EXAMPLES AND TASKS
Авторы:
1.
Mendeleev University of Chemical
Россия
Россия
Тип:
Глава
Страницы с - по
74 - 91
Издатель:
AUS PUBLISHERS
Язык материала:
английский
Ключевые слова:
digital economy , economic development, digital technologies, engineering applications, dissipative functions
Аннотация (русский):
The monograph presents the results of a study of the digital economy as a new paradigm of economic development, a system of economic relations implemented through the use of digital information computer technologies. It is noted that the main problem in the formation of sustainable economic growth and the successful introduction of digital technologies are the challenges of digitalization of the economy. New digital technologies, innovative business models penetrate into all spheres of the economic life of society, influencing the very essence of the economy, forming qualitative structural changes in it. As a result, a digital economy is being formed as a subsystem of the traditional economy, characterized by the active use of digital technologies and the circulation of specific electronic goods. The monograph is intended for researchers, teachers, graduate students, undergraduates, as well as a wide range of readers interested in topical issues of digitalization of the economy.
The monograph presents the results of a study of the digital economy as a new paradigm of economic development, a system of economic relations implemented through the use of digital information computer technologies. It is noted that the main problem in the formation of sustainable economic growth and the successful introduction of digital technologies are the challenges of digitalization of the economy. New digital technologies, innovative business models penetrate into all spheres of the economic life of society, influencing the very essence of the economy, forming qualitative structural changes in it. As a result, a digital economy is being formed as a subsystem of the traditional economy, characterized by the active use of digital technologies and the circulation of specific electronic goods. The monograph is intended for researchers, teachers, graduate students, undergraduates, as well as a wide range of readers interested in topical issues of digitalization of the economy.
Ключевые слова:
digital economy , economic development, digital technologies, engineering applications, dissipative functions
digital economy , economic development, digital technologies, engineering applications, dissipative functions
4.1 Examples.
Task 4.1. H_2 enters the isobaric heat exchanger at a flow rate m ̇_(H_2 )=0.1 ks⁄s. Initial pressure of hydrogen P_(1H_2 )=3∙10^5 Pa and temperature T_1=448 К. The heat flow removed from hydrogen in the heat exchanger is Q=145.5 kW. After the heat exchanger, the gas enters the adiabatic compressor, where the pressure H_2 increases to P_(3H_2 )=9∙10^5 Pa . The exergy loss rate in the compressor is 〖inD〗_ех=12.8 кВт. The ambient temperature is T_ос=298 К , the gas is calculated according to the MIG with a constant heat capacity, с_p^0 (T)=const.
To define:
Draw up an installation plan.
Compressor technical power, W_23^тex,kW;
η_S- adiabatic compressor efficiency;
η_ex- exergy efficiency of the compressor;
T_2,К- gas temperature at the compressor inlet; T_3,К- gas temperature at the outlet of the compressor.
Solution:
We will draw up an installation scheme
Fig.19 Technological site for preliminary preparation of raw materials.
1-heat exchanger;
2-turbocharger;
m ̇_(H_2 )-flow H_2, entering the heat exchanger.
T_1,P_1- temperature and pressure H_2 at the inlet from the heat exchanger;
T_2,P_2=P_1- temperature and pressure H_2 at the outlet of the heat exchanger;
T_3,〖 P〗_3- temperature and pressure H_2 at the outlet of the compressor;
Q ̇_12- heat flow removed from H_2 in the heat exchanger.
Let us represent the total energy balance equations (2.2) for the compression zone
∂Э/∂τ=∑_i▒m ̇_i (h_i+(v ̅_i^2)/2+gφ_i )+Q ̇+H ̇-W ̇^tex=0
Let us analyze equation (2.2) for the considered example:
а) Components (m_i v_i^2)/2=0, ∑_i▒〖m_i gφ_i 〗=0- since there is no change in kinetic and potential energy;
b) Q ̇=0 compressor is adiabatic, there is no heat exchange with the environment;
c) H ̇=0- there is no diffusion flow, no mass transfer transformations occur
Therefore, equation (2.2) takes the following form:
m ̇(h_2-h_3 )-W ̇_23^tex=0, whence the power consumed by the compressor is equal to the loss of gas enthalpy flow:
W ̇_23^tex=m ̇(h_2-h_3 )=m ̇c_p^0 (T_2-T_3 )
where:
T_2-gas temperature H_2 at the compressor inlet, К.
T_3- gas temperature H_2 at the outlet of the compressor, К.
Cp_(H_2)^0=3.5 Rm (MIG - model of an ideal gas,i.e.the number of degrees of freedom H_2 is 5 (j=5),because it is a diatomic gas).
To determine the temperature T_2 we compose the total energy balance equation (2.2) for the heat exchange zone:
∂Э/∂τ=∑_i^n▒m ̇_i (h_i+(v ̅_i^2)/2+gφ_i )+Q ̇+H ̇^ml-W ̇^tex=0
H ̇^ =0 since there is no heat transfer;
W ̇^tex=0- since the technical work in the heat exchanger is not consumed, since the process is isobaric;
∑_i^n▒〖m_i (v ̅_i^2)/2〗=0,∑_i^n▒〖m_i gφ_i 〗=0- there is no change in kinetic and potential energy;
Q ̇=-145.5 kW, since the heat flow is removed from H_2.
Taking into account the above remarks, we obtain the expression for the total energy balance:
m ̇(h_2-h_1 )+Q ̇_12=0
m ̇с_p^0 (T_2-T_1 )=-Q ̇_12
T_2=T_1-Q ̇_12/(m ̇∙с_p^0 )=T_1-Q ̇_12/(m ̇∙2.5R_M )
R_(M_H2 )=R ̃/(MM_H2 )=8314/2.016=4.124 kJ/(kg∙К)
Cp_(H_2)^0=3.5 R_M=3.5∙4.124=14.434 kJ/(kg∙К)
MM_(H_2 )=2.016 kg/kmol
Substituting the data for section 1 (heat exchange) of the technological scheme for T_1 and Q ̇_12, we obtain the value of the temperature H2 at the outlet from the heat exchange zone:
T_2=448-145.5/(0.1∙14.434)=347.19 К.
Now we determine the temperature H_2 at the outlet of the adiabatic compressor stage, T_3S.
T_3S=T_2 (p_3/p_2 )^((k-1)/k)=347.196((9∙10^5)/(3∙10^5 ))^((1,4-1)/1,4)=475.21 К,
where k=1.4 (adiabatic exponent for 2^x atomic gases)
In order to determine the actual gas temperature at the compressor outlet T_3, we write the entropy balance equation (2.5) for the gas compressed in the turbocharger stage:
∂S/∂τ=∑_(i=1)^n▒〖m_i S ̅_i+S ̇^Q+S ̇^mc+inS ̇_23 〗=0
Let's analyze:
Since the compression in the compressor is carried out in the adiabatic mode ( S) ̇^Q=0, i.e. there is no change in entropy due to heat transfer.
S ̇^mc=0- there is no mass transfer in the compressor, i.e. there is no change in entropy due to diffusion.
As a result of the above,expression (2.5)takes the following form:
m ̇(S_2-S_3 )+inS ̇_23=0
Whence the entropy growth rate due to irreversibility for the compression zone can be determined by the following formula:
inS ̇_23=m ̇(C ̇_P ln T_3/T_2 -R_M ln〖p_3/p_2 〗 )
Since the exergy loss rate in the compressor stage is D ̇_23=12.8kW, and the calculation formula for inD ̇_23 is as follows:
(inD) ̇_23=T_(о.с.) m ̇(C_P ln T_3/T_2 -R_M ln〖p_3/p_2 〗 ),
The gas temperature in the actual compression process T3 can be determined by the expression:
C_P ln T_3/T_2 =(inD ̇_23)/(T_(о.с.) m ̇ )+R_M ln〖p_3/p_2 〗=12.8/(298∙0.1)+4.124 ln(9/3)=4.9605 kJ/(kg∙К)
ln〖T_3/T_2 =4.9605/14.434=0.3437〗
Final answer T_3=489.54 К- temperature H_2 at the compressor outlet.
We determine the adiabatic efficiency of the compressor stage (3.1):
η_S=(W ̇_23S^TK)/(W ̇_23^вн )=(m ̇C_p^0 (T_2-T_3s ))/(m ̇C_p^0 (T_2-T_3 ) )
η_S=(T_2-T_3s)/(T_2-T_3 )=(347.196-475.213)/(347.196-489.546)=0.899
We determine the exergy efficiency of the compressor stage:
η_ex=((〖E ̇_x 〗_3-〖E ̇_x 〗_2))/(E ̇_x2 -W ̇_23^вн-E ̇_x2 )
η_ex=(m ̇(e_xвых-e_xтр))/(m ̇(e_xвх-e_xтр)),
where:
e_xвых=e_x3-specific exergy at the compressor outlet
e_тр=me_x2- transit exergy H_2; i.e. that part of the input exergy with which the gas enters the compressor and passes through the compressor zone unchanged.
η_ex=(m ̇(e_x3-e_x2 ))/(-W ̇_13^вн )=(m ̇((h_3-h_2 )-T_(о.с.) (S_3-S_2 )))/(-W ̇_23^вн )
η_ex=(m ̇(C_P (T_3-T_2 )-T_(0.e.) (C_p^ ln〖T_3/T_2 〗-R_M ln〖p_3/p_2 〗 )))/(-m ̇C_p^ (T_3-T_2))
η_ex=1-(inD) ̇_ex23/(-W ̇_23^вн )=1-12.8/(0.1∙14.434(488.546-347.146) )=0.937
The technical power of the compressor W ̇_23^тех, kW is determined by the following relation:
W ̇_23^тех=m ̇C_p^ (T_2-T_3 )=0.1∙14.434∙(347.196-489.546)==-20.402 кВт
Task 4.2.
In an isobaric heat exchanger, a heat flow of 1600 kW is supplied to the air. Air consumption 4 kg/s. Heat exchanger pressure
5∙10^5 Pa. After the heat exchanger, the air enters the adiabatic turbine, where it expands to 1∙10^5 Pa. The air temperature after the turbine is 693 K. The exergy loss rate in the turbine is 150 kW. Ambient temperature T_(о.с.)=298 К. Air according to MIGS with С_p^0 (T)=const.
To define:
To draw up an installation diagram;
Heat exchanger air temperature T_1,К;
W ̇_23^тex,kW- technical power of the turbine kW;
η_s,η_ex- adiabatic, exergy efficiency of the turbine;
inS ̇_(23 ),kW/К- rate of entropy production in the turbine.
inD ̇_23 ,kW- exergy loss due to irreversibility of the process
Solution:
Fig. 20. Technological section 1-isobaric heat exchanger 2 - turbine stage
3- electric generator
1.) Let's write the total energy balance equation for the heat exchanger:
∂Э/∂τ=∑_i^n▒m ̇_i (h_i+(v ̅_i^2)/2+gφ_i )+Q ̇+H ̇^ml-W ̇^tex=0
H ̇=0; W ̇^tex=0; ∑_i^n▒〖m_i (v ̅_i^2)/2〗=0; ∑_i^n▒〖m_i gφ_i 〗=0
m ̇C_p^0 (T_1-T_2 )+Q ̇_12=0
∆T_12=T_2-T_1=Q ̇_12/(m ̇C_p^0 )=1600/(4*1.29)=310,08К
where C_p^0 в-х=4.5∙R_M=4.5∙8.314/29=1.29 kJ/(kg∙К)
∆T_12=-310.08 К
T_1=T_2-∆T_12
(inD) ̇_23=T_(о.с.)∙inS ̇_23
inS ̇_23=150/298=0.503355=503.355 Вт/К- entropy production rate
In the turbine:
in S ̇_23=m ̇[C_p^0 ln〖T_3/T_2 〗-R_M ln〖p_3/p_2 〗 ]
where T_3,p_3- pressure and air temperature at the outlet of the turbine; T_2;p_2- pressure and air temperature at the turbine inlet.
C_p^0 ln〖T_3/T_2 〗-R_M ln〖p_3/p_2 〗=(inS ̇_23)/m ̇ ,
C_p^0 ln〖T_3/T_2 〗=(inS ̇_23)/m ̇ +R_M ln〖p_3/p_2 〗,
finally we have:
ln〖T_3/T_2 =(inS ̇_23+m ̇R_M ln〖p_3/p_2 〗)/(m ̇C_p^0 )〗
ln〖T_3/T_2 =(0.503+40.287 ln〖1/5〗)/(4∙1.29)=-0.2606〗
T_3/T_2 =0.7706
T_2=T_3/0.7706=693/0.7706=899.3 К- air temperature at the inlet to the expansion turbine, i.e. at the outlet of the heat exchanger, so
T_1=T_2+∆T_12=899.3-310.08=589.22 К
Let's define η_S:
η=(W ̇_23^вн)/W_23s =(m ̇C_p^0 (T_2-T_3 ))/(m ̇C_p^0 (T_2-T_3s ) )=(T_2-T_3)/(T_2-T_3s )
Therefore, it is necessary to calculate
T_3S- air temperature at the outlet of the turbine in the equilibrium adiabatic process of air expansion:
T_3s=T_2 (p_3/p_2 )^((k-1)/k)=899.3(1/5)^((1.29-1)/1.29),
where k=1.29 (because the number of degrees of freedom of air is j=7)
T_3s=626.29 К,
η_s=(T_2-T_3)/(T_2-T_3s )=(899.3-693)/(899.3-626.29)=0.756
η_ex=1-(inD) ̇_23/W ̇_23 =1-150/(4∙1.29(899.3-693) )=0.86,
where W ̇_p23=m ̇C_p^0 (T_2-T_3 )=4∙1.29(899.3-693)=1064.51 kW-
technical power of the turbine.
Task 4.3.
He (helium) flows in a horizontal pipe (d=0,05 m) MM_He=4.003 kg/mol initial He parameters at the pipe inlet:
T_1=298 К; p_1=1.2∙10^5 Pa
Final parameters:
T_2=473 К; p_2=1.0∙10^5 Pa
(pressure drops due to hydraulic resistance)
Mass flow He: m ̇_He=0.1 kg⁄s
T_(о.с.)=298 К (ambient temperature)
To calculate the gas according to MIG with constant heat capacity.
To define:
η_пол- polytropic efficiency of the process;
exergy loss rate (inD) ̇_12 (кВт) due to irreversibility of the process;
dissipation energy ψ ̇_12 (kJ⁄kg);
heat flux supplied to the gas Q ̇_12 (kW).
Solution:
Fig.21 Section of a horizontal pipe of constant diameter.
Let's make the equation of balance of kinetic energy:
∂/∂τ (K+π)=∑_(i=1)^n▒〖m ̇_i ((v_i^2)/2+gh) 〗-W ̇_12^tex+m ̇∫_1^2n▒(-dp)/ρ-ψ ̇_σ=0
Let's analyze these equations:
W ̇_12^tex=0- no technical work;
∆m ̇gh- there is no change in potential energy;
m ̇∫_1^2n▒(-dp)/ρ=m ̇ n/(n-1) p_1/ρ_1 [1-(p_2/p_1 )^((n-1)/n) ]-
polytropic power i.e. simulated equilibrium process,
where n=ln(p_2⁄p_1 )/ln(ρ_2⁄ρ_1 ) ;
ρ_1=p_1/(R_M T_1 )=0.194 kg/m^3 ;
ρ_2=p_2/(R_M T_2 )=0.102 kg/m^3 ;
where n=ln(1⁄1.2)/ln(0.102⁄0.194) =0.283.
The power of the polytropic pressure change process:
W ̇_12^пол=m ̇ n/(n-1) p_1/ρ_1 [1-(p_2/p_1 )^((n-1)/n) ]
W ̇_12^пол=0.1 0.283/(0.283-1) (1.2∙10^5)/0.194 [1-(1/1.2)^((0.283-1)/0.283) ]=14.335 kW
Loss of kinetic energy due to friction:
ψ ̇_σ=m ̇ n/(n-1) p_1/ρ_1 [1-(p_2/p_1 )^((n-1)/n) ]-m ̇((v_2^2)/2-(v_1^2)/2);
wheree v_1,v_2- the He speed at the pipe inlet and outlet.
v_1=(4∙m ̇)/(πd_1^2 ρ_1 )=(4∙0.1)/(3.14∙(0.05)^2∙0.194)=262.66 m/s
v_2=(4∙m ̇)/(πd_2^2 ρ_2 )=(4∙0.1)/(3.14∙(0.05)^2∙0.102)=499.56 m/s
ψ ̇_σ=0.1∙143.35-0.1∙10^(-3) (〖499.56〗^2/2-〖262.66〗^2/2)
ψ ̇_σ=5.307 kW- kinetic energy leakage for friction.
We determine the polytropic efficiency of the gas flow.
η_пол=(W ̇_12^необр)/(W ̇_12^обр )=(m ̇ n/(n-1) p_1/ρ_1 [1-(p_2/p_1 )^((n-1)/n) ]-ψ ̇_σ)/(m ̇ n/(n-1) p_1/ρ_1 [1-(p_2/p_1 )^((n-1)/n) ] )
η_пол=(14.335-5.307)/14.335=0.63
Q ̇_12-heat flux supplied to the gas.
Let us write the total energy balance equation for the pipe:
∂Э/∂τ=∑_i▒m ̇_i (h_i+(v ̅_i^2)/2+gφ_i )+Q ̇_12+H ̇-W ̇_12^tex=0
m ̇(h_1-h_2 )+m ̇((v_2^2)/2-(v_1^2)/2)+Q ̇_12=0
Q ̇_12=m ̇(h_2-h_1 )+m ̇((v_2^2)/2-(v_1^2)/2)
Q ̇_12=m ̇C_p^0 (T_2-T_1 )+m ̇((v_2^2)/2-(v_1^2)/2)
Q ̇_12=9.028+90.822=99.85 kW.
V. (inD) ̇_12=T_(о.с.)/Tтд*ψ ̇_σ=298/378.8*5.307=4.17кВт,where Ттд=(Т1-Т2)/(ln Т1/Т2)=378.8 К – average thermodynamic temperature of the gas flow process.
Task 4.4. To determine the dissipation of kinetic energy during the movement of ammonia in the heat exchanger, ammonia moves from top to bottom with a flow rate of kg/s. The cross-sectional diameter of the inlet pipe м, outlet pipe m, the difference in the levels of the location of the pipes is 6 m. Parameters of ammonia at the inlet to the heat exchanger; ; Р1 = 1,5 bar; parameters at the outlet of the heat exchanger Т2 = 300 K; Р2 = 1,3 bar.
Solution:
We find the dissipation of kinetic energy from the balance equation of kinetic and potential energy in the integral form (4.1):
.
To calculate the kinetic energy sink functional, we use the model of an equilibrium polytropic process, the initial and final states of which completely coincide with the real process:
, (1.31)
where – average value of the polytropic index for the initial and final parameters characterizing the state of the gas in the real process.
To determine the density of ammonia, we use the equation of state of an ideal gas, due to a slight deviation of the parameters from standard conditions:
kg/m3;
kg/m3
Let's determine the value of the polytropic index: :
Let's calculate the kinetic energy sink functional:
The gas velocities at the inlet and outlet of the heat exchanger are determined by the equation:
,
where: – the gas velocity at the inlet to the heat exchanger, m/s;
– gas velocity at the outlet from the heat exchanger, m/s.
m/s,
m/s.
Let's determine the dissipation of kinetic energy:
.
Task 4.5. In a cooled compressor, N_2 is compressed in the state of an ideal gas with a constant heat capacity. Compressor initial gas parameters
T_1=293 К,P_1=1∙10^5 Pa, gas mass flow rate m ̇_N2=1 kg⁄s. The compressor gas pressure is P_2=5∙10^5 Pa. The compressor is cooled by circulating water entering through the compressor jacket, cooling water consumption m ̇_(H_2 O)=3.58 kg⁄s,C_P^(H_2 O)=4.19 kJ/(kg∙К). The compressor temperature of the chilled water increases
by 10 К. The technical work of the compressor is W_12^tex=-200 кДж/кг. The compression process is non-equilibrium.
To define:
T_2- temperature N_2 at the outlet from the cooled compressor, К;
ψ ̇_12-dissipation of kinetic energy into friction,kW;
η_пол- polytropic efficiency of the process;
To present (PV), (TS) the charts.
η_ex- exergy efficiency of the process;
Fig. 22. Functional diagram of gas compression in the compressor stage.
Solution:
1.) Let's determine the final temperature of nitrogen N2 at the outlet of the cooling stage of the compressor T2, К.
Let us compose the total energy balance equation (2.2) for the zone of the cooled stage with respect to nitrogen. When analyzing the compression process, we do not take into account changes in the kinetic and potential energy of the gas.
m ̇_(N_2 ) (h_1-h_2 )-W ̇_12^tex-Q ̇_12=0,
where: W ̇_12^tex=-200 kW, the available work flow supplied to the gas in the nominal operating mode.
Q ̇_12=m ̇_H2O·〖Cp〗_H2O·∆T_H2O=-150 kW – the heat flow removed from the gas in the cooling system.
We express the value of the final temperature of nitrogen at the outlet of the compressor stage:
T_2=T_1-(Q_12+W ̇_12^tex)/(m ̇_(N_2 )·〖Cp〗_(N_2 ) )
T_2=293-(150+(-200))/(1·1,038)=341,17 К
2.) Let's determine the dissipation of kinetic energy on friction , kW.
Calculation of dissipation of kinetic energy on friction .
We estimate dissipation using the balance equation for kinetic and potential energy for a nonequilibrium process (2.4).
,
where kW – internal power of the nonequilibrium process;
– the power of the equilibrium polytropic process, kW,
– average value of the polytropic index for the initial and final parameters characterizing the state of the gas at the inlet and outlet of the compressor.
The results of calculations of viscous dissipation are presented in Table 4.1.
Table 4.1
Analysis of the polytropic equilibrium process of nitrogen compression
kg/m3
kg/m3
1,15 341,17 4,94 1,161 –151,12 –200 48,88
Let us represent the equilibrium polytropic process of nitrogen compression (1–2 ) in coordinates (PV), (TS) (Fig. 23).
Fig. 23. Polytropic equilibrium process of nitrogen compression with heat removal in the coordinates PV(a), TS(b).
We determine the polytropic efficiency, the value of which characterizes the degree of energy perfection of the gas compression process .
.
3.) Exergy analysis of the nitrogen compression process with heat removal.
The exergy analysis of the process of nitrogen compression with heat removal is carried out on the basis of the exergy balance equation (3.7). Heat flow exergy is calculated based on the equation:
,
where – heat flow removed from the gas, kW,
То.с.= 298,15 K – ambient temperature, under standard conditions,
– average thermodynamic temperature of the process, K,
The exergy balance equation (3.7), taking into account the ratio, takes the following form:
,
where – exergy flow due to the convective movement of the gas flow, kW,
– heat flow exergy, kW,
– technical power of the compressor, kW, where - the internal efficiency of the compressor stage,
– internal exergy losses due to irreversibility of the process, kW.
Calculation of internal losses of exergy , due to the irreversibility of the process itself is carried out on the basis of the following relation (3.16):
,
where – the dissipation of kinetic energy due to friction, kW,
– ambient temperature under standard conditions,
– average thermodynamic temperature of the process, K.
The results of calculations of the exergy flux of nitrogen due to convection are presented in Table 4.2.
Table 4.2
Calculation of nitrogen specific exergy loss due to convection
kJ/kg
kJ/(kg∙K) W_12^tex e_x1-e_x2
kJ/kg
kJ/kg
293 341,17 50 0,158 –200 147,13
The calculation of the average thermodynamic temperature of the process is carried out on the basis of the exergy balance equation (3.15).
.
The calculation of exergy losses due to the irreversibility of the process is carried out according to the calculated ratio (3.16):
.
The value is the final loss of exergy, which can only be compensated by external energy carriers.
The difference between the values of dissipation of kinetic energy and internal losses of exergy, due to the irreversibility of the process, is that part of the exergy that can still be usefully used in the future.
.
In the event that the thermal exergy of the cooling agent is not used, since its value is very small, then the exergy losses are calculated according to the equation:
,
where – the total exergy losses, kW,
– internal exergy losses, kW,
– external exergy losses, kW.
The calculation of the exergy efficiency of the compression process with heat removal in this case is carried out according to the following relation:
.
The results of calculations of the exergy efficiency of the ammonia compression process without utilization of the removed heat are presented in Table. 4.3.
Table 4.3
Calculation of the exergy efficiency of the process without utilization of the removed heat flow
–150 –200 5,92 46,95 310,4 0,74
To increase the efficiency of the process, it is advisable to provide a deeper cooling of the compressed gas. To improve the cooling system, air or oil cooling can be used.
4.2. Tasks for solitary work.
Task 4.6. Reciprocating refrigerated compressor 1 sucks in nitrogen at a temperature Т1 = 306 K and pressure Р1 = 1 bar and pumps compressed gas into cylinders 3 with a volume of 50 l (Fig. 23). The gas pressure at the outlet of the compressor 1 is maintained constant and equal to Р2 = 5 bar due to the installation of a control valve 2 on the discharge lines. The compression process is non-equilibrium, it is known that the ratio of the removed heat and internal work is = 0,6. Gas parameters at the outlet of the compressor: Р2 = 5 bar. The compressor capacity is equal to = 1,2 m3/min in terms of compressed gas parameters. To calculate the parameters and functions of the state of nitrogen according to the Bogolyubov-Mayer virial equation of a non-ideal gas in a truncated form [1]. Critical parameters and heat capacity constants of nitrogen are presented in Annex (P-3, P-4). The internal isothermal efficiency of the compressor is taken equal to = 0,7.
Fig. 23. Schematic diagram of the installation:
1 - piston compressor; 2 - control valve;
3 - cylinder for compressed gas
Let's determine: 1) nitrogen temperature at the compressor outlet Т2, K;
2) compressor internal power , kW; 3) dissipation of kinetic energy on friction, , kW; 4) compressor operation time for filling thirty cylinders , s; 5) the rate of exergy loss due to the irreversibility of the process, , kW; 6) exergy efficiency of the compressor unit, provided that the removed heat flow is usefully used, .
Answer: 1) Т2 = 386,67 K; 2) = –18,16 kW; 3) = 3,83 kW; 4) = 75 с; 5) = 3,315 kW; 6) = 0,82.
Task 4.7. The turbocharger is used to compress carbon dioxide to a final pressure of Ркон = 20 bar. The state of the gas during suction is determined by the pressure Рнач = 1 bar and temperature Тнач = 280 K. The compression process is adiabatic, non-equilibrium. The compressor capacity is 30 m3/h of compressed gas, the degree of pressure increase in one stage is not allowed more than = 4. Between the stages, the compressed gas is cooled isobarically to the initial temperature Тнач = 280 K in heat exchangers. For cooling, water from the circulating water supply is used, the heating of the cooling water is = 7 K. The value of the adiabatic efficiency of all stages of the compressor unit is assumed to be the same and equal to = 0,83. In calculations, it is permissible to use the model of an ideal gas with a constant heat capacity.
Let's determine: 1) the number of stages of the compressor unit, n; 2) internal power of each stage , kW; 3) heat flow removed in intermediate coolers , kW; 4) exergy efficiency of the compressor unit .
Answer: 1) n = 3; 2) kW; 3) = –17,44 kW; 4) = 0,78.
Task 4.8. Oxygen is compressed in the adiabatic stage of an uncooled compressor. Oxygen parameters at the compressor inlet: Т1 = 300 K,
Р1 = 1 bar, compressor outlet pressure: Р2 = 4 bar. After compression, the gas is sent to the refrigerator, where it is cooled isobarically to the initial temperature Т1 = 300 K. The gas flow rate is = 1 kg/s, the compression process is nonequilibrium. Calculation of the parameters and functions of the state of oxygen is carried out according to the virial equation of a nonideal Bogolyubov-Mayer gas in a truncated form [1]. The critical parameters and heat capacity constants of oxygen are given in [1, 4] and presented in Annex (P-1, P-2). The adiabatic internal efficiency of the compressor is = 0,8.
To determine: 1) oxygen temperature at the outlet of the compressor
Т2, K; 2) compressor internal power , kW; 3) polytropic compressor efficiency пол; 4) heat flow removed from oxygen in the heat exchanger , kW; 5) exergy efficiency of the compressor unit, provided that the removed heat flow is not usefully used, .
Answer: 1) Т2 = 481,55 K; 2) = –163,18, kW; 3) пол = 0,847;4) = –162,27 kW; 5) = 0,87.
Task 4.9. Oxygen is compressed in an adiabatic compressor, gas parameters at the compressor inlet: Т1 = 300 K, Р1 = 1 bar, gas pressure at the compressor outlet: Р2 = 4 bar. The compression process is non-equilibrium, the gas flow rate is = 1 kg/s. It is known that the internal work of the compressor is 28 kJ/kg greater in absolute value than in a reversible adiabatic process proceeding up to the same final pressure. The functions and parameters of the state of oxygen are calculated according to the Bogolyubov-Mayer virial equation of a non-ideal gas in a truncated form [1]. The critical parameters and heat capacity constants of oxygen are given in [1, 4] and presented in Annex (P-1, P-2).
To determine: 1) the final temperature of oxygen in the irreversible process of compression Т2, K; 2) compressor internal power , kW; 3) loss of exergy due to the irreversibility of the process , kW; 4) dissipation of kinetic energy on friction , kW; 5) exergy efficiency of the compressor .
Answer: 1) Т2 = 476,15 K; 2) = –161,03 kW; 3) = 16,0063 kW; 4) = 23,73 кВт; 5) = 0,901.
Task 4.10. To determine the savings in internal power that can be obtained by switching from single-stage to two-stage compression with intermediate isobaric cooling in a heat exchanger to the initial temperature of a mixture of oxygen and nitrogen in equal mole fractions. The parameters of the mixture at the inlet to the compressor: Т1 = 306 K, Р1 = 1 bar, outlet pressure: Р2 = 7 bar. The compression process is non-equilibrium, adiabatic, mixture flow rate is = 1 kg/s. The degree of pressure increase in both stages is considered the same, the internal adiabatic efficiency of each stage is taken equal to = 0,8. A mixture of gases can be considered ideal with a constant heat capacity.
Answer: 10,6∙10–2.
Task 4.11. To determine the energy dissipation during the movement of nitrogen into the reactor along the horizontal section of the gas pipeline. Inlet diameter
d1 = = 0,04 m, outlet diameter d2 = 0,03 m. Nitrogen parameters at the gas pipeline inlet: , bar; parameters at the outlet of the gas pipeline; , bar. Gas consumption kg/s. The process is unbalanced. Nitrogen can be considered an ideal gas with a constant heat capacity due to a slight deviation of the parameters from the standard ones.
Answer: kW.
Task 4.12. The compressor sucks in air at a pressure of 1 atm and a temperature of 20°C and compresses it isothermally up to 8 atm. (compressor internal efficiency = 0,85). To determine the capacity of the compressor m3/h, if the theoretical engine power to drive the compressor is 40 kW. Also determine the hourly consumption of cooling water if its temperature rises by 10°C when the compressor cylinder is cooled. The heat capacity of water is taken equal to 4.19 kJ/kg K.
1. Calculation and analysis of thermodynamic processes of expansion (compression) of a non-ideal gas / comp. N. Z. Khabibova. – M.: RCTU named after D.I. Mendeleev 2005. – 48 p.
2. Low-temperature processes of expansion, liquefaction and separation of gases in chemical technology / L.V. Ravichev, N.Z. Khabibova – M.: RCTU named after D.I. Mendeleev, 2021. – 72 p.
3. Analysis of non-equilibrium processes of compression of a non-ideal gas. Examples and tasks. / N.Z. Khabibova – M.: RCTU named after D.I. Mendeleev, 2007. – 52 p.
4. Modern thermodynamics from heat engines to dissipative structures / I. Prigozhin, D. Kondepudi. – M.: Mir, 2002. – 461 p.
5. Examples and tasks in the course of processes and apparatuses of chemical technology / K.F. Pavlov, P.G. Romankov, A.A. Noskov. 11th ed. – M.: "Rusmedia Consult" LLC, 2004. – 576 p.
6. Properties of gases and liquids / R. Reed, J. Prausnitz, T. Sherwood. – L.: Khimiya, 1982. – 534 p.
7. Basic processes and devices of chemical technology. Design manual / ed. Yu.I. Dytnersky. – 2nd ed. – M.: Khimiya, 1991. – 496 p.
8. Heat engineering / V.I. Krutov. – M.: Mashinostroenie, 1986. – 432 p.
9. Secondary energy resources and energy-technological combination in industry / L.I. Kuperman, S.A. Romanovsky, L.N. Sidelkovsky. – 2nd ed. – Kyiv: Vishcha school, 1986. – 303 p.
10. Basic processes and devices of chemical technology / N.I. Gelperin. – M.: Khimiya, 1981. – 812 p.
11. Fundamentals of energy-resource-efficient environmentally safe technologies of oil refining / V.P. Meshalkin, L.L. Tovazhnyansky, P.A. Kapustenko. – Kharkov: NTU "KPI", 2011. – 616 p.
12. Convective heat and mass transfer / V.M. Case. – M.: "Energiya", 1972. – 448 p.
