Production of Low Temperature

aProduction of Low TemperatureRefrigerator Any device of removing heat from a cold place and adding it to a hotter place is called refrigerator. it is essentially a heat engine running backward. In refrigerator, a working substance takes in heat from a body at a lower temperature by an external agent and gives out a large amount of heat to a hot body as shown in figure. Thus, Refrigerator continuously transfers heat from a cold to a hot body at the expense of mechanical energy supplied to it by an external work done. Thus cold body is cooled more and more. The working substance is called a refrigerant.

Let Q2 be the heat removed from the cold body at temperature T2. w is the amount of work performed by the working substance of the refrigerator through compressor. Q1 of the heat delivered to the hot body at temperature T1. Then,a

Q1=Q2+W

W=Q1-Q2

∴W=Q2( Q1 Q2-1)……(i)

If we use an ideal gas as a working substance. We can show that, Q1

Q2=T1

T2 then equation (i) can be written as,∴w=Q2(T1

T2-1) This is the expression for the work done that must be supplied to run the refrigerator.The cooling process is based on the principle of Joule-Thomson expansion. In this process, the gas is practically cooled which becomes a source of cooling for the remaining amount of the gas. The gas passes through cycles along regenerative pipe and finally attains the temperature of liquification. For most of the gases the Joule-Thomson cooling is very small. However, the cooling effect can be intensified by employing the process called the regenerative cooling.Coefficient of performanceDefine coefficient of performance of a refrigerator. How is it related to the efficiency of a Carnot's engine operating between the same temperatures? Coefficient of performance of refrigerator is defined as the ratio of amount of heat absorbed from the cold reservoir to the work done in running the machinery.i. e. 𝛽= Amount of heat absorved from the cold reservied

Work done in running the machine We know,𝛽=Q2

W=Q2

Q1-Q2In Carnot's cycle, Q1

Q2=T1

T2So, 𝛽=T2

T1-T2It 𝜂 is efficiency of a Carnot's engine operating between same temperature.a

	𝜂=1-T2 T1
⇒	1 𝜂=T1 T1-T2
⇒	1 n-1=T1 T1-T2-1
⇒	1-𝜂 𝜂=T1- T1+T2 T1-T2
⇒	1-𝜂 𝜂=T2 T1-T2=𝛽
∴	𝛽=1-𝜂 𝜂

Derive expression for Boyle temperature and critical temperature of a real gas. Explain the significance of these two temperature.Boyles temperatureThe temperature at which the product of pressure and volume of certain amount of perfect gas remains fairly constant up to a high value of pressure is called the boyle temperature for that gas.

Now a Vander wall's equation of state for a real gas is,(P+a

v2)(V-b)=RT ...............(i )where a and b are called Vander walls' constants. PV-Pb+a

V-ab

V2=RT PV=RT+Pb-a

V+ab

V2But, PV=RT or, V=RT

P or, PV=RT+Pb-aP

RT+abP2

RTT2 PV=RT+(b-a

RT)P+ab

R2T2P2.............................(ii)The Boyle law (PV = constant) is suitable only under ideal condition (i.e. high temperature and low pressure). It is found experimentally in certain gas N2,O2,CO2. that the values of the product PV decreases initially with increase in pressure, becomes minimum to a particular pressure and then increases with the increase in pressure, But for hydrogen the product PV increases steadily with increase in pressure. Hence a modification was necessary in Boyle's formula. After a long investigation, the following expression is given for certain temperature.PV=A+BP+CP2+DP3+…………..(iii)Where A,B,C and D etc are constant for a given temperature depending upon the nature of gas.Comparing this with equation (ii) we get,A=RT,B=b-a

RT,C=ab

R2T2At Boyle temperature T=TB and B=0a

O=b-a RT

TB=a Rb

This is required expression for Boyle temperature in terms of Vander wall's constants.Critical temperatureThe temperature below which a gas can be liquefied by the application of pressure alone is called critical temperature and is denoted by TC The corresponding value of pressure and volume are called critical pressure (PC) and volume (Vc) respectively. From Vander Waal's equation of state, we have:(P+a

V2)(V-b)=RTWhere, a and b are Vander Waal's constants.or, P=RT

V-b-a

V2...... (i) Differentiating with respect to V, we get:or, dP

dV=- RT

(V-b)2+2a

V3 ..............(ii)Again, differentiating,d2P

dV2=2RT

(V-b)3-6a

V4 .....................(iii)At critical point, (i.e., at point of inflexion):|a

dP dV=0 and d2P dV2=0

with T=TC,P=PC and V=VC

}............... (iv)So, from equation (ii), (iii) and (iv), we get:RTC

(VC-b)2=2a

VC3 ..........(v) and 2RTC

(VC-b)3=6a

VC4 ..........(vi)Dividing equation (v) and (vi), we get:VC=3b ..................(vii)Then, from equation (v), using (vii), we get:

TC=8a

27Rb

...........(viii)SignificanceThe significance of Boyles temperature is that at this temperature any real gas behaves like a perfect gas up to a wide range of pressure. If the temperature be higher than the Boyle temperature, the product of pressure and volume of certain amount of a gas continuously increase as its pressure is gradually increases. On the other hand, if the temperature be below the Boyles temperature. Such product initially decreases gradually a minimum then continuously increases with increase in pressure. Thus, it is the temperature which gives rise to an inversion of the value of the second coefficient for a gas, from negative to positive, as its temperature gradually change from lower the higher values.The significance of critical temperature is that at this temperature the liquid and gaseous state of a matter can co-exist and pass form one state to another without any heterogeneity in the whole mass. At the critical temperature the surface tension of the liquid state becomes minimum and the compressibility of the gaseous state becomes maximum Moreover, it is the temperature which demarcates in the gaseous state when a matter is to be called as gas and when a vapour. Joules Thomson Cooling for Vander wall's Gas.Deduce an expression for Joule Thomson cooling for Vander Wall's gas. Discuss the importance of temperature of inversion. When a gas under constant pressure is forced through a porous plug to a region of a lower constant pressure. It's temperature changes. This effect is called the Joule-Thomson effect. To express the expression for Joule-Thomson cooling for Vander Wall's gas, let us consider a thermally insulated cylinder contains 1 mole of gas allowed to expand from pressure P1 and volume V1 to pressure P 2 & volume V2 through the porous plug. During the expansion of the gas some external as well as some internal work is done by the internal energy of gas.The work done on the gas =P1V1and the work done by the gas =P2V2∴The net external work done = P2V2 - P1V1 i.e Wexternal= P2V2 - P1V1 .................(i)As some work is done of inter molecular force of attraction and according to vander-wall gas equation, this work is equivalent to the pressure a

V2 ∴ internal work done for one mole of gas from volume V1 to V2 is, Winternal =v2∫V1pdV =v2∫V1a

V2dV =a[-1

V]V2 V1 ∴Winternal =a

V1-a

V2 ................(ii) ∴ Total work donea

W=Wexternal +Winternal

W=P2V2-P1V1+a V1-a V2 ⋯⋯⋯⋯(iii )

for Vander wall's gas equation(p+a

V2)(v-b)=RTWhere the symbols have their usual meaning.a

PV-Pb+a V-ab V2=RT

PV=RT+Pb-a V+ab V2

As a and b are very small. So we have to consider ab

V2→0 then,PV=RT+Pb-a

V .................(iv)Putting the value of PV from (iv) in equation (iii) with appropriate suffixes.W=(RT2+P2b-a

V2)-(RT1+P1b-a

V1)+a

V 1-a

V2=R(T2-T1)-b(P1-P2)+2a

V1-2a

V2 ………..(v)As a and b are very small so equations can be written as,a

PV=RT

V=RT P

V1=RT1 P1 and V2=RT 2 P2

on substituting the value of V1 and V2 in equation (v) we get,a

W	=R(T2-T1)-b(P1-P2)+2aP1 RT1-2aP2 RT2
	=R(T2-T1)-b(P1-P2)+ 2a R(P1 T1-P2 T2)

As, change in temperature is very small in Joules- Thomsons expansion,a

T1=T2=T,T1-T2=dT

W=-RdT-b(P1-P2)+2a RT(P1- P2)

W=-RdT+(P1-P2){2a RT-b}

It Cv be the specific heat capacity at constant volume at temperature dT, therefore heat supplied is equal to work done. i.e. CvdT=-RdT+(P1-P2){2a

RT-b} (Cv+R)dT=(P1-P2)(2a

RT-b} CpdT=(P1-P2){2a

RT-b} ∵Cp-Cy=R ∴dT= P1-P2

Cp{2a

RT-b} As P1>P2,S0(P1-P2) is a+Ve quantity. Now dT varies with {2a

RT-b}.Case : I a

When dT is +Ve

a 2a RT>b i.e. T<2a Rb

This implies that the temperature of gas is smaller than 2a

RbCase : IIfor dT=02a

kT=bi.e. T=2a

RbThis temperature is called temperature of inversion. All this temperature no heating or cooling effect is seen. The case II implies the temperature of inversion.Temperature of inversionThe temperature at which Joule-Thomson effect changes this is know as the temperature of inversion. It is denoted by Ti and at this temperature. 2a

RTi=b or, Ti=2a

RbWhen the room temperature is above the temperature of inversion, the gas will show a heating effect. For example, the temperature of inversion for hydrogen and helium are much below the ordinary room temperature and they show heating effect at ordinary room temperature. Regenerative CoolingRegenerative cooling is a method of cooling gases in which compressed gas is cooled by allowing it to expand and thereby take heat from the surroundings. The cooled expanded gas then passes through a heat exchanger where it cools the incoming compressed gas. It is based upon the principle of Joules - Thomson expansion. Cooling in a single step is very less so this process is repeated for several times to attain desirable low temperature. Cascade CoolingThe Cascade refregeration system is a freezing system that uses two kinds of refrigerants having different boiling points, which run throught their own independnt freezing cycle and are joined by a heat exchanger.

This system is employed to obtain temperature of -400 to -800 C or ultra-low temperature lower than that. At such ultra-low rtemperatures, a common single refrigerant two stage compression system limits the low temperature characteristics of the refregerant to a cinsiderable poor level, making the system significantly inefficient. The efficiency is improved by combining two kinds of refrigerants having different temperature characteristics.Advantage of Cascade Cooling1. Energy is saved because the system allows use of refrigerants that have suitable temperature characteristics for each of the higher-temperature side and the lower-temperature side.2. It allows stable ultra-low-temperature operation.3. The running cost is inexpensive.4. Repair and maintance is easy. Numerical#Calculate the change in temperature when helium gas suffers Joule Thomson expansion at - 173∘C, the pressure difference on the two sides of the plug being 20 atmospheres. Does the gas show a heating effect or a cooling effect in this expansion at -173∘C ?Given: R=8.3 Joules/mole- K and for helium Vander Waal's constants a=0.0341 litre 2 atoms, mole -2 and b=0.0237 litre mole -1. Here, T=-173∘C=100K a

a	=0.0341 libe 2 atom mole-2=0.0341×10-1Nm4me-
b	=0.0237 litre mole -1 =0.0237×10-3m3mole-1
	cp=5 2R=5 2×8.3 Joule/mole is

Hence, Temperature of Joule's-Thomnn's expansion is given as,a

dT	=1 CP(2a RT-b)dp
	=1 5×4.1502×0.0341×10-1-0.237×10-3 8.3×100.20×105
dT	=-1.5K

since dT is negative, i.e. as -173∘C is above the temperature of inression of Helicem (i.e. -240∘C ), it will show heating effect at this temperature.# An ideal refrigerator is run by a 40W electric motor with efficiency of 85%. If the temperature of the hot and the cold reservoirs are 0∘C and 30∘C. Calculate the time required by it to freeze 1kg of water at 0∘C. Latent heat of fusion of ice =334 kJ/kg-1Given,Temperature of cold reservoir T1=0∘C=273K Temperature of hot reservoir T2=30∘C=303KThe quantity of heat taken from 1kg of water at 0∘Ca

Q1=mL	=1×334×103𝚥
	=3.34×103J

If Q, is the quantity of heat rejected to the surrounding then,Q1

Q3=T1

T2a

Q2=T2 TQ1=303 273×3.34×103

-3.70×108J

Hence, energy supplied to the refrigerator,a

W	=Q2-Q1
	=3.70×103-3.34×105
	=0.36×108 J

Now, the refrigerator is being driven by a motor of 40 W with 85% efficieney.∴ rate of chergy input =40×85

100=34 Whence, the time required to supply 0.36×103 J energy is,a

0.36×103 34

=1088.2sec

=18.1 min

# Calculate the Boyle's temperature and temperature of inversion for nitrogen gas whose critical temperature is 126k.Given,Critical temperature of nitrogen gas =126kWe know,Boyle's temperature TD=a

RbTemperature of invertion (Ti)=2a

Rb=2×425.25=850.5k4. How much work must be done to extract 10.Oj of heat (a) from a reservoir at 2∘C and transfer it to one at 27∘C by means of a refrigerator using a carnot cycle; (b) from one at -73∘C to one at 27∘C; (c) from one at -173∘C to one at 27∘C; and (d) from one at - 223∘C to one at 27∘C ? (TU 2074 Q. No. 13)Sol n. Extract heat |(Q2)=10⌋⌋(a) Temperature of resorvoir = 7∘C=273+7280~KTemperature (T1)=27∘C=27+273=300~KWe know that,Work done (𝜎)=Q2 (T1-T2)

T2=10(300-280)

280=0.7JJ(b) Extract heat (Q2)=10jTemperature of resorvoir (T2)=-73∘C*-73+273=200~KTemperature (Tt)=27∘C=(27+273)K=300~K We know that,Work done (𝜔)=Q2(T1-T2)

T2 =10(300-200)

200=10×100

200=5J(c) Hear extract (Q2)=10JTemperature of resorvoir (T2)=-173∘C=-173+273=100~K Temperature (T1)=27∘C=(273+27)K=300~KWe know that,Work done (𝜔)=Q2T1-T2

T2=10300-100

100=20~J(d) Heat extract (Q2)=10 ~JTemperature (T2)=27∘C=(27+273)K=300~KTemperature (T1)=223∘C=(223+273)K=496~KWe know that,a=Q2T1-T2

T2=10496-300

300=6.5~J