30°C and 150 Kpa?
1171(0) = (l/R)RT A. I ml/kg C. 12 m 3 /kg
1'3 = 319°[( B. 0 m3/kg D. 0 mJ/kg
SOLUTION:
ermodynamics - 55
PV = rnRT
e temperature of an ideal gas remains constant while the absolute v = Vim
.ssure changes from 103 Kpa to 827 Kpa. If the initial volume is v = RTfP
liters, what is the final volume?
A. 100 Ii c. 8 Ii 0(30 + 273)
B. 10 Ii D. 1000 Ii v
ISO
SOLUTION
v = 0 m 3 / kg
PI VI ~c P 2 V 2
1 034(80) . 827(VJ)
Thermodynamics - 58
V, •• 10 liters
The compression ratio of an Otto cycle is 5. If initial pressure is 100
Kpa, determine the final pressure.
-rmodynarnics - 56 A. 1000 Kpa C. 300 Kpa
B. 952 Kpa P 3 D. 100 Kpa
at is the density of air under standard condition: I
A. j] kg/m ' C. 1 kg/rn 3 f~ SOLUTION
~
.
B. I kg/rn ' D. I KN/m' . P1V 1 M = P"V,/'
PI V/ ~ P 2 V2k
SOLUTION P 2 / PI - (V I/V 2 )k
P •.• 14 psi P 2 / P I ~ (rK/ P,=100kPa
P = 101 Kpa P 2 / 100 = (5)
t c. 70°F P 2 . 951 Kpa
_v,- v, v
t ~ 21 °C
PV rnRT
w - m/R Thermodynamics - 59
w P/RT
101J25 How much work is done when 20 fe of an air initially at a pressure of
V,';::-- --.-- 15 psia and a temperature of 40°F experience an increase of pressure
0(2 LlI + 273) to 80 psi while the volume remains constant.
w .c 1,g/m 3 A 1000 Btu C zero
B 3000 Btu 0 2000 Btu
30 Thermodynamics
Thermodynamics 31
m 427 kg
SOLUTION.
c, 0 171 Kcal/kg-OC
After 10 minutes:
7500(10/60) - 427(0)(tz- 16) Since the molecular weight of ammonia is 17, then
R 8/M
t 2 = 33 cr R = 8/
R 0 KJlkg-OK
T 1 38-1-
Thermodynamics - 64 (ME Bd. Oct. 1994)
T! 311°K
The compression ratio of an ideal Otto cycle is 6:1. Initial conditions
T 2 = 100 + 273
are 101 Kpa and 20 c C. Find the pressure and temperature at the end
of adiabatic compression. T 2 = 373°K
A 1244 Kpa. 599°K PIV
B 12445 Kpa, gage, 60°C 1 '. mRT I
413(V I) = 22(0)(311)
C 1244 Kpa, 60°C VI = 8 m'
D. 1244 Kpa, 599°C
P2 V 2 = mRT2
SOLUTION: (413)(V 2) = 22(0)(373)
V 2 c 9 rrr'.
Compression ratio .C V IIV
Compression ratio = 6 W = P(V,-V
(P/P 2)
j ) = (V/V,;' W == 413(9 - 8(1)
(P/I013) = (6)14 W = 667 KJ
P 2 = /244 Kpa
T/T J = (V IIV 2 ) k. 1
T 2 /(20 + 273)= (60)14. Thermodynamics - 66 (Power-ME Bd Oct. 1997)
T 2 = 599 oK
Determine the average C, value in KJ/kg-K of a gas if 522 KJ/kg of
heat is necessary to raise the temperature from 300 OK to 8000 K
making the pressure constant:
Thermodynamics - 65 (ME Bd. Apr. 1996) A. 1 C. 1038
B. 1 D. 1026
Ammonia weighing 22 kgs is confirmed inside a cylinder equipped
with a piston has an initial pressure of 413 Kpa at 38°C. If 2900 KJ of SOLUTION
heat is added to the ammonia until its final pressure and temperature
are 413 Kpa and 100°C, respectively, what is the amount of work done For constant pressure process,
by the fluid in KJ?
A. 630 C 420 Q .• m cp (t 2 - td
B 304 D. :'102 522 = I (c p ) (800 - 3(0)
cpc UJ44 IU/kg-"J(
32 Thermodynamics Thermodynamics 33 Thermodynamics - 67 (ME Bd. Oct. 1993) Thermodynamics - 69 (Power-ME Bd Oct. 1(97) A tank contains 80 fe of air at a pressure of 350 psi; if the air is cooled A large mining company was provided with a 3 m 3 of compressed air until its pressure and temperature decreases to 200 psi and 70°F tank. Air pressure in the tank drops from 700 kpa to 180 kpa while respectively, what is the decrease in internal energy? the temperature remains constant at 28°C. What percentage has the A. +4575 C. 5552 mass orair in the tank been reduced? B. -5552 D. 0 A. 74 C. 76 8 D. 78 SOLUTION: SOLUTION: G) Q) m = PV/RT m = (200 + 14)(144)(80)/(53)(70 + 460) Solving for m., PI =700Kpa P,= 180 K;~ m = 87 lbs PI VI = m, R T I V, =3m' V, =3m' I 1,= 28°1£ _ t.,;:: 2S°t: I' I m,. 700(3) = ill] (0)(28 + 273) ill I = 24 kg ill, For constant volume process: PI/T] = P2/T 2 Before use Afterl1se T 2 = 70 + 460 Solving for rn-; T 2 = 530 0 R P 2 V 2 = m R T 2 (350 + 14) (200 + 14) 180(3) = m2 (0)(28 + 273) rn, = 6 kg T 1 530 T I = 900 0 R Percent of mass reduced: 6U = mc v (T 2 - T I ) 24 i -- 6. 6U = 87(0)(530-900) 24. 6U = -5544 Btu 74% Thermodynamics - 70 Thermodynamics - 68 (ME Bd. Oct. 1993) In a diesel cycle, the air is compressed to one-tenth of its uriginal If 10 Ibs of water are evaporated at atmospheric pressure until a volume. If the initial temperature of the air is 27"C, what is the final volume of 288 fe is computed, how much work is done? tern peratu re? A. 1680 Btu C. -610,000 ft-lb A. 420°C C. 460 ~C B. no work D. 550 ft-lb B. 440°C D. 480"C SOLUTION: SOLUTION: VI = 10/62. V 3 V 2 '=1/10 V I I=0 Vi/Vz=lO W = P(V 2 - VI) T: / T W (14 x 144)(288 - 016) I = (VI / V2) k. l W -r- 610,358 ft-lb cc
Alcorcon-pipe - nothing
Course: BS Chemical Engineering (ITE 0001-9)
University: Pamantasan ng Lungsod ng Maynila
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