If a throttling valve is connected to the outlet for one of these tanks, whereby air leaves this valve steadily at 305K with a pressure drop of 50bar, apply an exergy balance across the valve and calculate the exergy change (units: kJ/kg). CAES-Background Info to (Calculated) QuestionsA small-scale Compressed Air Energy Storage (CAES) system uses gas tank storage,with any single tank sized to carry a maximum 2000kg air. The air storage helpssupport various fluid power (pneumatics) requirements, but also serves for energystorage (at 100% charge).The CAES system sits in a plant room at a geographical location with an environmentaltemperature of 300K and ambient pressure of 1 atmosphere (101.325kPa). Due to thepresence of other equipment in the plant room and its built-up nature, is theimmediate surroundings from which tanks are filled are normally a little warmercompared to the outside ambient air (but same ambient pressure). The plant roomitself is at 310K.The specification for a single tank stipulates it can only be filled with(dry) air to a maximum (gauge) pressure of 500bar (at typical operating conditions).When left inside the plant room, the CAES system eventually attains the plant roomtemperature.You may assume:Steady state conditions applyNegligible changes in kinetic and potential energy• Thermal mass of the tank (metal) is negligible• Air behaves as an ideal gas●A gas tank is rigid

given; inlet temperature(T1)=310Kinlet pressure(P1)=500barexit temperature (T2)=305Kexit…

Answered: If a throttling valve is connected to…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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If a throttling valve is connected to the outlet for one of these tanks, whereby air leaves this valve steadily at 305K with a pressure drop of 50bar, apply an exergy balance across the valve and calculate the exergy change (units: kJ/kg).

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the pressure dropped by 50 bar and the initial was 500 bar so P2 should be 450 bar. Also, how did u get this ln50500

Step 1
given;
inlet temperature(T1)=310Kinlet pressure(P1)=500barexit temperature (T2)=305Kexit pressure(P2)=50ba
renvironment temperature(TO)=300K
change of exergy(41-42)=?
Step 2
solution;
we have used for exergy balance equation for flowing of fluid after valve open;change of exergy(41-42)=
(h1-h2)-TO*(S1-S2)(or) change of exergy(41-42)=Cp*(T1-T2)-T0*(S1-
eq.1where;h1=initial enthalpy of airh2=exit enthalpy of airS1=inlet entropyS2-exit entropy
TO=enviornment temperaturespecific heat of air at constant pressure(Cp)=1.005 KJ/kgkgas constant for
air(R)=0.287KJ/kgknow;chnage of entropy(S2-S1)=Cp*lnT2T1-
S2)
R*InP2P1
0.287 50500
S2)
(-0.644)
=1.005*In305310-
=-0.0163+0.6608
ntropy(S2-S1)=0.644KJ/kgk(or)chnage of entropy(S1-
S2)=-0.644KJ/kgkput all values in eq.1; change of exergy(41-q2)=Cp* (T1-T2)-T0*(S1-
=1.005* (310-305)-300*
=198.225KJ/kg change of exergy(p1-q2)=198.225 KJ/kg
Step 3
=0.644KJ/kgkchnage of e
answer;
change of exergy(p1-p2)=198.225KJ/kg

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