FE/EIT sample problems from EIT/FE exam for dynamics
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SAMPLE PROBLEMS:
Exergy Calculations
1. A parallel-shaft gearbox is planned to be installed on a new exercise machine. This gearbox receives work through a high-speed shaft and then delivers work through a low-speed shaft. The gearbox is cooled on its outer surface by convection with the surrounding air. A test is needed to exam the performance of this gearbox. The gearbox receives 60 k through a high-speed shaft. The outer surface of the gearbox is 0.! m ". #eat transfer coefficient between the gearbox and its surrounding air is 1.! k$m"-% and the temperature of the outer surface of the gearbox is &0 o'. The ambient pressure is 1 bar( and the ambient temperature is ") o'. *etermin *etermine e the work delivered by this gearbox. +xplore the exergy transfer of the gearbox and the exergy destruction. SOLUTION: To complete the test on the gear box( the work output from the gear box( the exergy transfer( and the exergy destruction need to be determined. Assumptions, The temperature on the outer surface is uniform. The test process is steady. steady. • •
1 *etermine the work delivered by the gearbox Take Take the gear box as a closed system. The gearbox satisfies the energy balance( which is Ein - Eout = ΔU where +in / total energy transferred into the system +out / total energy transferred out of the system / 2nternal change of the gearbox 3ince the test process is steady( properties inside the gearbox do not change. ork in is received received through the high-speed high-speed shaft( shaft( work out is delivered delivered through the low-speed low-speed shaft( and heat is transferred from the outer surface of the gearbox to the ambient air. Therefore( Therefore( the energy balance becomes( Win - Wout - Q = 0 where in( out( and 4 are positive numbers. 2n order to determine out( heat transferred from the gearbox 4 needs to be determined first. 5or convection heat transfer( 4 euals( Q = Ah(T b - T0) = 0.(!"00)(#0 - $%) = $!.& 'W in is given as 60 k. 3ubstitute 4 and in into the energy balance gives( &0 - Wout - $!.& = 0 Wout = .# 'W " +xplore the exergy transfer of the gearbox and the exergy destruction +xergy can be transferred by heat( work( and mass. 3ince this system is a closed system( exergy of this gearbox is only transferred by heat and work. +xergy transferred by heat is given as 74 / 1 - T0$T4 #eat is transferred from the system to the surroundings( hence( 4 is negative in the above euation.
#eat transferred from the system results in exergy transferred from the system. +xergy transfer by the shaft work is eual to the work itself. #ence( 7 w(in / 60.0 k and 7w(out / 8!.& k +xergy destruction is given as 7destroyed / T03gen
where 3gen / entropy generation during the test process Tb / ambient temperature( is given as ")o' The entropy generation during the test process can be determined by the entropy balance of the gearbox. That is( Sin - Sout S*+n = ΔS,,t+ 3ince the test process is steady( no entropy change occurs inside the system. ΔS,,t+ = 0 Then the entropy balance for the gearbox is simplified to, S*+n = Sout - Sin 9ote that entropy cannot be transferred by work. 2t can only be transferred by heat and mass. 3ince the gearbox is a closed system( entropy is only transferred by heat. S*+n = Sout - 0 = Q/T b where 4 / heat transfer from the gearbox( 4 / "1.6 k Tb / temperature at the outer surface( is given as &0o' 3ubstitute 4 and Tb to the entropy balance yields( 3gen / "1.6$&0:";8 3ince 7destroyed / T03gen( +xergy destruction can be determined as 7destroyed / T03gen / ") : ";8"1.6$&0 : ";8 / "0.6 k Another way to determine the exergy destruction is, exergy destruction euals the difference between the exergy in and exergy out( which is 7destroyed / 7in - 7out / 60.0 - 8!.& - 1.0 / "0.6 k The exergy analysis is summari =ate of +xergy out, low-speed shaft 8!.& k 6&> heat transfer 1.0 k 1.;> =ate of +xergy *estruction, "0.6 k 8&.8> ". ?at is taking a thermodynamics course this semester and he tries to connect his daily life with what he learned in class. =ecently( he has calculated how much entropy is generated when boiling eggs. 9ow( he wanders how much exergy is destroyed when the eggs are boiled. hat is known, The 6-cm-diameter spherical egg at 10 o' is dropped into boiling water. The boiling water is at 100o'. The egg has a density of 1(000 kg$m8 and a specific heat of 8.& k@$kg-%. The ambient pressure is 1 bar and the ambient temperature is ") o'. #ow much exergy is destroyed by boiling one egg to 100 o' •
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SOLUTION: +xergy destroyed by boiling an egg from 10o' to 100o' in 100o' water needs to be determined. Assumptions, Bodel both the egg and the water as incompressible substance. 'onstant-specific-heat assumption is valid for both the water and the egg. %inetic and potential energy changes of the egg are negligible. 9o work interaction is involved. ater remains 100 o' during the cooking process. • • • •
Take the egg as a system. 2t is a closed system since n o mass flows in or out of the egg. The exergy balance for a closed system is,
where 1 denotes the initial state and " denotes the final state. The assumption states that no work interaction is involved( and the volume of the egg does not change during the cooking process since eggs are incompressible. Thus( the exergy balance can be simplified as
=earranging the above euation gives the expression of the exergy destruction. That is(
The first term on the right hand in the above euation represents the exergy transfer by heat transfer. The egg is cooked from 10 o' to 100o' in 100o' water. Assuming the boundary of the egg remains at 100o' during the whole process then this term can be simplified to
where T0 is the ambient temperature and is given as ") o'. T is the temperature at the boundary of the egg where heat transfer occurs. #ence(T is eual to 100o'. 4 is the heat transferred to the egg from the water. 4 can be determined by the energy balance of the egg. The egg is approximated as an incompressible substance and no work interaction is involved. ith constantspecific-heat assumption( the energy balance is Q = 12+ (T$ - T!) The mass of the egg is = 34 = #/(.!#)(0.0)(!"000) = 0.!! '* The average specific heat is given as 8.& k@$kg-%. 3ubstituting all the data into the expression of energy balance yields Q = 0.!!("#00)(!00 - !0) = #"%5 6 3ubstituting 4( T 0 and T to the expression of exergy transfer by heat transfer gives(
The second term on the right hand side of the exergy destruction is the exergy change in the egg during the cooking process. The exergy change of a closed system without kinetic and potential energy changes is, 7$ - 7! = (U$ - U!) P0(3$ - 3!) - T0(S$ - S!) 3ince the egg is incompressible( the second term in the right hand of the above euation euals 0. Also( the constant-specific-heat assumption is valid. The exergy change in the egg can be determined from 7$ - 7! = 812+(T$ - T!) - T0812+9n(T$ /T!) T1 is the initial temperature and T " is the final temperature of the egg. They are given as T1 / 10o' / "!8 % T" / 100o' / 8;8 % 3ubstituting all the data into the expression of exergy change in the egg yields 7$ - 7! = 0.!!("#00)(5 - $) - ($% $5)(0.!!)("#00) 9n(5/$) = $"&0.$ 6 After calculating the exergy transfer by heat and the exergy change in the egg( the exergy destruction can be determined. 7;+,t
8. hen a train in a subway approaches the station( it releases heat to its surrounding air when it brakes to stop. The heated air is exhausted through the vents under the station island. An ideal to reuse this waste heat is to design a waste heat recovery system. 2n this system( cold water is heated by the exhausted air. To evaluate the recovery system( exergy analysis of this system should be done. hat is known, "00 kg$s air at 60 o' enters the heat recovery system. Air leaves the heat recovery system at 8)o'. 'old water at ")o' is sent to the heat recovery system and leaves at )0o'. Ambient temperature is ")o'. •
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*etermine the mass flow rate of the cold water. #ow much exergy is destroyed during the heat exchange process in the heat recovery system *etermine the second-law efficiency of the heat recovery system. SOLUTION: A heat recovery system is designed to heat cold water using the exhausted hot air from a subway station. Bass flow rate of the cold water( exergy destruction( and the second-law efficiency of the heat recovery system needs to be determined. Assumptions, Bodel the water as incompressible substance with a density C / 1(000 kg$m8 and a specific heat cw/ &.1! k@$kg-%. The exhausted air is modeled as an ideal gas with a density C / 1.! kg$m8 and a specific heat c?a / 1.00) k@$kg-%. The heat recovery system is well insulated that the heat exchange process is adiabatic. 9o work interaction is involved. 9eglect the kinetic and potential energy changes. The heat exchange process is an isobaric steady process. •
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'onsider the heat recovery system as a control volume and denote the hot air inlet as 1( hot air exit as "( cold water inlet as 8( and hot water exit as & ( shown on the left. 1 *etermine the mass flow rate of the cold water The energy balance for a control volume is,
According to the assumptions( the energy balance can be simplified to ater is modeled as an incompressible substance. Thus( its enthalpy euals h& / cwT& and h8 / cwT8 +xhausted air is modeled as an ideal gas( its enthalpy euals h" / c?aT" and h1 / c?aT1 where T1(T"( T8 ( T& and T1 / 60o' / 888 % T& / )0o' / 8"8 %
are given as T" / 8)o' / 80! % / "00 kg$s
T8 / ")o' / "D! %
3ubstituting all the data to the energy balance gives the mass flow rate of cold water.
" *etermine the exergy destruction of the heat recovery system The exergy balance of control volume undergoing a steady-flow process is
3ince the heat recovery system is an adiabatic heat exchanger( it becomes
where E is the flow exergy. it is expressed as Therefore( the exergy destruction can be rewritten as
= h - T 0,
3ubstitute all the data given to the above euation gives
8 *etermine the second-law efficiency of the heat recovery system The definition of the second-law efficiency of a heat exchanger is
