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9/12/2020PPI Learning Hub Admin : QuestionsThis problem corresponds toMEHRPE2 problem number 16.Test Bank Question Return to Questions (/admin/questions/0?sfield magento id&stext 0000004586&sdka &stype &sdiff )previewQUESTION DATAQuestionA solid copper sphere at 852 F is dropped into a large tank of oil at 167 F. The sphere has a diameter of7.88 inches and a thermal conductivity of 2628 Btu-in/(hr-ft2- F). The average convective heat transfercoefficient is 155 Btu/(hr-ft2- F) and the oil is stirred uniformly at all times. Most nearly, how long aftersubmersion does the sphere reach a temperature of392 F?Answers(A) 5.3 sec(B) 140 sec(C) 910 sec(D) 4700 secThe answer is (B).SolutionContent in blue refers to the NCEES Handbook.Since the problem involves transient heat flow, determine if the lumped capacitance approximation isvalid. The average temperature of the copper is T Ts Tl 852 F 392 F2The sphere’s specific heat and density are found in a table of properties of metals. [Properties of Metals- I-P Units]oc p 0.09 Btu/lbm- F 557 eated On03/25/2020 01:02:18AMPublished On03/25/2020 01:02:18AMModified On09/10/2020 11:11:03PMOTHER VERSIONS10/23/2019 08:12:52 P(/admin/questions/pre2 622 FρVendor0000004586Solving Time3Determine the Biot number.03/25/2020 01:02:18 A(/admin/questions/preDISCIPLINESFE Other Disciplines(/admin/questions/indesfield discipline&stextOther Disciplines)Transient Conduction Using the Lumped Capacitance ModelFE Chemical(/admin/questions/indesfield discipline&stextChemical)FE Environmental(/admin/questions/indesfield ss.com/admin/questions/0/preview/482231/3
9/12/2020PPI Learning Hub Admin : QuestionshVBi PE Chemical(/admin/questions/indesfield discipline&stextChemical)kAh( 43πr3)2k (4πr )hr PE Mechanical: HVAC aRefrigeration(/admin/questions/indesfield discipline&stextMechanical: HVAC andRefrigeration)3kBtu(1557.88 in)(2)2hr-ft - F Btu-in(3) (2628in) (122)fthr-ft - F 0.0065Because the Biot number is much less than 0.1, the internal thermal resistance of the sphere isnegligible compared to the external thermal resistance in the oil bath. Therefore, the lumped parametermethod can be used to solve for the time constant.Constant Fluid Temperature βtTt T (T0 T )e βt eKNOWLEDGE AREASTt T Thermodynamics and(/admin/questions/indesfield area&stext Theand Heat Transfer)T0 T βt lnTt T T0 T t β lnTt T T0 T Heat Transfer(/admin/questions/indesfield area&stext HeaTransfer)The exponent β isβ PE Mechanical: Thermaand Fluid Systems(/admin/questions/indesfield discipline&stextMechanical: Thermal aFluid Systems)hAsρV c pThermodynamics(/admin/questions/indesfield area&stext The2h (4πr ) ρ(43πr3) cp3hMechanisms(/admin/questions/indesfield area&stext Mec ρrc pBtu(3) (155in) (122 lbm(55737.88 in)(Btu) (0.092ft 0.00785 sec)fthr-ft - Fsec) (3600lbm- F)hr 1Solve for t.Heat Transfer Principle(/admin/questions/indesfield area&stext HeaTransfer Principles)PRODUCTS USED IN1t lnβPEMEHV2EXPEMEHVQBPEMETSQBT T Ti T 1 0.00785 141.82 sec1sec ln ( 392 F 167 F)852 F 167 F (140 review/482232/3
s/0/preview/48223PPI Learning Hub Admin : Questions3/3
10/3/2020PPI Learning Hub Admin : QuestionsThis problem corresponds toMEHRPE2 problem number 38. Return to Questions (/admin/questions/0?sfield magento id&stext 0000004612&sdka &stype &sdiff )Test Bank QuestionpreviewQUESTION DATAQuestionA single-stage chiller circulates 117,000 lbm/hr of refrigerant-22 (R-22) and operates with a 90 Fcondensing temperature and 10 F evaporating temperature. Saturated refrigerant vapor enters thecompressor with no superheat, and saturated liquid refrigerant leaves the condenser with nosubcooling. Heat is rejected to condenser water that enters the condenser at 85 F and leaves at 95 F.The rated coefficient of performance is 5.5 under these conditions. The condenser water flow neededfor heat rejection is most nearlyVendor0000004612Solving tiveCreated On06/09/2020 04:44:49PMPublished On06/09/2020 04:44:49PMModified On09/29/2020 08:53:21PMContent in blue refers to the NCEES Handbook.OTHER VERSIONSAnswers(A) 560 gpm(B) 1600 gpm(C) 1950 gpm(D) 2200 gpmThe answer is (C).Since the temperatures of water entering and leaving the condenser are fixed, the required flow isdetermined by the total load to be rejected, which includes both the load absorbed by the evaporatorand the heat added by the compressor. The relationship for the condenser water temperature rise is q rejected mcondenser, water c p ΔTThe ΔT term represents the water temperature change, and the load to be rejected isq rejected q refrigeration q comp, heatThe required condenser water can be determined as mcondenser, water qrefrigeration qcomp, heatcp ΔTThe refrigeration capacity of the chiller is the product of the refrigerant mass flow rate and therefrigeration effect.q refrigeration mrefrigerant (hevap, leaving hevap, entering ) mrefrigerant (hcomp, entering hevap, entering )The enthalpy of the refrigerant mixture entering the chiller evaporator is equal to that of the saturatedliquid leaving the chiller condenser since there is no change in enthalpy when the refrigerant goesthrough the throttling valve. Refrigerant enthalpies are found in an R-22 property table. Find eview/5023210/23/2019 08:12:53 P(/admin/questions/pre06/09/2020 04:44:49 P(/admin/questions/preDISCIPLINESPE Mechanical: HVAC aRefrigeration(/admin/questions/indsfield discipline&stextMechanical: HVAC andRefrigeration)KNOWLEDGE AREASSystems and Compone(/admin/questions/indsfield area&stext Sysand Components)PRODUCTS USED INPEMEHV2EX1/2
10/3/2020PPI Learning Hub Admin : Questionsenthalpy of the saturated liquid at 90 F. [Pressure Versus Enthalpy Curves for Refrigerant 22]PEMEHVQBhevap, entering 36.391 Btu/lbmSince there is no superheat, the enthalpy of the refrigerant entering the compressor is equal to theenthalpy of the saturated vapor at 10 F. The enthalpy of the saturated vapor at 10 F ishcomp, entering 105.58 Btu/lbmThe useful refrigeration is q refrigeration mrefrigerant (hcomp, entering hevap, entering ) (117,000lbm) (105.58hrBtu 36.391lbmBtu)lbm 8,095,113 Btu/hrBased on the definition of coefficient of performance (COP), the compressor work can be determined.q comp q refrigerationCOPBtu8,095,113hr 5.5 1619022.6 Btu/hrThe required rate of heat removal isq rejected q refrigeration q compBtu 8,095,113Btu 1,619,022.6hrhr 9,714,135.6 Btu/hrFind the condenser water flow that is needed to remove the heat. [Measurement Relationships] mcondenser ρQ condenserq rejected c p ΔTq rejectedQ condenser ρc p ΔTBtu9,714,135.6hr lbm(8.34Btu) (1.0gal)lbm- Fmin (95 F 85 F) (60)hr 1941.27 0/preview/50232(1950 gpm)2/2
10/3/2020PPI Learning Hub Admin : QuestionsThis problem corresponds toMEHRPE2 problem number 37.Test Bank Question Return to Questions (/admin/questions/0?sfield magento id&stext 0000004613&sdka &stype &sdiff )previewQUESTION DATAQuestionA simple Rankine cycle operates between superheated steam entering a turbine at 1200 F and 700psia and entering a pump at 2 psia. The cycle’s maximum possible efficiency is most nearlyAnswers(A) 27%(B) 31%(C) 39%(D) 43%The answer is (C).SolutionContent in blue refers to the NCEES Handbook.From steam tables, the entropy of the steam entering the turbine at 1200 F and 700 psia, s3, is 1.769Btu/lbm- R. [Properties of Superheated Steam - I-P Units]For maximum efficiency, the entropy of the steam entering the condenser, s4, has the same value. Thequality can be found from the equationProperties for Two-Phase (Vapor-Liquid) Systems s3 sfsg sfsf and sg are the entropies of the saturated liquid and vapor, respectively, and are taken from asaturated steam table for a pressure of 2 psia. [Properties of Saturated Water and Steam (Pressure) - IP Units]Btux s3 sfsg sf1.769Btu 0.1749lbm- R lbm- RBtu1.9195Btu 0.1749lbm- R eated On10/23/2019 08:12:53PMPublished On10/23/2019 08:12:53PMModified On09/29/2020 08:53:17PMOTHER VERSIONS10/23/2019 08:12:53 P(/admin/questions/preDISCIPLINESs3 s4 sf xsf gxVendor0000004613Solving Timelbm- RPE Mechanical: HVAC aRefrigeration(/admin/questions/indsfield discipline&stextMechanical: HVAC andRefrigeration)KNOWLEDGE AREASEquipment and Comp(/admin/questions/indsfield area&stext Equand Components)The enthalpy of the steam entering the condenser isProperties for Two-Phase (Vapor-Liquid) Systemsh4 hf xhf iew/45266Systems and Compone(/admin/questions/indsfield area&stext Sysand Components)1/2
10/3/2020PPI Learning Hub Admin : Questionshf and hg are the enthalpies of the saturated liquid and vapor, respectively, and are taken from asaturated steam table for a pressure of 2 psia. [Properties of Saturated Water and Steam (Pressure) - IP Units]PRODUCTS USED INPEMEHV2EXPEMEHVQBh4 hf x (hg hf ) 94.00Btulbm (0.9137) (1115.74Btulbm 94.00Btulbm) 1027.6 Btu/lbmFrom steam tables, the enthalpy between the turbine and boiler, h3, for 1200 F and 700 psia is 1625.9Btu/lbm. The enthalpy of the saturated 2 psia liquid, h1, is 94.00 Btu/lbm. [Properties of SuperheatedSteam - I-P Units]The efficiency of the cycle isη WQ h3 h4h3 h1Btu1625.9Btu 1027.6lbm lbmBtu1625.9Btu 94.00lbm 0.391lbmReplace this equationwith the one ons/0/preview/452662/2
10/9/2020PPI Learning Hub Admin : QuestionsThis problem corresponds toMEHRPE2 problem number 34.Test Bank Question Return to Questions (/admin/questions/0?sfield magento id&stext 0000004618&sdka &stype &sdiff )previewQUESTION DATAQuestionA store in a shopping mall is to be maintained at 75 F and 45% relative humidity with supply air at 55 Fand 30% relative humidity. The space cooling load is 73,000 Btu/hr sensible and 26,000 Btu/hr latent atoutdoor design conditions of 94 F dry-bulb temperature and 72 F wet-bulb temperature. Theventilation requirement is 850 ft3/min. The coil load due to the ventilation air is most nearlyAnswers(A) 30,000 Btu/hr(B) 60,000 Btu/hr(C) 80,000 Btu/hr(D) 100,000 Btu/hrThe answer is (C).Solutionmove this section downContent in blue refers to the NCEES Handbook.The coil load due to the ventilation air is the rate at which sensible and latent heat must be removedfrom the ventilation air to reduce it to the return state.Vendor0000004618Solving ted On03/31/2020 06:23:09PMPublished On03/31/2020 06:23:09PMModified On09/29/2020 08:53:20PMOTHER VERSIONS10/23/2019 08:12:53 P(/admin/questions/preMoist-Air Sensible Heating or Cooling q total mΔh Qρ (houtdoor hsupply )From the psychrometric chart, at 94 F dry-bulb and 72 F wet-bulb, the enthalpy of the outside air,houtdoor, is 35.6 Btu/lbm. [ASHRAE Psychrometric Chart No. 1 - Normal Temperature at Sea Level]At 55 F and 30% relative humidity,01/15/2020 10:36:26 P(/admin/questions/pre03/31/2020 06:23:09 P(/admin/questions/prehsupply 16.2 Btu/lbmInterpolating from a table of properties of air at atmospheric pressure, for air at 55 F, the density is0.0769 lbm/ft3. [Properties of Air at Atmospheric Pressure]The coil load isq total Qρ (houtdoor hsupply )3 (850ft) (60min (35.6min) (0.0769lbm 75,495 /preview/48581 16.2)PE Mechanical: HVAC aRefrigeration(/admin/questions/indsfield discipline&stextMechanical: HVAC 000 Btu/hr)PE Mechanical: Thermaand Fluid Systems(/admin/questions/indsfield discipline&stextMechanical: Thermal aFluid Systems)1/2
This problem corresponds toMEHRPE2 problem number 48. Return to Questions (/admin/questions/0?sfield magento id&stext 4621&sdka &stype &sdiff )Test Bank QuestionpreviewQUESTION DATAQuestionAn east-facing vertical window at a latitude of 40 degrees north has an area of 12 ft2. The solar heat gaincoefficient for the window is 0.87. The overall heat transfer coefficient is 1.2 Btu/hr-ft2- F. The table showngives the incident total irradiance for 40 degrees north.Maysolartimeincident total 02146651023880357199175174Vendor0000004621Solving ted On05/06/2020 01:28:31PMPublished On05/06/2020 01:28:31PMModified On09/10/2020 11:11:06 PMOTHER VERSIONShalf daytotalJune10/23/2019 08:12:54 P(/admin/questions/prev05/06/2020 01:28:31 203187186PE Mechanical: HVACand Refrigeration(/admin/questions/indexsfield discipline&stext Mechanical: HVAC andRefrigeration)PE Mechanical: Thermand Fluid Systems(/admin/questions/indexsfield discipline&stext Mechanical: Thermal aFluid Systems)KNOWLEDGE AREAShalf daytotalHeating/Cooling Loads(/admin/questions/index
On a day in May at solar time 0800, the indoor temperature is 75 F, and the outdoor temperature is 42 F.The total instantaneous heat gain for the window is most nearlyAnswerssfield area&stext ield area&stext Coo(A) 480 Btu/hr(B) 1800 Btu/hr(C) 2300 Btu/hrPRODUCTS USED IN(D) 2800 Btu/hrPEMETSQBPEMEHVQBPEMEHV2EXThe answer is (D).BSolutionContent in blue refers to the NCEES Handbook.From the table, the incident total irradiance for an east-facing window in May at 0800 isEt 218 Btu/hr-ft2The solar heat gain coefficient, SHGC, is given as 0.87, and the overall heat transfer coefficient, U, is 1.2Btu/hr-ft2- F. The instantaneous heat gain, where the direction of heat flow is from inside to outside, isFenestrationq U A pf (Tin Tout ) (S H C G)A pf ET C (AL)A pf ρCp (Tin Tout )No air leakage is given in the problem; assume it to be 0. The total instantaneous heat gain for the windowisq U A pf (Tin Tout ) (S H C G)A pf ET C (AL)A pf ρCp (Tin Tout ) (1.2Btu2) (12 ft )(75 F 42 F)2hr-ft - F2Btu (0.87)(12 ft ) (2182hr-ft - F 2751Btuhr(2800Btuhr)) 0
9/24/2020PPI Learning Hub Admin : QuestionsThis problem corresponds toMEHRPE2 problem number 45.Test Bank Question Return to Questions (/admin/questions/0?sfield magento id&stext 0000004623&sdka &stype &sdiff )previewQUESTION DATAQuestionA sun room has an exposed 6 in concrete floor slab measuring 20 ft 30 ft. By nightfall, the room is heatedpassively by solar radiation to an average temperature of 82 F. The density and specific heat of the concreteis 140 lbm/ft3 and 0.22 Btu/lbm- F, respectively. The room thermostat setting is kept at 60 F. Assume a floorslab convective heat transfer coefficient of 16.8 Btu/hr-ft2- F. The average temperature change of the slabafter 1 hr of cooling is most nearlyAnswers(A) 10 F(B) 12 F(C) 13 F(D) 15 FThe answer is (D).Vendor0000004623Solving ted On09/23/2020 06:24:11 PMPublished On09/23/2020 06:24:11 PMModified On09/23/2020 06:24:11 PMOTHER VERSIONSSolution10/23/2019 08:12:53 PM(/admin/questions/previContent in blue refers to the NCEES Handbook.The thermal energy that can be stored in the slab depends on the mass, specific heat, and the temperaturedifference between the slab and the room. The temperature variation of the slab with time isConstant Fluid TemperatureT T (Ti T ) e βt03/25/2020 01:02:21 AM(/admin/questions/previCalculate the time constant, β.Constant Fluid Temperature09/23/2020 06:24:11 PM(/admin/questions/previBtu(16.8β hAsρVc p) (20 ft) (30 ft)2DISCIPLINEShr-ft - F lbm(140ft302/20/2020 12:58:34 AM(/admin/questions/previ) (0.50 ft) (20 ft)Btu (30 ft) (0.22)lbm- FFE Other Disciplines(/admin/questions/indexsfield discipline&stext Other Disciplines) 1 1.091 hrRearrange the equation for temperature variation so that the body temperature, T, represents thetemperature as a function of time.T (t) (Ti T ) eThe temperature after 1 hr view/52667 βt T FE Chemical(/admin/questions/indexsfield discipline&stext Chemical)FE Environmental(/admin/questions/index1/2
9/24/2020PPI Learning Hub Admin : QuestionsT (1 hr) (Ti T ) e βt T 1 (82 F 60 F) e 67.39 F( 1.091 hr)(1 hr) 60 F(67.4 F)To find the temperature change in the slab, take the difference between the initial temperature and thetemperature after 1 hr.sfield discipline&stext Environmental)PE Chemical(/admin/questions/indexsfield discipline&stext Chemical)Ti T (1 hr) 82 F 67.4 F 14.6 F (15 F)PE Mechanical: HVACand Refrigeration(/admin/questions/indexsfield discipline&stext Mechanical: HVAC andRefrigeration)PE Mechanical: Thermaand Fluid Systems(/admin/questions/indexsfield discipline&stext Mechanical: Thermal anFluid Systems)KNOWLEDGE AREASThermodynamics and HTransfer (/admin/questiosfield area&stext Therand Heat Transfer)Heat Transfer(/admin/questions/indexsfield area&stext sfield area&stext TherMechanisms(/admin/questions/indexsfield area&stext MechHeat Transfer Principles(/admin/questions/indexsfield area&stext HeatTransfer Principles)PRODUCTS USED /questions/0/preview/526672/2
9/13/2020PPI Learning Hub Admin : QuestionsThis problem corresponds toMEHRPE2 problem number 63.Test Bank Question Return to Questions (/admin/questions/0?sfield magento id&stext 0000142298&sdka &stype &sdiff )previewQUESTION DATAQuestionA scuba diver carries an oxygen tank holding 80 ft3 of air at 70 F and atmospheric pressure. Thecylindrical oxygen tank has an inner diameter of 7.25 in and a height of 30 in. The diver is performingmoderate work to repair an underwater vessel. The tank is considered empty once it reaches 700 psig.Based on the diver’s level of exertion, the number of hours of oxygen in the tank is most nearlyAnswers(A) 10 hr(B) 14 hr(C) 23 hr(D) 45 hrThe answer is (C).SolutionContent in blue refers to the NCEES Handbook.The volume of the tank isVendor0000142298Solving ted On07/31/2020 07:14:06PMPublished On07/31/2020 07:14:06PMModified On09/10/2020 11:11:07PMOTHER VERSIONSV2 πr H (π2) (7.25 in)4 30 inin(123)ft10/23/2019 08:12:55 P(/admin/questions/pre01/15/2020 10:36:34 P(/admin/questions/pre 3 0.717 ft05/13/2020 06:19:39 P(/admin/questions/preThe initial mass of oxygen in the tank islbm3(80 ft ) (0.0753) 6 lbm07/31/2020 07:14:06 P(/admin/questions/preftThe mass of oxygen when there is 700 psig isDISCIPLINESIdeal GaspV mRTpVm RTlbf(700 2lbf 14.7in2inin) (122)ft-lbf(53.3) (70 F 460 )lbm- R 2.61 eview/516223(0.717 ft )ftPE Mechanical: HVAC aRefrigeration(/admin/questions/indsfield discipline&stextMechanical: HVAC andRefrigeration)PE Mechanical: Thermaand Fluid Systems(/admin/questions/ind1/2
9/13/2020PPI Learning Hub Admin : QuestionsThe diver can breathe 3.39 lbm of oxygen before the tank reaches low pressure and is consideredempty. To find the minimum hours of oxygen the diver needs to perform the work, use the maximumrecommended value for the level of exertion. From an ASHRAE table of oxygen consumption atdifferent activity levels, moderate work consumes 1–2 ft3/hr of oxygen. [Heart Rate and OxygenConsumption at Different Activity Levels]The mass of oxygen the diver consumes per hour is3ft(2lbm) (0.075hr3) 0.15 lbm/hrftThe number of hours of oxygen in the tank for performing moderate work is6 lbm 2.61 lbm 22.6 hr (23 hr)lbm0.15hrsfield discipline&stextMechanical: Thermal aFluid Systems)KNOWLEDGE AREASHeating/Cooling Loads(/admin/questions/indsfield area&stext ld area&stext CooPRODUCTS USED com/admin/questions/0/preview/516222/2
This problem corresponds toMEHRPE2 problem number 75. Return to Questions (/admin/questions/0?sfield magento id&stext 0000142301&sdka &stype &sdiff )Test BankQuestion previewQUESTION DATAQuestionA local health club with a floor area of 8000 ft2 uses air to control the humidity in the building. Air at 95 Fand 80% humidity enters an air cooling coil and leaves at 65 F and 20% humidity. The mass flow rate ofcondensate leaving the airstream is most nearlyAnswers(A) 650 lbm/hr(B) 700 lbm/hr(C) 760 lbm/hr(D) 800 lbm/hrThe answer is (C).SolutionContent in blue refers to the NCEES Handbook.The mass flow rate of water is related to the mass flow rate of the air and the entering and exiting humidityratios.Q m w m da (W1 W2 ) (W1 W2 )vThe required volumetric flow rate of dry air for a health club can be found from a table of air and ventilationrequirements for various facilities. A health club requires a combined outdoor air rate of 22 ft3/person.[Minimum Ventilation Rates in the Breathing Zone, Based on ANSI/ASHRAE Standard 62.1-2007]The total flow rate of air for the health club is ft3 replace with this equation40 persons min 2Q (8000 ft ) () 22 2 person1000 ft 3 7040 ft / minFrom the psychrometric chart, the specific volume of the dry air is approximately 14.7 ft3/lbm. Thehumidity ratios of the entering and exiting streams are 0.0292 lbm/lbm and 0.0026 lbm/lbm, respecitvely.[ASHRAE Psychrometric Chart No. 1 - Normal Temperature at Sea Level]The mass flow rate of the condensate is ft7040 mw 14.73 lbmlbmminmin 0.0026) (60) (0.02923 lbmlbmhrftlbm 764.3 lbm/hrVendor0000142301Solving ted On06/02/2020 10:11:55PMPublished On06/02/2020 10:11:55PMModified On09/29/2020 08:53:21PMOTHER VERSIONS10/23/2019 08:12:56 P(/admin/questions/prev06/02/2020 10:11:55 P(/admin/questions/prevDISCIPLINESPE Mechanical: HVACand Refrigeration(/admin/questions/indesfield discipline&stext Mechanical: HVAC andRefrigeration)KNOWLEDGE AREASSupportive Knowledge(/admin/questions/indesfield area&stext SupKnowledge)PRODUCTS USED IN(760 lbm/hr)PEMEHV2EXPEMEHVQB
9/30/2020PPI Learning Hub Admin : QuestionsThis problem corresponds toMEHRPE2 problem number 76. Return to Questions (/admin/questions/0?sfield magento id&stext 142303&sdka &stype &sdiff )Test Bank QuestionpreviewQUESTION DATAQuestionA building’s dedicated outdoor air system (DOAS) is designed to handle the entire latent load requirementfor both the outside air and the building. For winter heating, the supply air condition for the building is20,000 cfm at 105 F dry-bulb temperature and 25% relative humidity. The inside design condition is 75 Fdry-bulb temperature and 45% relative humidity. The outside air requirement is 4000 cfm with winter designconditions of 35 F dry-bulb and 20% relative humidity. The dry-bulb temperature of the supply air for theDOAS is 75 F. The needed relative humidity of the supply air for the DOAS is most nearlyVendor0000142303Solving tiveCreated On05/05/2020 08:27:35PMPublished On05/05/2020 08:27:35PMModified On09/29/2020 08:53:20PMContent in blue refers to the NCEES Handbook.OTHER VERSIONSAnswers(A) 60%(B) 70%(C) 80%(D) 90%The answer is (D).Plotting the supply air condition, the outside air condition, and the indoor design condition on apsychrometric chart, the humidity ratio for the supply air is 0.012 lbm/lbm, for the outside air is 0.00085lbm/lbm, and for the indoor design conditions is 0.0084 lbm/lbm. [ASHRAE Psychrometric Chart No. 1 Normal Temperature at Sea Level]Find the rate of moisture that is needed by the building based on supply air conditions with a flow rate of20,000 cfm. From a psychrometric chart, the specific volume of the supply air (105 F db, 25% rh) is 14.5ft3/lbm dry air. [ASHRAE Psychrometric Chart No. 1 - Normal Temperature at Sea Level]The total rate of moisture added by the supply air isMoist-Air Cooling and Dehumidification m w m DA (WSA WIDC ) ρQ (WSA WIDC ) QvSA (WSA WIDC )ft3(20,000min 14.5min)hrft03/17/2020 07:04:21 P(/admin/questions/prev05/02/2020 12:57:19 P(/admin/questions/prev05/02/2020 02:23:14 P(/admin/questions/prev05/05/2020 08:27:35 P(/admin/questions/prev) (60 10/23/2019 08:12:56 P(/admin/questions/prev3lbm 298 /preview/49736 lbmlbm (0.012 0.0084) lbmlbm DISCIPLINESFE Chemical(/admin/questions/indexsfield discipline&stext Chemical)1/2
9/30/2020PPI Learning Hub Admin : QuestionsThe outside air will supply only 4000 cfm. From a psychrometric chart, the specific volume of the outsideair (35 F db and 20% rh) is 12.5 ft3/lbm dry air. [ASHRAE Psychrometric Chart No. 1 - Normal Temperatureat Sea Level]Find the needed increase in specific humidity to achieve the same humidification for the entire building. mw WDOAS QvOA(WDOAS WOA ) m w vOAQKNOWLEDGE AREAS WOAlbm(298ft3) (12.5hr)lbm 0.00085ft3(4000lbmlbmmin) (60minPE Mechanical: HVACand Refrigeration(/admin/questions/indexsfield discipline&stext Mechanical: HVAC andRefrigeration))hr 0.0164 lbm/lbmFrom the psychrometric chart, for a temperature of 75 F and a humidity ratio of about 0.0164 lbm/lbm, therelative humidity needed for the supply air for the dedicated outdoor air system (DOAS) is 88% (90%).[ASHRAE Psychrometric Chart No. 1 - Normal Temperature at Sea Level]Mass Transfer andSeparation(/admin/questions/indexsfield area&stext MasTransfer and ld area&stext PsycPRODUCTS USED n/questions/0/preview/497362/2
10/3/2020PPI Learning Hub Admin : QuestionsThis problem corresponds toMEHRPE2 problem number 74.Test Bank Question Return to Questions (/admin/questions/0?sfield magento id&stext 0000142304&sdka &stype &sdiff )previewQuestionAn ideal, vapor compression refrigeration cycle using refrigerant-134a operates with a 90 F saturationcondensing temperature and a 40 F saturation evaporator temperature. There is 10 F subcooling inthe condenser and 20 F superheat at the end of the evaporator. The ideal horsepower needed for thecompressor to achieve 3 tons of cooling is most nearlyAnswers(A) 0.80 hp(B) 1.3 hp(C) 2.4 hp(D) 3.5 hpThe answer is (B).SolutionContent in blue refers to the NCEES Handbook.Use the pressure-enthalpy (p-h) diagram for refrigerant-134a and the single-stage refrigeration cycle todetermine the mass flow rate. The illustration shows the p-h diagram for this cycle, including the 20 Fsuperheat and 10 F subcooling. [Pressure Versus Enthalpy Curves for Refrigerant 134a]QUESTION DATAVendor0000142304Solving ted On10/23/2019 08:12:56PMPublished On10/23/2019 08:12:56PMModified On09/29/2020 08:53:19PMOTHER VERSIONS10/23/2019 08:12:56 P(/admin/questions/preDISCIPLINESFE Chemical(/admin/questions/indsfield discipline&stextChemical)PE Mechanical: HVAC aRefrigeration(/admin/questions/indsfield discipline&stextMechanical: HVAC andRefrigeration)PE Mechanical: Thermand Fluid Systems(/admin/questions/indsfield uestions/0/preview/452971/2
10/3/2020PPI Learning Hub Admin : QuestionsMechanical: Thermal aFluid Systems)KNOWLEDGE field area&stext FluiMechanics/Dynamics)Equipment and Comp(/admin/questions/indsfield area&stext Equand Components)Hydraulic and FluidEquipment(/admin/questions/indsfield area&stext Hydand Fluid Equipment)Conditions at the evaporator, as denoted in the P-h diagram above, arehin 38 Btu/lbmPRODUCTS USED IN112.5hout 115 Btu/lbmThe capacity in British thermal units per hour isBtuq (3 tons) (12,000PEMEHV2EXPEMEHVQBPEMETSQB)hr-ton 36,000 Btu/hrThe mass flow rate isBtu(36,000 m )hrq hout hinBtu(115Btu 38lbmsec) (3600lbm)hr 0.13070 lbm/ secThe conditions at the compressor, as denoted in the P-h diagram above, arehin 115 Btu/lbmhout 122 Btu/lbmThe power needed for the compressor isCompressors P m (hin hout )lbm(0.130Btu) (122secBtu 115lbmft-lbf) (778lbm)Btu ft-lbf550hp-sec 1.29 hp(1.3 eview/452972/2
9/20/2020PPI Learning Hub Admin : QuestionsThis problem corresponds toMEHRPE2 problem number 70.Test Bank Question Return to Questions (/admin/questions/0?sfield magento id&stext 0000142308&sdka &stype &sdiff )previewQUESTION DATAQuestionA round duct industrial ventilation system is shown. The static pressure at the branch is 0.10 in wg,and the exit pressure at each diffuser is 0.03 in wg. The straight branch is 20 ft long with a diameter of11 in, and the take-off branch is 35 ft long. The flow rate of air needed in each branch is 1000 ft3/min.The friction factor for the ducts is 0.02. Assume standard dry air conditions at sea level, and that theflow rate is not impacted by the 90 turn. Dynamic losses are negligible.The duct diameter needed for the take-off branch is most nearlyVendor0000142308Solving ted On05/05/2020 08:27:36PMPublished On05/05/2020 08:27:36PMModified On09/10/2020 11:11:06PMOTHER VERSIONSAnswers10/23/2019 08:04:58 P(/admin/questions/pre(A) 8 in(B) 10 in05/02/2020 12:57:26 P(/admin/questions/pre(C) 13 in(D) 18 in05/05/2020 08:27:36 P(/admin/questions/preThe answer is (C).SolutionDISCIPLINESContent in blue refers to the NCEES Handbook.The only reason for pressure drop in the ductwork is friction. From the Darcy-Weisbach equation,Hea
PE Mechanical: HVAC a. Refrigeration (/admin/questions/ind sfield discipline&stext Mechanical: HVAC and Refrigeration) KNOWLEDGE AREAS. Equipment and Comp (/admin/questions/ind sfield area&stext Equ and Components) Systems and Compone (/admin/questions/ind sfield area&stext Sys and Components) replace with s 4. Neglecting the efficiency of .
PE Mechanical: HVAC and Refrigeration (/admin/questions/inde sfield discipline&stext Mechanical: HVAC and Refrigeration) PE Mechanical: Therm . The rate of heat loss from the bare pipe is most nearly. Answers. 14,000 Btu/hr 18,000 Btu/hr 19,000 Btu/hr 21,000 Btu/hr The answer is (A). Solution. Content in blue refers to the NCEES Handbook .
Bigtable [19], and Megastore [8] all run on Borg. For this paper, we classify higher-priority Borg jobs as “production” (prod) ones, and the rest as “non-production” (non-prod). Most long-running server jobs are prod; most batch jobs are non-prod. In a representative cell, prod jobs are allocated about 70% of the total CPU resources and .
Dodger Theatrical/Araca Group, Prod. 42 ND STREET 2001 Revival Mark Bramble, Dir. Randy Skinner, Choreo. Stage Holding/Dodger Theatrical, Prod. STONES IN HIS POCKETS Ian McElhenny, Dir. Emanuel Azenberg, Prod. JUDGEMENT AT NUREMBERG John Tillinger, Dir. National Actors Theatre, Prod. A
- An example of a process for a SOC would be to follow a naming convention like appending [pre-prod] and [prod] to policy names to differentiate policies - Follow a naming convention like [pre-prod] and [prod] policy names to differentiate policies and filter active channels Monitoring Phase Part 2 - Involve Key Departments 12 TSM
From Hyper Converged Infrastructure to Hybrid Cloud Infrastructure Karan Gupta, Principal Architect, Nutanix. Dheeraj Pandey . 250 300 350 Prod Wkld 1 Prod Wkld 2 KOPS RocksDB TRIAD 0 2 4 6 8 10 Prod Wkld 1 Prod Wkld 2 Write Amplification RocksDB TRIAD uniform skewed TRIAD: low and uniformWA. 4x
2 - the library building is a public library recognized by the state library agency as a public library; 3 - the library building serves an area of greater than 10 percent poverty based on U.S.Census . Falmouth Area Library 5,242.00 Fennville District Library 16,108.00 Ferndale Public Library 16,108.00 Fife Lake Public Library 7,054.00 Flat .
3 07/2021 Dublin Public Library – SW f Dudley-Tucker Library – See Raymond Gilsum Public library [via Keene] Dummer Public Library [via White Mountains Community College, Berlin] NE t,r Dunbar Free Library – See Grantham Dunbarton Public Library – SW f Durham Public Library – SW w, f East Andover (William Adams Batchelder Library [via
The Baldrige Education Criteria for Performance Excellence Framework Empirical test and validation Masood Abdulla Badri and Hassan Selim Department of Business Administration, College of Business .
