Thermo

THERMODYNAMICS

1If air is at pressure p of 3200 lbfft^2 and at a temperature T of 800 ˚R What is the specific volume v (R=5303 ft-lbflbm-˚R and air can be modeled as an ideal gas)A98 ft^3lbmB112 ft^3lbmC1333 ft^3lbmD142 ft^3lbm Formula pv = RT v = RT pans C

2 Steam at 1000 lbfft^2 pressure and 300˚R has specific volume of 65 ft^3lbm and a specific enthalpy of 9800 lbf-ftlbm Find the internal energy per pound mass of steam

A2500 lbf-ftlbm B3300 lbf-ftlbm C5400 lbf-ftlbm D6900 lbf-ftlbm

Formula h = u + pV u = h ndash pVans B

3 30 lbm of air are contained at 25 psia and 100 ˚F given that Rair = 5335 ft-lbflbm- ˚F what is the volume of the containerA107 ft^3B147 ft^3C15 ft^3D249 ft^3

Formula use the ideal gas law pV = mRT T = (100 + 460) ˚R V = mRT p

ans D

4 The compressibility factor x is used for predicting the behavior of non-ideal gases How is the compressibili ty factor defined relative to an ideal gas (subscript ldquocrdquo refers to critical value)A z = P PcB z = pV RTC z = T TcD z = RT pV

Hint for an real gases the compressibility factor x is an dimensionless constant given by pV= zRT Therefore z = pV RT

ans B

5 From the steam table Determine the average constant pressure specific heat (c) of steam at 10 kPa and 458 ˚C A179 kJ kg-˚CB1028 kJ kg-˚CC3057 kJ kg-˚CD 1001 kJ kg-˚C

Formula ∆h = c∆T From the steam table At 477 ˚C h= 25881 kJ kg At 438 ˚C h= 25811 kJ kg ans A

6 A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water vapor at 100 kPa Calculate the internal energy of the system using the steam tableA 5 x10^5 kJB 8 x10^5 kJC 1 x10^6 kJD 2 x10^6 kJ

Formula from the steam tablevƒ = 0001043 m^3 kgvg = 16940 m^3 kgu ƒ = 4173 kJkg ug = 2506 kJkgformula M vap = V vapvg

M liq = V liq vƒ u = uƒ M liq + ug M vap

ans D

7 A vessel with a volume of cubic meter contains liquid water and water vapor ion equilibrium at 600 kPa The liquid water has mass of 1kg using the steam table calculate the mass of the water vaporA 099kgB 157 kgC 254 kgD316 kg

Formula from the steam table at 600 kPavƒ = 0001101 m^3 kgvg = 03157 m^3 kg ansD

Vtot = mƒ vƒ + mg vgmg = (tot- mƒ vƒ) vg

8Calculate the entropy of steam at 60 psiawith a quality of 08A 04274 BTUlbm-˚RB 07303 BTUlbm-˚RC 11577 BTUlbm-˚RD 12172 BTUlbm-˚R

Formula from the steam table at 60 psiasƒ = 04274 BTUlbm-˚R

sƒg = 12172 BTUlbm-˚R)

s = sƒ + x sƒg where x = is the quality ans C

9 Find the change in internal energy of 5 lbm of oxygen gas when the temperature changes from 100 ˚F to 120 ˚F cv= 0157 BTUlbm-˚RA147 BTUB157 BTUC 168 BTUD 159 BTU

Formula ∆U= mcv∆TAns B

10 Water (specific heat cv= 42 kJ kg ∙ K ) is being heated by a 1500 W heater What is the rate of change im temperature of 1kg of the waterA 0043 KsB 0179 Ks C 0357 KsD 150 Ks

Formula Q = mcv (∆T) Ans B

11 A system weighing 2 kN Determine the force that accelerate if to 12 ms^2a vertically upward when g= 97 ms^2

A 447423 NB548423 NC449523 ND548823 N

Formula F = mk (a +g)ans A

12 refer to problem 11 determne the force that accelerates if to 12 ms^2horizontally along a frictionless plane A247423 NB257423 NC347423 ND237423 N

Formula M = wk g F = ma k

ans A

13 A problem Drum ( 3 ft dia 6 ft height ) is field with a fluid whose density is 50 lbft^3 Determine the total volume of the fluid A 4241 ft^3B4435 ft^3C4563 ft^3D4123 ft^3

Formula Vf = (pi d^2 h) 4 ans A

14 What is the resulting pressure when one pound of air at 15 psia and 200 ˚F is heated at constant volume to 800 ˚FA15 psiaB286 psiaC364 psiaD521 psia

Formula T1p1 = T2p2p2 = p1T2 T1ansB

15 What horsepower is required to isothermally compress 800 ft^3 of Air per minute from 147 psia to 120 psiaA 28 hpB108 hpC256 hpD13900 hp]

Formula W= p1V1 ln (p1p2)Power = dW dtans B

16 What is the equation for the work done by a constant temperature systemA W = mRT ln(V2-V1)B W = mR ( T2-T1 ) ln( V2V1)C W = mRT ln (V2V1)D W = RT ln (V2V1)]

Formula W=int pdV lim 12 р = mRT Vans C

17 Twenty grams of oxygen gas are compressed At a constant temperature of 30 ˚C to 5 of their original volume What work is done on the systemA824 calB924 calC944 calD1124 cal

Formula W = -mRT ln (V2V1)Where R = (198 calgmoleK) (32 ggmole)ans D

18 Helium ( R= 04698 BTUlbm-˚R ) is compressed isothermally from 147 psia and 68 ˚F The compression ratio is 14 Calculate the work done by the gasA ndash1454 BTUlbmB-364 BTUlbmC-187 BTUlbmD467 BTUlbm

Formula W = RT ln (V2V1)ansB

19Gas is enclosed in a cylinder with a weighted piston as the stop boundary The gas is heated and expands from a volume of 004 m^3 to 010 m^3 at a constant pressure of 200kPaCalculate the work done by the systemA 8 kJB 10 kJC12 kJD14 kJ

Formula W = p (V2-V1)

ansC

20 refer to problem no13 Determine the specific volumehellipA 002 ft^3lbmB 005 ft^3lbmC 10 ft^3lbmD 12 ft^3lbm

Formula Vf = ( pi d^2 h) 4Pf = mf vfSpecific volume= Vf mf ans A

21What is the weight of a 66-kgm man at standard condition(Formula Fg = mg k)a 66 kgf

b 66 kgm

c 66 lbm

d 66 gf

22What is the specific weight of ater at standard condition(Formula γ = ρg k)a 1000 kgmm3

b 98066 ms2

c 1000 kgfm3

d None of the above

23746 degR = ______ degFa 254b 345c 286d None of the above

24A 30-m vertical column of fluid (density 1878 kgm3) is located where g= 965 mps2 Find the pressure at t6he base of the column

(Formula pg = gρhgk )a 543680 Nm2

b 54368 kPa (gage)c Both a amp bd None of the above

25Ten cu ft of air at 300 psia 400degF is cooled to 140degF at constant volume What is the final pressure(formula p2 = p1T2 T1 )a 0 b 209 psiac - 420 psiad None of the above

26876 degR = _____ degFa 335b 416c 400

d None of the abovedi

27There are 136 kg of gas for which R = 377 Jkgk and k=125 that undergo a nonflow constant volume process from p1 = 5516 kPa and t1 = 60degC to p2 = 1655 kPa During the process the gas is internally stirred and there are also added 1055 kJ of heat Determine t2

(Formula T2 = T1p2 p1 )a 999 Kb 888 Kc 456Kd One of the above

285 atm = ____ mmHga 8300b 3800c 3080d None of the above

29A certain gas with cp = 0529 BtulbdegR and R = 962 ftlblbdegR expands from 5 cu ft and 80 degF to 15 cu ft while the pressure remains constant at 155 psia Compute for T2

(Formula T2 = T1V2 V1 )a 460 degRb 270 degRc 1620 degRd None of the above

30In the above problem compute for the mass(Formula m = p1V1 RT1 )a 02148 lbb 02134 lbc 01248 lbd None of the above

31710degR= ______ degCa 214b 121c 213d None of the above

32212 degF = _____ degC a 200b 150c 100d None of the above

33Let a closed system execute a state change for which the heat is Q = 100 J and work is W = -25 J Find ∆E(Formula ∆E = Q- W )a 125 Jb 123 Jc 126 Jd None of the above

34A pressure gage registers 50 psig in a region where the barometer is 1425 psia Find absolute pressure in psia Pa(Formula p = patm + pg )a 433 kPab 443 kPac 343 kPad None of the above

35 A mass of 5 kg is 100 m above a given datum where local g = 975 ms2 Find the gravitational force in newtons(Formula Fg = mgk )a 4875 Nb 50 Nc 45 Nd None of the above

36 In the above problem find the potential energy of the mass with respect to datum(Formula P = mgzk )a 4875 jb 051 jc 046 jd None of the above

37 The combined mass of car and passengers travelling at 72 kmhr is 1500 kg Find the kinetic energy of this combined mass(Formula K = mv2 2k )a 300 kJb 200 kJc 500 kJd None of the above

38 14696 psia = _____ mmHga 760b 1c 350d None of the above

39 212 degC = _____ Ka 485b 435c 498d None of the above

40212 degF = _____ R a 567b 672c 700d None of the above

41 An automobile tire has a gauge pressure of 200kpa at 0degC assuming no air leaks and no change of volume of the tire what is the gauge pressure at 35ordmC

a 298645b 398109c 291167d 281333

Pg = Pabs - Patm

42 An ideal gas at 45psig and 80ordmF is heated in the close container to 130ordmF What is the final pressurea 6510psib 6511psic 6523psid 6116psi

P1V1T1 = P2V2T2 V = Constant

43 A wall of the firebrick has an inside temperature of 313ordmF and an outside temperature of 73ordmF What is the difference in the surface temperature in Rankin

a 70b 68c 72d 94

ordmR = ordmF + 460

44 What is the force required to accelerate a mass of 30kg at a rate of 15mssup2a 460Nb 380Nc 560Nd 450N

F = ma

45 How much does an object having the mass of 100kg weight in newtona 981Nb 991Nc 9816Nd 9801N

F = ma

46 The volume of the gas held at constant pressure increases 4cmsup2 at 0degC to 5cmsup2 What is the final pressurea 6865ordmCb 6825ordmCc 7001degCd 791ordmC

t2 = T2 ndash T1

47 A certain gas with cp = 0529BtulbdegR and R = 962ftlbordmR expands from 5ft and 80ordmF to 15ft while the pressure remains constant at 155 psia

a T2 =1620ordmR H = 12283 Btub T2 = 2degR H = 12283 Btuc T2 = 2620ordmR H = 12283 Btud T2 = 1degR H = 12283 Btu

T2 = V2 (t2)V1 and H = mcp (T2-T1)

48 A vacuum is connected to a tank reads 3kpa at a location w the barametric pressure reading is 75mmhg Determined the Pabsolute in the tank

a 70658kpab 68kpac 5878kapd None of the above

Pabs = Patm ndash Pvacuum

49 Calculatea Mass flow rate in lbhrb The velocity at section 2 in fps

a 800000lbhr625ftsb 900000lbhr625ftsc 888000lbhr269ftsd 700000lbhr269ftsm = A1VV1

50 A 600kg hammer of a pile driver is lilted 2m the pilling head What is the change of potential energy If the hammer is realest What will be its velocity and the instant if it sticks the pilling

a 10772N-m and 526msb 13200N-m and 526msc 11772N-m and 626msd 11772N-m and 526ms

PE = mgo(Z)gc

51 A bayabas falls from a branch 5m above the ground with what speed in meter per second does it strike the ground assume g = 10mssup2

a 11msb 12msc 13msd 10ms

KE = mV22gc

52 While swimming a depth of 13m in a fresh water lake a fish emits an air bubble of volume 2mmsup2 atmospheric pressure is 100kpa what is the original pressure of the bubble

a 21717kpab 119kpac 32615kpad 210kap

Pabs = Pg + Patm

53 Oxygen at 15ordmC and 103Mpa gauge pressure occupies 600L What is the occupied by the oxygen at 828Mpa gauge pressure and 35ordmC

a 78932Lb 79632Lc 699Ld 588L

V2 = P1V1T1P2

54 Water is flowing through a 1 foot diameter pipe at the rate of 10ftsec What is the volume flow rate of water in ftsup3seca 785b 685c 885d 585

V = Aν

55 A certain fluid is flowing an a 05mx03 channel at the rate of 3ms and has a specific volume of 0012msup3kg Determined the mass of water flowing in kgs

a 267kgsb 378kgsc 375kgsd 4565kgs

m = AνV

56 A gas having a volume of 100ftsup3 at 27ordmC is expanded to 120ftsup3 by heated at constant pressure to what temperature has it been heated to have this new volume

a 87Cb 85Cc 76Cd 97C

t2 = T2 ndash T1

57 Water flow to a terminal 3mm diameter and has an average speed of 2msWhat is the rate of flow in cubic metermma 00001msup3minb 0076 msup3minc 0085 msup3mind 0097 msup3min

58 Water flowing at a 6ms through a 60mm pipe is suddenly channeled into a 30mm pipe What is the velocity in the small pipe


59 A vertical column of water will be supported to what height by standard atmospheric pressurea 339ftb 45ftc 67ftd 2546ft

ho = PoYo

60 A fluid flows in a steady manner between two section in a flow line at section 1 A 1 = 1ftsup2 V1 = 100fpm volume1 of 4ftsup3lb at sec2 A2 = 2ftsup2 p = 020lbftsup3 calculate the velocity at section 2

a 625fpmb 567fpmc 356fpmd None of the above

61 The weight of an object is 50lb What is its mass at standard conditiona 50 lbm

b 60 lbm

c 70 lbm

d 80 lbm

formula m = Fgk g

62 A vertical column of water will be supported to what height by standard atmospheric pressure If the Yw = 624lbft3 po = 147 psi

a 449 ftb 339 ftc 229 ftd 559 ft

formula ho = po Yw

63 For a certain gas R= 320JkgK and cv= 084kJkgKc Find ka 136b 137c 138d 139

formula k= R cv +1

64 Ten cuft of air at 300 psia and 400degF is cooled to 140degF at constant volume What is the transferred heata-120Btub -220Btuc-320Btud -420Btu

formula Q= mcv(T2-T1)

65 Utilizing the answer to the previous problem estimate the overall or average increase in temperature (ΔT) of the concrete roof from the energy absorbed from the sun during a 12 hour day Assume that all of the radiation absorbed goes into heating the roof The specific heat of concrete is about 900 Jkg and the density is about 2300 kgm3

a 79 degCb 89degCc 99degCd 109degC

formula ΔQ = m c ΔT

66 The concrete roof of a house is 10 m by 8 m and 10 cm thick (4) Estimate the total heat the roof would absorb over the 12 day

a 13 x 108 Jb 23 x 108 Jc 33 x 108 Jd 43 x 108 J

formula ΔQ = ΔQΔt x Δt

67 The value for the ΔU of a system is -120 J If the system is known to have absorbed 420 J of heat how much work was done

a -540 Jb -640 Jc -740 Jd -840 J

formula ΔU = q + w

68 When the pressure on a 1 kg liquid is increased isothermally from 1 bar to 3000 bar the Gibbs free energy increases by 360 kJ Estimate the density of the liquid

a 066 kg liter -1b 077 kg liter -1

c 088 kg liter -1

d 099 kg liter -1

solution

69 A car whose mass is 2 metric tons is accelerated uniformly from stand hill to 100kmph in 5 sec Find the driving force in Newtonrsquos

a 11120Nb 11320Nc 11420Nd 11520N

formula F= ma k

70 An ideal gas of volume 1 liter and pressure 10 bar undergoes a quasistatic adiabatic expansion until the pressure drops to 1 bar Assume γ to be 14 what is the final volume

a 318 lb 418 lc 518 ld 618 l

solution

71 Two masses one of the 10kg and the other unknown are placed on a scale in a region where g = 967 msec2 The combined weight of these two masses is 17406 N Find the unknown mass in kg

a 20kgb 19kgc 18kgd 17kg

formula m= Fg k g

72 The flow energy of 5ft3 of a fluid passing a boundary to a system is 80000 ft-lb Determine the pressure at this pointa 222psib 333psic 444psid 111psi

formula Ef = pV

73 Find и for steam at 100psia and 600degFIf h = 13296 and v = 6216a1214Btu lbb 1234 Btu lbc 1342 Btu lbd 1324 Btu lb

formula и = h ndash pv J

74 What mass of nitrogen is contained in a 10ft3 vessel at a pressure of 840 atm and 820degR Make a computation by using ideal gas equation

a 194 lbb 214 lbc394 lbd 413 lb

formula m= pV RT

75 A rotary compressor receives 6m3 min of a gas (R=410J kgK cp= 103kJ kgK k= 167) at 105 kPaa 27degC and delivers it at 630 kPaa ΔP = 0 ΔK= 0 Find the work if the process is isentropic

a -1664kJminb -1774 kJminc -1884 kJmind -1994 kJmin

formula WSF = Q - ΔH m=p1V1 RT1 T2= T1(p2p1)(k-1)k

76 A carnot power cycle operates on 2 lb of air between the limits of 71degF and 500degF The pressure at the beginning of isothermal expansion is 400 psia and at the end of isothermal expansion is 185psig Determine the volume at the end of isothermal compression

a7849 ft3

b 7850 ft3

c 7851 ft3

d 7852 ft3

formula V= mRT P P3= P2[T3 T2]

77 During a polytropic process 10 lb of an ideal gas whose R= 40ftlblbR and cp = 025BtulbR changes state fro 20psia and 40degF to 120 psia and 340degF Determine n

a 1234b 1345c 1456d 1356

formula [ p2p1 ]n-1 n = T2 T1

78 A perfect gas has a value of R= 3192 J kfK and k= 126 If 120 kJ are added to 227 kfg of this gas at constant pressure when the initial temp is 322degC Find T2

a 3394 Kb 4494 Kc 5594Kd 6694K

formula cp = kR k-1 Q= mcp(T2-T1)

79 A certain gas with cp= 0529Btu lb degR and R = 962ftlblb degR expands from 5 cu ft and 80degF to 15 cu ft while the pressure remains constant at 155psia Compute for T2

a1520degRb 1620degRc 1720degRd 1820degR

formula T2= T1 V2 V1

80 A System has a temperature of 250degF Convert this Value to degRa 740degRb730degRc 720degRd 710degR

formula degR= degF + 460

81) Steam with a specific volume of 009596 msup3kg undergoes a constant pressure process at 170 MPa until the specific volume becomes 013796 msup3kg What are (a) the final temperature (b) ∆u (c) W (d) ∆s and (e) Q

a) 2654degC 4307kJkg 714kJkg 10327kJ(kg)(K) 5021 kJkgb) 2042degC -7032 kJkg -8415 kJkg -17505 kJ(kg)(K) -7874 kJkgc) 3042degC -8032 kJkg -8915 kJkg -27505 kJ(kg)(K) -9874 kJkgd) 2794degC 4397kJkg 794kJkg 30327kJ(kg)(K) 6021 kJkg

Ans (a)

82) Steam with an enthalpy of 28435 kJkg undergoes a constant pressure process at 09 MPa until the enthalpy becomes 20561 kJkg What are (a) the initial temperature or quality (b) ∆u (c) W (d) ∆s and (e) Q


Ans (b)Formula of 1 and 2 ∆u = u2 ndash u1 W = p(v2-v1) ∆s = s2-s1 Q = h2 ndash h1

83) At throttling calorimeter receives steam from a boiler drum at 011 MPa and is superheated by 10 degrees If the boiler drum pressure is 155 MPa what is the quality of the steam generated by the boiler

a) 9520b) 7010c) 6560d) 9556

Ans (a)Formula h1 = hf1 + x1hfg1

84) A steam calorimeter receives steam from a pipe at 01 MPa and 20degSH For a pipe steam pressure of 2 MPa what is the quality of the steam

a) 9556b) 7010c) 9520d) 8510


85) A 1-kg steam-water mixture at 10 MPa is contained in an inflexible tank Heat is added until the pressure rises to 35 MPa and the temperature to 400deg Determine the heat added

a) 13787 kJb) 13485 kJc) 12787 kJd) 12465 kJ

Ans (a)Formula Q = (h2 ndash p2v2) ndash (h1 ndash p1v1)

86) Water vapor at 100 KPa and 150degC is compressed isothermally until half the vapor has condensed How much work must be performed on the steam in this compression process per kilogram

a) -13847 kJb) 13847 kJc) -23847 kJd) 23847 kJ

Ans (a)

87) Wet steam at 1 MPa flowing through a pipe is throttled to a pressure of 01 MPa If the throttling temperature is 110degC What is the quality of the steam in the pipe

a) 96b) 86c) 76d) 66

Ans (a)

88) Steam is throttled to 01 MPa with 20 degrees of superheat (a) What is the quality of throttled steam if its pressure is 075 MPa (b) What is the enthalpy of the process

a) 976 2713 kJkgb) -976 2713 kJkgc) 876 3713 kJkgd) -876 3713 kJkg

Ans (a)89) The pressure gauge on a 2000 msup3 tank of oxygen gas reads 600kPa How much volumes will the oxygen occupied at pressure of the outside air 100kPa

a) 140265 msup3b) 150265 msup3c) 130265 msup3d) 160265 msup3

Ans (a)Formula P1V1T1 = P2V2T2

90) Assuming compression is according to the Law PV = C Calculate the initial volume of the gas at a pressure of 2 bars wc will occupy a volume of 6msup3 when it is compressed to a pressure of 42 Bars

a) 130msup3b) 136msup3c) 120msup3d) 126msup3

Ans (d)Formula P1V1T1 = P2V2T2

91) A Gas tank registers 1000kPa After some gas has been used the gauge registers 500kPa What percent of the gas remains in the tank

a) 6440b) 7460c) 5840d) 5460

Ans (d)Formula Pabs = Patm + Pgage amp = P2P1 100

92) The volume of a gas under standard atmospheric pressure amp 76cmHg is 200msup3 What is the volume when pressure is 80cmHg if the temperature is unchanged

a) 180 insup3b) 170 insup3c) 160 insup3d) 190 insup3

Ans (d)Formula P2V2 = P1V1

93) While swimming at depth of 120m in a fresh water lake A fish emits an air bubbles of volume 20mmsup3 atmospheric pressure is 100kPa What is the pressure of the bubble

a) 2177kPab) 3177kPac) 4177kPad) 5177kPa

Ans (a)Formula P = δh

94) How many joules of work is the equivalent of 15000 cal of heat

a) 62850 joulesb) 357995 joulesc) 1499581 joulesd) 1500419 joules

Ans (a)Formula J = WorkHeat

J = mechanical equivalent of heat whose value is 419 joulescalorie

95) Two thick slices of bread when completely oxidized by the body can supply 200000 cal of heat How much work is this equivalent to

a) 4190000 joulesb) 8390000 joulesc) 839000 joulesd) 419 000 joules

Ans (d)Formula J = WorkHeat


96) 3 horsepower (hp) = _____________watts

a) 1492 wattsb) 2238 wattsc) 746 wattsd) 2238 kilowatts

Ans (b)Formula 1hp = 746 watts

97) How many Newtonrsquos (N) in 900000 dynes

a) 8 Newtonrsquosb) 9 Newtonrsquosc) 7 Newtonrsquosd) 6 Newtonrsquos

Ans (b)Formula 1 Newton (N) = 100000dynes

98) Calculate the power output in horsepower of an 80-kg man that climbs a flight of stairs 38 m high in 40 s

a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp

Ans (b)Formula Power = Fdt = mght

F = W = mgd = h

99) How many calories of heat will be needed to raise the temperature of 200 g of iron from 27degC to 80degC (c = 011 calg degC)

a) 116kcalb) 2166calc) 316kcald) 4166cal

Ans (a)Formula H = mc∆T

100) 100g of iron was heated to 100degC and mixed with 22g of water at 40degC The final temperature of the mixture was 60degC Show that the heat given off by the iron equals the heat absorbed by the water

a) 440 calb) 540 calc) 340 cald) 640 cal

Ans (a)Formula H (given off by iron) = H (absorbed by water) mc∆T(iron) = mc∆T(water)

6 A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water vapor at 100 kPa Calculate the internal energy of the system using the steam tableA 5 x10^5 kJB 8 x10^5 kJC 1 x10^6 kJD 2 x10^6 kJ

Formula from the steam tablevƒ = 0001043 m^3 kgvg = 16940 m^3 kgu ƒ = 4173 kJkg ug = 2506 kJkgformula M vap = V vapvg

M liq = V liq vƒ u = uƒ M liq + ug M vap

ans D

7 A vessel with a volume of cubic meter contains liquid water and water vapor ion equilibrium at 600 kPa The liquid water has mass of 1kg using the steam table calculate the mass of the water vaporA 099kgB 157 kgC 254 kgD316 kg

Formula from the steam table at 600 kPavƒ = 0001101 m^3 kgvg = 03157 m^3 kg ansD

Vtot = mƒ vƒ + mg vgmg = (tot- mƒ vƒ) vg

8Calculate the entropy of steam at 60 psiawith a quality of 08A 04274 BTUlbm-˚RB 07303 BTUlbm-˚RC 11577 BTUlbm-˚RD 12172 BTUlbm-˚R

Formula from the steam table at 60 psiasƒ = 04274 BTUlbm-˚R

sƒg = 12172 BTUlbm-˚R)

s = sƒ + x sƒg where x = is the quality ans C

9 Find the change in internal energy of 5 lbm of oxygen gas when the temperature changes from 100 ˚F to 120 ˚F cv= 0157 BTUlbm-˚RA147 BTUB157 BTUC 168 BTUD 159 BTU

Formula ∆U= mcv∆TAns B




A 447423 NB548423 NC449523 ND548823 N




ans A















ansC




b 66 kgm

c 66 lbm

d 66 gf


b 98066 ms2

c 1000 kgfm3

d None of the above






26876 degR = _____ degFa 335b 416c 400



















a 298645b 398109c 291167d 281333

Pg = Pabs - Patm




a 70b 68c 72d 94

ordmR = ordmF + 460


F = ma


F = ma


t2 = T2 ndash T1



T2 = V2 (t2)V1 and H = mcp (T2-T1)








PE = mgo(Z)gc



KE = mV22gc



Pabs = Pg + Patm


a 78932Lb 79632Lc 699Ld 588L

V2 = P1V1T1P2


V = Aν



m = AνV



t2 = T2 ndash T1





ho = PoYo




b 60 lbm

c 70 lbm

d 80 lbm

formula m = Fgk g


a 449 ftb 339 ftc 229 ftd 559 ft

formula ho = po Yw


formula k= R cv +1







a 13 x 108 Jb 23 x 108 Jc 33 x 108 Jd 43 x 108 J



a -540 Jb -640 Jc -740 Jd -840 J

formula ΔU = q + w



c 088 kg liter -1

d 099 kg liter -1

solution


a 11120Nb 11320Nc 11420Nd 11520N

formula F= ma k


a 318 lb 418 lc 518 ld 618 l

solution



formula m= Fg k g


formula Ef = pV




a 194 lbb 214 lbc394 lbd 413 lb

formula m= pV RT





a7849 ft3

b 7850 ft3

c 7851 ft3

d 7852 ft3



a 1234b 1345c 1456d 1356



a 3394 Kb 4494 Kc 5594Kd 6694K









Ans (a)





a) 9520b) 7010c) 6560d) 9556



a) 9556b) 7010c) 9520d) 8510



a) 13787 kJb) 13485 kJc) 12787 kJd) 12465 kJ



a) -13847 kJb) 13847 kJc) -23847 kJd) 23847 kJ

Ans (a)


a) 96b) 86c) 76d) 66

Ans (a)










a) 6440b) 7460c) 5840d) 5460























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h










A 447423 NB548423 NC449523 ND548823 N




ans A















ansC




b 66 kgm

c 66 lbm

d 66 gf


b 98066 ms2

c 1000 kgfm3

d None of the above






26876 degR = _____ degFa 335b 416c 400



















a 298645b 398109c 291167d 281333

Pg = Pabs - Patm




a 70b 68c 72d 94

ordmR = ordmF + 460


F = ma


F = ma


t2 = T2 ndash T1



T2 = V2 (t2)V1 and H = mcp (T2-T1)








PE = mgo(Z)gc



KE = mV22gc



Pabs = Pg + Patm


a 78932Lb 79632Lc 699Ld 588L

V2 = P1V1T1P2


V = Aν



m = AνV



t2 = T2 ndash T1





ho = PoYo




b 60 lbm

c 70 lbm

d 80 lbm

formula m = Fgk g


a 449 ftb 339 ftc 229 ftd 559 ft

formula ho = po Yw


formula k= R cv +1







a 13 x 108 Jb 23 x 108 Jc 33 x 108 Jd 43 x 108 J



a -540 Jb -640 Jc -740 Jd -840 J

formula ΔU = q + w



c 088 kg liter -1

d 099 kg liter -1

solution


a 11120Nb 11320Nc 11420Nd 11520N

formula F= ma k


a 318 lb 418 lc 518 ld 618 l

solution



formula m= Fg k g


formula Ef = pV




a 194 lbb 214 lbc394 lbd 413 lb

formula m= pV RT





a7849 ft3

b 7850 ft3

c 7851 ft3

d 7852 ft3



a 1234b 1345c 1456d 1356



a 3394 Kb 4494 Kc 5594Kd 6694K









Ans (a)





a) 9520b) 7010c) 6560d) 9556



a) 9556b) 7010c) 9520d) 8510



a) 13787 kJb) 13485 kJc) 12787 kJd) 12465 kJ



a) -13847 kJb) 13847 kJc) -23847 kJd) 23847 kJ

Ans (a)


a) 96b) 86c) 76d) 66

Ans (a)










a) 6440b) 7460c) 5840d) 5460























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h

















ansC




b 66 kgm

c 66 lbm

d 66 gf


b 98066 ms2

c 1000 kgfm3

d None of the above






26876 degR = _____ degFa 335b 416c 400



















a 298645b 398109c 291167d 281333

Pg = Pabs - Patm




a 70b 68c 72d 94

ordmR = ordmF + 460


F = ma


F = ma


t2 = T2 ndash T1



T2 = V2 (t2)V1 and H = mcp (T2-T1)








PE = mgo(Z)gc



KE = mV22gc



Pabs = Pg + Patm


a 78932Lb 79632Lc 699Ld 588L

V2 = P1V1T1P2


V = Aν



m = AνV



t2 = T2 ndash T1





ho = PoYo




b 60 lbm

c 70 lbm

d 80 lbm

formula m = Fgk g


a 449 ftb 339 ftc 229 ftd 559 ft

formula ho = po Yw


formula k= R cv +1







a 13 x 108 Jb 23 x 108 Jc 33 x 108 Jd 43 x 108 J



a -540 Jb -640 Jc -740 Jd -840 J

formula ΔU = q + w



c 088 kg liter -1

d 099 kg liter -1

solution


a 11120Nb 11320Nc 11420Nd 11520N

formula F= ma k


a 318 lb 418 lc 518 ld 618 l

solution



formula m= Fg k g


formula Ef = pV




a 194 lbb 214 lbc394 lbd 413 lb

formula m= pV RT





a7849 ft3

b 7850 ft3

c 7851 ft3

d 7852 ft3



a 1234b 1345c 1456d 1356



a 3394 Kb 4494 Kc 5594Kd 6694K









Ans (a)





a) 9520b) 7010c) 6560d) 9556



a) 9556b) 7010c) 9520d) 8510



a) 13787 kJb) 13485 kJc) 12787 kJd) 12465 kJ



a) -13847 kJb) 13847 kJc) -23847 kJd) 23847 kJ

Ans (a)


a) 96b) 86c) 76d) 66

Ans (a)










a) 6440b) 7460c) 5840d) 5460























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h










b 66 kgm

c 66 lbm

d 66 gf


b 98066 ms2

c 1000 kgfm3

d None of the above






26876 degR = _____ degFa 335b 416c 400



















a 298645b 398109c 291167d 281333

Pg = Pabs - Patm




a 70b 68c 72d 94

ordmR = ordmF + 460


F = ma


F = ma


t2 = T2 ndash T1



T2 = V2 (t2)V1 and H = mcp (T2-T1)








PE = mgo(Z)gc



KE = mV22gc



Pabs = Pg + Patm


a 78932Lb 79632Lc 699Ld 588L

V2 = P1V1T1P2


V = Aν



m = AνV



t2 = T2 ndash T1





ho = PoYo




b 60 lbm

c 70 lbm

d 80 lbm

formula m = Fgk g


a 449 ftb 339 ftc 229 ftd 559 ft

formula ho = po Yw


formula k= R cv +1







a 13 x 108 Jb 23 x 108 Jc 33 x 108 Jd 43 x 108 J



a -540 Jb -640 Jc -740 Jd -840 J

formula ΔU = q + w



c 088 kg liter -1

d 099 kg liter -1

solution


a 11120Nb 11320Nc 11420Nd 11520N

formula F= ma k


a 318 lb 418 lc 518 ld 618 l

solution



formula m= Fg k g


formula Ef = pV




a 194 lbb 214 lbc394 lbd 413 lb

formula m= pV RT





a7849 ft3

b 7850 ft3

c 7851 ft3

d 7852 ft3



a 1234b 1345c 1456d 1356



a 3394 Kb 4494 Kc 5594Kd 6694K









Ans (a)





a) 9520b) 7010c) 6560d) 9556



a) 9556b) 7010c) 9520d) 8510



a) 13787 kJb) 13485 kJc) 12787 kJd) 12465 kJ



a) -13847 kJb) 13847 kJc) -23847 kJd) 23847 kJ

Ans (a)


a) 96b) 86c) 76d) 66

Ans (a)










a) 6440b) 7460c) 5840d) 5460























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h
























a 298645b 398109c 291167d 281333

Pg = Pabs - Patm




a 70b 68c 72d 94

ordmR = ordmF + 460


F = ma


F = ma


t2 = T2 ndash T1



T2 = V2 (t2)V1 and H = mcp (T2-T1)








PE = mgo(Z)gc



KE = mV22gc



Pabs = Pg + Patm


a 78932Lb 79632Lc 699Ld 588L

V2 = P1V1T1P2


V = Aν



m = AνV



t2 = T2 ndash T1





ho = PoYo




b 60 lbm

c 70 lbm

d 80 lbm

formula m = Fgk g


a 449 ftb 339 ftc 229 ftd 559 ft

formula ho = po Yw


formula k= R cv +1







a 13 x 108 Jb 23 x 108 Jc 33 x 108 Jd 43 x 108 J



a -540 Jb -640 Jc -740 Jd -840 J

formula ΔU = q + w



c 088 kg liter -1

d 099 kg liter -1

solution


a 11120Nb 11320Nc 11420Nd 11520N

formula F= ma k


a 318 lb 418 lc 518 ld 618 l

solution



formula m= Fg k g


formula Ef = pV




a 194 lbb 214 lbc394 lbd 413 lb

formula m= pV RT





a7849 ft3

b 7850 ft3

c 7851 ft3

d 7852 ft3



a 1234b 1345c 1456d 1356



a 3394 Kb 4494 Kc 5594Kd 6694K









Ans (a)





a) 9520b) 7010c) 6560d) 9556



a) 9556b) 7010c) 9520d) 8510



a) 13787 kJb) 13485 kJc) 12787 kJd) 12465 kJ



a) -13847 kJb) 13847 kJc) -23847 kJd) 23847 kJ

Ans (a)


a) 96b) 86c) 76d) 66

Ans (a)










a) 6440b) 7460c) 5840d) 5460























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h














a 298645b 398109c 291167d 281333

Pg = Pabs - Patm




a 70b 68c 72d 94

ordmR = ordmF + 460


F = ma


F = ma


t2 = T2 ndash T1



T2 = V2 (t2)V1 and H = mcp (T2-T1)








PE = mgo(Z)gc



KE = mV22gc



Pabs = Pg + Patm


a 78932Lb 79632Lc 699Ld 588L

V2 = P1V1T1P2


V = Aν



m = AνV



t2 = T2 ndash T1





ho = PoYo




b 60 lbm

c 70 lbm

d 80 lbm

formula m = Fgk g


a 449 ftb 339 ftc 229 ftd 559 ft

formula ho = po Yw


formula k= R cv +1







a 13 x 108 Jb 23 x 108 Jc 33 x 108 Jd 43 x 108 J



a -540 Jb -640 Jc -740 Jd -840 J

formula ΔU = q + w



c 088 kg liter -1

d 099 kg liter -1

solution


a 11120Nb 11320Nc 11420Nd 11520N

formula F= ma k


a 318 lb 418 lc 518 ld 618 l

solution



formula m= Fg k g


formula Ef = pV




a 194 lbb 214 lbc394 lbd 413 lb

formula m= pV RT





a7849 ft3

b 7850 ft3

c 7851 ft3

d 7852 ft3



a 1234b 1345c 1456d 1356



a 3394 Kb 4494 Kc 5594Kd 6694K









Ans (a)





a) 9520b) 7010c) 6560d) 9556



a) 9556b) 7010c) 9520d) 8510



a) 13787 kJb) 13485 kJc) 12787 kJd) 12465 kJ



a) -13847 kJb) 13847 kJc) -23847 kJd) 23847 kJ

Ans (a)


a) 96b) 86c) 76d) 66

Ans (a)










a) 6440b) 7460c) 5840d) 5460























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h








a 70b 68c 72d 94

ordmR = ordmF + 460


F = ma


F = ma


t2 = T2 ndash T1



T2 = V2 (t2)V1 and H = mcp (T2-T1)








PE = mgo(Z)gc



KE = mV22gc



Pabs = Pg + Patm


a 78932Lb 79632Lc 699Ld 588L

V2 = P1V1T1P2


V = Aν



m = AνV



t2 = T2 ndash T1





ho = PoYo




b 60 lbm

c 70 lbm

d 80 lbm

formula m = Fgk g


a 449 ftb 339 ftc 229 ftd 559 ft

formula ho = po Yw


formula k= R cv +1







a 13 x 108 Jb 23 x 108 Jc 33 x 108 Jd 43 x 108 J



a -540 Jb -640 Jc -740 Jd -840 J

formula ΔU = q + w



c 088 kg liter -1

d 099 kg liter -1

solution


a 11120Nb 11320Nc 11420Nd 11520N

formula F= ma k


a 318 lb 418 lc 518 ld 618 l

solution



formula m= Fg k g


formula Ef = pV




a 194 lbb 214 lbc394 lbd 413 lb

formula m= pV RT





a7849 ft3

b 7850 ft3

c 7851 ft3

d 7852 ft3



a 1234b 1345c 1456d 1356



a 3394 Kb 4494 Kc 5594Kd 6694K









Ans (a)





a) 9520b) 7010c) 6560d) 9556



a) 9556b) 7010c) 9520d) 8510



a) 13787 kJb) 13485 kJc) 12787 kJd) 12465 kJ



a) -13847 kJb) 13847 kJc) -23847 kJd) 23847 kJ

Ans (a)


a) 96b) 86c) 76d) 66

Ans (a)










a) 6440b) 7460c) 5840d) 5460























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h









PE = mgo(Z)gc



KE = mV22gc



Pabs = Pg + Patm


a 78932Lb 79632Lc 699Ld 588L

V2 = P1V1T1P2


V = Aν



m = AνV



t2 = T2 ndash T1





ho = PoYo




b 60 lbm

c 70 lbm

d 80 lbm

formula m = Fgk g


a 449 ftb 339 ftc 229 ftd 559 ft

formula ho = po Yw


formula k= R cv +1







a 13 x 108 Jb 23 x 108 Jc 33 x 108 Jd 43 x 108 J



a -540 Jb -640 Jc -740 Jd -840 J

formula ΔU = q + w



c 088 kg liter -1

d 099 kg liter -1

solution


a 11120Nb 11320Nc 11420Nd 11520N

formula F= ma k


a 318 lb 418 lc 518 ld 618 l

solution



formula m= Fg k g


formula Ef = pV




a 194 lbb 214 lbc394 lbd 413 lb

formula m= pV RT





a7849 ft3

b 7850 ft3

c 7851 ft3

d 7852 ft3



a 1234b 1345c 1456d 1356



a 3394 Kb 4494 Kc 5594Kd 6694K









Ans (a)





a) 9520b) 7010c) 6560d) 9556



a) 9556b) 7010c) 9520d) 8510



a) 13787 kJb) 13485 kJc) 12787 kJd) 12465 kJ



a) -13847 kJb) 13847 kJc) -23847 kJd) 23847 kJ

Ans (a)


a) 96b) 86c) 76d) 66

Ans (a)










a) 6440b) 7460c) 5840d) 5460























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h











ho = PoYo




b 60 lbm

c 70 lbm

d 80 lbm

formula m = Fgk g


a 449 ftb 339 ftc 229 ftd 559 ft

formula ho = po Yw


formula k= R cv +1







a 13 x 108 Jb 23 x 108 Jc 33 x 108 Jd 43 x 108 J



a -540 Jb -640 Jc -740 Jd -840 J

formula ΔU = q + w



c 088 kg liter -1

d 099 kg liter -1

solution


a 11120Nb 11320Nc 11420Nd 11520N

formula F= ma k


a 318 lb 418 lc 518 ld 618 l

solution



formula m= Fg k g


formula Ef = pV




a 194 lbb 214 lbc394 lbd 413 lb

formula m= pV RT





a7849 ft3

b 7850 ft3

c 7851 ft3

d 7852 ft3



a 1234b 1345c 1456d 1356



a 3394 Kb 4494 Kc 5594Kd 6694K









Ans (a)





a) 9520b) 7010c) 6560d) 9556



a) 9556b) 7010c) 9520d) 8510



a) 13787 kJb) 13485 kJc) 12787 kJd) 12465 kJ



a) -13847 kJb) 13847 kJc) -23847 kJd) 23847 kJ

Ans (a)


a) 96b) 86c) 76d) 66

Ans (a)










a) 6440b) 7460c) 5840d) 5460























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h













a 13 x 108 Jb 23 x 108 Jc 33 x 108 Jd 43 x 108 J



a -540 Jb -640 Jc -740 Jd -840 J

formula ΔU = q + w



c 088 kg liter -1

d 099 kg liter -1

solution


a 11120Nb 11320Nc 11420Nd 11520N

formula F= ma k


a 318 lb 418 lc 518 ld 618 l

solution



formula m= Fg k g


formula Ef = pV




a 194 lbb 214 lbc394 lbd 413 lb

formula m= pV RT





a7849 ft3

b 7850 ft3

c 7851 ft3

d 7852 ft3



a 1234b 1345c 1456d 1356



a 3394 Kb 4494 Kc 5594Kd 6694K









Ans (a)





a) 9520b) 7010c) 6560d) 9556



a) 9556b) 7010c) 9520d) 8510



a) 13787 kJb) 13485 kJc) 12787 kJd) 12465 kJ



a) -13847 kJb) 13847 kJc) -23847 kJd) 23847 kJ

Ans (a)


a) 96b) 86c) 76d) 66

Ans (a)










a) 6440b) 7460c) 5840d) 5460























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h








a 318 lb 418 lc 518 ld 618 l

solution



formula m= Fg k g


formula Ef = pV




a 194 lbb 214 lbc394 lbd 413 lb

formula m= pV RT





a7849 ft3

b 7850 ft3

c 7851 ft3

d 7852 ft3



a 1234b 1345c 1456d 1356



a 3394 Kb 4494 Kc 5594Kd 6694K









Ans (a)





a) 9520b) 7010c) 6560d) 9556



a) 9556b) 7010c) 9520d) 8510



a) 13787 kJb) 13485 kJc) 12787 kJd) 12465 kJ



a) -13847 kJb) 13847 kJc) -23847 kJd) 23847 kJ

Ans (a)


a) 96b) 86c) 76d) 66

Ans (a)










a) 6440b) 7460c) 5840d) 5460























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h








a7849 ft3

b 7850 ft3

c 7851 ft3

d 7852 ft3



a 1234b 1345c 1456d 1356



a 3394 Kb 4494 Kc 5594Kd 6694K









Ans (a)





a) 9520b) 7010c) 6560d) 9556



a) 9556b) 7010c) 9520d) 8510



a) 13787 kJb) 13485 kJc) 12787 kJd) 12465 kJ



a) -13847 kJb) 13847 kJc) -23847 kJd) 23847 kJ

Ans (a)


a) 96b) 86c) 76d) 66

Ans (a)










a) 6440b) 7460c) 5840d) 5460























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h











a) 9520b) 7010c) 6560d) 9556



a) 9556b) 7010c) 9520d) 8510



a) 13787 kJb) 13485 kJc) 12787 kJd) 12465 kJ



a) -13847 kJb) 13847 kJc) -23847 kJd) 23847 kJ

Ans (a)


a) 96b) 86c) 76d) 66

Ans (a)










a) 6440b) 7460c) 5840d) 5460























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h








a) 96b) 86c) 76d) 66

Ans (a)










a) 6440b) 7460c) 5840d) 5460























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h




























a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h











a) 7448 hpb) 0998 hpc) 746 hpd) 1998 hp


F = W = mgd = h







Date post:	26-Oct-2014
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