Gas Power Cycles

Our study of gas power cycles will involve the study of those heat engines in which the working fluid remains in the gaseous state throughout the cycle. We often study the ideal cycle in which internal irreversibilities and complexities (the actual intake of air and fuel, the actual combustion process, and the exhaust of products of combustion among others) are removed.

We will be concerned with how the major parameters of the cycle affect the performance of heat engines. The performance is often measured in terms of the cycle efficiency.

Wnet th = Qin

Carnot Cycle

The Carnot cycle was introduced in Chapter 5 as the most efficient heat engine that can operate between two fixed temperatures TH and TL. The Carnot cycle is described by the following four processes. Carnot Cycle

Process Description 1-2 Isothermal heat addition 2-3 Isentropic expansion 3-4 Isothermal heat rejection 4-1 Isentropic compression

Note the processes on both the P-v and T-s diagrams. The areas under the process curves on the P-v diagram represent the work done for closed systems. The net cycle work done is the area enclosed by the cycle on the P-v diagram. The areas under the process curves on the T-s diagram represent the heat transfer for the processes. The net heat added to the cycle is the area that is enclosed by the cycle on the T-s diagram. For a cycle we know Wnet = Qnet; therefore, the areas enclosed on the P-v and T-s diagrams are equal.

TL = 1− th, Carnot TH

We often use the Carnot efficiency as a means to think about ways to improve the cycle efficiency of other cycles. One of the observations about the efficiency of both ideal and actual cycles comes from the Carnot efficiency: Thermal efficiency increases with an increase in the average temperature at which heat is supplied to the system or with a decrease in the average temperature at which heat is rejected from the system.

Air-Standard Assumptions

In our study of gas power cycles, we assume that the working fluid is air, and the air undergoes a thermodynamic cycle even though the working fluid in the actual power system does not undergo a cycle.

To simplify the analysis, we approximate the cycles with the following assumptions:

• The air continuously circulates in a closed loop and always behaves as an ideal gas.

• All the processes that make up the cycle are internally reversible.

• The combustion process is replaced by a heat-addition process from an external source.

• A heat rejection process that restores the working fluid to its initial state replaces the exhaust process.

• The cold-air-standard assumptions apply when the working fluid is air and has constant specific heat evaluated at room temperature (25oC or 77oF). Terminology for Reciprocating Devices

The following is some terminology we need to understand for reciprocating engines—typically piston-cylinder devices. Let’s look at the following figures for the definitions of top dead center (TDC), bottom dead center (BDC), stroke, bore, intake valve, exhaust valve, clearance volume, displacement volume, compression ratio, and mean effective pressure.

The compression ratio r of an engine is the ratio of the maximum volume to the minimum volume formed in the cylinder. V max V r = = BDC V min VTDC

The mean effective pressure (MEP) is a fictitious pressure that, if it operated on the piston during the entire power stroke, would produce the same amount of net work as that produced during the actual cycle.

W w MEP = net = net Vmax −Vmin vmax − vmin Otto Cycle: The Ideal Cycle for Spark-Ignition Engines

Consider the automotive spark-ignition power cycle.

Processes Intake stroke Compression stroke Power (expansion) stroke Exhaust stroke

Often the ignition and combustion process begins before the completion of the compression stroke. The number of crank angle degrees before the piston reaches TDC on the number one piston at which the spark occurs is called the engine timing. What are the compression ratio and timing of your engine in your car, truck, or motorcycle?

c c a a

b d d b The air-standard Otto cycle is the ideal cycle that approximates the spark- ignition combustion engine.

Process Description 1-2 Isentropic compression 2-3 Constant volume heat addition 3-4 Isentropic expansion 4-1 Constant volume heat rejection

The P-v and T-s diagrams are

Air Otto Cycle P - v Diagram

3

s = co ns t 2

P

4 T 3

1 T 1

v Air Otto Cycle T- s Diagram

P 3

3 P 1 T

4

2 v = c ons t 1

s Thermal Efficiency of the Otto cycle:

W Q Q − Q Q  = net = net = in out = 1 − out th Q Q in in Qin Qin

Now to find Qin and Qout.

Apply first law closed system to process 2-3, V = constant.

= U Qnet , 23 − Wnet , 23 23

Wnet , 23 = Wother , 23 + Wb, 23 = 0 + P dV = 0 23 z Thus, for constant specific heats,

Qnet , 23 = U23

Qnet , 23 = Qin = mCv (T3 − T2 )

Apply first law closed system to process 4-1, V = constant.

= U Qnet , 41 − Wnet , 41 41

Wnet , 41 = Wother , 41 + Wb, 41 = 0 + P dV = 0 41 z Thus, for constant specific heats,

Q = U net , 41 41

Qnet , 41 = −Qout = mCv (T1 − T4 )

Qout = −mCv (T1 − T4 ) = mCv (T4 − T1 ) The thermal efficiency becomes

Q  = 1− out th, Otto Q in mC (T − T ) = 1− v 4 1 mCv (T3 − T2 )

(T − T )  = 1− 4 1 th, Otto (T − T ) 3 2 T (T / T − 1) = 1− 1 4 1 T2 (T3 / T2 − 1)

Recall processes 1-2 and 3-4 are isentropic, so

k −1 k −1 T2 = GF V 1 JI and T3 = GF V 4 J I T1 H V2 K T4 H V3 K

Since V3 = V2 and V4 = V1, we see that T T 2 = 3 T1 T4 or T T 4 = 3 T1 T2

The Otto cycle efficiency becomes

T1 = 1− th, Otto T2

Is this the same as the Carnot cycle efficiency?

Since process 1-2 is isentropic,

k −1 T F V I 2 = G 1 J T 1 HV 2 k −1 k −1 1 T FV 2 I 1 K F I = J = G J T2 GH V1 K H r K

where the compression ratio is r = V1/V2 and

1

 = 1− th, Otto r k −1 We see that increasing the compression ratio increases the thermal efficiency. However, there is a limit on r depending upon the fuel. Fuels under high temperature resulting from high compression ratios will prematurely ignite, causing knock.

Example 8-1

An Otto cycle having a compression ratio of 9:1 uses air as the working o fluid. Initially P1 = 95 kPa, T1 = 17 C, and V1 = 3.8 liters. During the heat addition process, 7.5 kJ of heat are added. Determine all T's, P's, th, the back work ratio, and the mean effective pressure.

Process Diagrams: Review the P-v and T-s diagrams given above for the Otto cycle.

Assume constant specific heats with C v = 0.718 kJ/kg K, k = 1.4. (Use the 300 K data from Table A-2)

Process 1-2 is isentropic; therefore, recalling that r = V1/V2 = 9, V1 k −1 T = T k −1 F I = T brg 2 1 G J 1 H V2 K 1.4−1 = (17 + 273) Kb9g = 698.4K

k V = P r P = P F 1 I k b g 2 1 G J 1 H V2 K 1.4 = 95kPab9g = 2059 kPa

The first law closed system for process 2-3 was shown to reduce to (your homework solutions must be complete; that is, develop your equations from the application of the first law for each process as we did in obtaining the Otto cycle efficiency equation)

Qin = mCv (T3 − T2 )

Let qin = Qin / m and m = V1 /v1

RT v = 1 1 P1 kJ 0.287

(290 K) kg  K m3kPa = 95 kPa kJ m3 = 0.875 kg

Qin v 1 qin = = Qin m V1 m3 0.875 kg = 7.5kJ −3 3 3.8 10 m kJ = 1727

kg Then,

qin T = T + 3 2 Cv kJ 1727 kg = 698.4K + kJ 0.718 kg  K = 3103.7 K

Using the combined gas law

T3 P = P = 9.15 MPa 3 2 T 2

Process 3-4 is isentropic; therefore,

k −1 3 k −1 T = T F V I = T3 F 1I 4 3 G J GH JK H V4 K r F 1I 1.4−1 = (3103.7) KG J H 9 K = 1288.8 K k k V 1 P F 3I = F I P = 3 G J P3 G J 4 H K H V4 K r F 1I 1.4 = 9.15 MPaG J H 9 K = 422 kPa

Process 4-1 is constant volume. So the first law for the closed system gives, on a mass basis,

Qout = mCv (T4 − T1 )

Qout − q = = C (T T ) out v 4 1 m kJ = 0.718 (1288.8 − 290) K kg  K kJ = 717.1 kg

For the cycle, u = 0, and the first law gives

w = q = q − net net in qout kJ = (1727 − 717.4) kg kJ = 1009.6 kg

The thermal efficiency is

kJ 1009.6

wnet kg th, Otto = = kJ qin 1727 kg = 0.585 or 58.5%

The mean effective pressure is

w MEP = Wnet net = Vmax −Vmin vmax − vmin w w w = net = net = net

v1 − v2 v1 (1− v2 / v1 ) v1 (1−1/ r) kJ 1009.6 kg m3kPa = = 1298 kPa 3 1 kJ 0.875 m (1− ) kg 9

The back work ratio is (can you show that this is true?)

w u BWR = comp = 12 wexp −u34

C (T − T ) (T − T ) = v 2 1 = 2 1 Cv (T3 − T4 ) (T3 − T4 ) = 0.225 or 22.5%

Air-Standard Diesel Cycle

The air-standard Diesel cycle is the ideal cycle that approximates the Diesel combustion engine

Process Description 1-2 Isentropic compression 2-3 Constant pressure heat addition 3-4 Isentropic expansion 4-1 Constant volume heat rejection

The P-v and T-s diagrams are

Thermal efficiency of the Diesel cycle

W Q  = net = 1− out th, Diesel Q in Qin

Now to find Qin and Qout.

Apply the first law closed system to process 2-3, P = constant.

Ein − Eout = E

Q − W = U net , 23 net , 23 23 3 Wnet , 23 = Wother , 23 + Wb, 23 = 0 + PdV z2

= P2 (V3 −V2 )

Thus, for constant specific heats

Qnet , 23 = U23 + P2 (V3 −V2 )

Qnet , 23 = Qin = mCv (T3 − T2 ) + mR(T3 − T2 )

Qin = mCp (T3 − T2 )

Apply the first law closed system to process 4-1, V = constant (just like the Otto cycle)

Ein − Eout = E = U Qnet , 41 − Wnet , 41 41

Wnet , 41 = Wother , 41 + Wb, 41 = 0 + PdV = 0 41 z Thus, for constant specific heats

Q = U net , 41 41

Qnet , 41 = −Qout = mCv (T1 − T4 )

Qout = −mCv (T1 − T4 ) = mCv (T4 − T1 )

The thermal efficiency becomes

Q  = 1− out th, Diesel Q in mC (T − T ) = 1− v 4 1 mCp (T3 − T2 )

C (T − T )  = 1− v 4 1 th, Diesel C (T − T ) p 3 2 1 = 1− T1 (T4 / T1 − 1)

k T2 (T3 / T2 − 1)

What is T3/T2 ? P3V3 P2V2 where P = P = 3 2 T3 T2

T3 V3 = = rc T2 V2 where rc is called the cutoff ratio, defined as V3 /V2, and is a measure of the duration of the heat addition at constant pressure. Since the fuel is injected directly into the cylinder, the cutoff ratio can be related to the number of degrees that the crank rotated during the fuel injection into the cylinder.

What is T4/T1 ?

P4V4 P1V1 where V = V = 4 1 T4 T1 T P 4 = 4 T1 P1

Recall processes 1-2 and 3-4 are isentropic, so

k k k k PV = PV and PV = PV 1 1 2 2 4 4 3 3

Since V4 = V1 and P3 = P2, we divide the second equation by the first equation and obtain k

P4 F V3 I k = G J = r

c P1 H V2 K

Therefore,

1 = 1 − T (T / T − 1)  1 4 1 th, Diesel k T (T / T − 1) 2 3 2 1 T r k − 1 1 c = 1 −

k T2 (rc − 1) 1 r k − 1 c

= 1 − k −1 r k (rc − 1)

What happens as rc goes to 1? Sketch the P-v diagram for the Diesel cycle and show rc approaching1 in the limit.

When rc > 1 for a fixed r, th, Diesel  th, Otto . But, since rDiesel  rOtto ,

th, Diesel  th, Otto .

Brayton Cycle

The Brayton cycle is the air-standard ideal cycle approximation for the gas- turbine engine. This cycle differs from the Otto and Diesel cycles in that the processes making the cycle occur in open systems or control volumes. Therefore, an open system, steady-flow analysis is used to determine the heat transfer and work for the cycle.

We assume the working fluid is air and the specific heats are constant and will consider the cold-air-standard cycle.

Brayton cycle

Qin

2 3

Comp Turb Wnet Wc

1 4

The closed cycle gas-turbine engine Closed Brayton cycle

Qin

2 3

Comp Turb Wnet Wc

1 4

Qout

Process Description 1-2 Isentropic compression (in a compressor) 2-3 Constant pressure heat addition 3-4 Isentropic expansion (in a turbine) 4-1 Constant pressure heat rejection

The T-s and P-v diagrams are

Thermal efficiency of the Brayton cycle

W Q  = net = 1− out th, Brayton Q in Qin

Now to find Qin and Qout.

Apply the conservation of energy to process 2-3 for P = constant (no work), steady-flow, and neglect changes in kinetic and potential energies.

˙ ˙ Ein = Eout ˙ m˙ 2h2 + Qin = m˙ 3h3

The conservation of mass gives

m˙ in = m˙ out

m˙ 2 = m˙ 3 = m˙

For constant specific heats, the heat added per unit mass flow is

˙ Qin = m˙ (h3 − h2 ) ˙ Qin = m˙ Cp (T3 − T2 ) Q˙ q = in = C (T − T )

in p 3 2 m˙

The conservation of energy for process 4-1 yields for constant specific heats (let’s take a minute for you to get the following result) ˙ Qout = m˙ (h4 − h1 ) ˙ Qout = m˙ Cp (T4 − T1 ) Q˙ q = out = C (T −

T ) out p 4 1 m˙

The thermal efficiency becomes

˙ Qout q  = 1 − = 1− out th, Brayton Q˙ in qin C (T − T ) = 1 − p 4 1 Cp (T3 − T2 )

(T − T )  = 1 − 4 1 th, Brayton (T − T ) 3 2

T (T / T − 1) = 1 − 1 4 1 T2 (T3 / T2 − 1)

Recall processes 1-2 and 3-4 are isentropic, so

( k −1)/ k ( k −1)/ k T F P I T F P I 2 and 3 = G 2 = G 3 T1 H P1J K T4 H P4J K

Since P3 = P2 and P4 = P1, we see that

T2 T3 T T = or 4 = 3

T1 T4 T1 T2

The Brayton cycle efficiency becomes

T1 = 1− th, Brayton T2

Is this the same as the Carnot cycle efficiency?

Since process 1-2 is isentropic, P ( k −1)/ k T 2 F 2 I = r ( k −1)/ k = J G p T1 H P1 K

T1 1 = ( k −1)/ k T2 rp

where the pressure ratio is rp = P2/P1 and

1

th, Brayton = 1− ( k −1)/ k rp

Extra Assignment

Evaluate the Brayton cycle efficiency by determining the net work directly from the turbine work and the compressor work. Compare your result with the above expression. Note that this approach does not require the closed cycle assumption.

Example 8-2

The ideal air-standard Brayton cycle operates with air entering the o compressor at 95 kPa, 22 C. The pressure ratio r p is 6:1 and the air leaves the heat addition process at 1100 K. Determine the compressor work and the turbine work per unit mass flow, the cycle efficiency, the back work ratio, and compare the compressor exit temperature to the turbine exit temperature. Assume constant properties.

Apply the conservation of energy for steady-flow and neglect changes in kinetic and potential energies to process 1-2 for the compressor. Note that the compressor is isentropic.

˙ ˙ Ein = Eout ˙ m˙ 1h1 + Wcomp = m˙ 2h2

The conservation of mass gives

m˙ in = m˙ out

m˙ 1 = m˙ 2 = m˙

For constant specific heats, the compressor work per unit mass flow is

W˙ = m˙ (h − h ) comp 2 1 W˙ = m˙ C (T − T ) comp p 2 1

W˙ comp w = = C (T − T )

comp p 2 1 m˙

Since the compressor is isentropic P ( k −1)/ k T 2 F 2 I = r ( k −1)/ k = J G p T1 H P1 K ( k −1)/ k T2 = Tr1 p (1.4−1)/1.4 = (22 + 273) K(6) = 492.5 K

wcomp = Cp (T2 − T1 ) kJ = 1.005 (492.5 − 295) K kg  K kJ = 198.15 kg

The conservation of energy for the turbine, process 3-4, yields for constant specific heats (let’s take a minute for you to get the following result)

W˙ = m˙ (h − h ) turb 3 4 W˙ = m˙ C (T − T ) turb p 3 4 W˙ turb w = = Cp (T3 − T4 ) turb m˙

Since process 3-4 is isentropic

T ( k −1)/ k P 4 F I = G 4 T3 H P3J K

Since P3 = P2 and P4 = P1, we see that

4 ( k −1)/ k T = GF 1 JI

T3 H rp K ( k −1)/ k F 1 I T = T 4 3 G J H rp K F 1I (1.4−1) /1.4 = 1100KG J H 6K = 659.1 K kJ w = C (T − T ) = 1.005 (1100 − 659.1) K turb p 3 4 kg  K kJ = 442.5 kg

We have already shown the heat supplied to the cycle per unit mass flow in process 2-3 is

m˙ 2 = m˙ 3 = m˙ ˙ m˙ 2h2 + Qin = m˙ 3h3 Q˙ q = in = h − h

in m˙ 3 2 kJ = C (T − T ) = 1.005 (1100 − 492.5) K p 3 2 kg  K kJ = 609.6 kg

The net work done by the cycle is

w = w − net turb wcomp kJ = (442.5 − 198.15) kg kJ = 244.3 kg

The cycle efficiency becomes

w  = net th, Brayton q in kJ 244.3 kg = kJ = 0.40 or 40% 609.6 kg

The back work ratio is defined as

w w BWR = in = comp wout wturb kJ 198.15 kg = kJ = 0.448 442.5 kg

Note that T4 = 659.1 K > T2 = 492.5 K, or the turbine outlet temperature is greater than the compressor exit temperature. Can this result be used to improve the cycle efficiency?

What happens to th, win /wout, and wnet as the pressure ratio r p is increased?

Let's take a closer look at the effect of the pressure ratio on the net work done.

wnet = wturb − wcomp

= Cp (T3 − T4 ) − Cp (T2 − T1 )

= CpT3 (1− T4 / T3 ) − CpT1 (T2 / T1 − 1) 1 = C T (1− ) − C T (r ( k −1)/ k − 1) p 3 ( k −1)/ k p 1 p rp

Note that the net work is zero when

3 ( k −1)/ k rp = 1 and rp = GF T JI

H T1 K

For fixed T3 and T1, the pressure ratio that makes the work a maximum is obtained from: dw net = 0 drp

(k-1)/k This is easier to do if we let X = r p

1 w = C T (1− ) − C T ( X − 1) net p 3 p 1 X

−2 dwnet

= C T [0 − (−1) X ] − C T [1− 0] = 0 p 3 p 1 dX

Solving for X

2 T3 2( k −1)/ k X = = drp i T1

Then, the r p that makes the work a maximum for the constant property case and fixed T3 and T1 is F T3 I k /[2( k −1)] r =

p, max work G J H T1 K

For the ideal Brayton cycle, show that the following results are true. • When rp = rp, max work, T4 = T2 • When rp < rp, max work, T4 > T2 • When rp > rp, max work, T4 < T2

The following is a plot of net work per unit mass and the efficiency for the above example as a function of the pressure ratio.

280 0.60

260 0.55

0.50 240 0.45 220

0.40 200 T = 22C 1 0.35 kJ/kg P = 95 kPa 180 1

0.30 th,Brayto

net T = 1100 K

n 3 

w 160 t= c = 100% 0.25

140 r 0.20 p,m ax 120 0.15 0 2 4 6 8 10 12 14 16 18 20 22

Pratio

Regenerative Brayton Cycle

For the Brayton cycle, the turbine exhaust temperature is greater than the compressor exit temperature. Therefore, a heat exchanger can be placed between the hot gases leaving the turbine and the cooler gases leaving the compressor. This heat exchanger is called a regenerator or recuperator. The sketch of the regenerative Brayton cycle is shown below.

Regenerative Brayton cycle

Regenerator

6 4 2 3 5 Qin Comp Turb Wnet Wc

1

We define the regenerator effectiveness regen as the ratio of the heat transferred to the compressor gases in the regenerator to the maximum possible heat transfer to the compressor gases.

q = h − h regen, act 5 2

qregen, max = h5' − h2 = h4 − h2

q h − h = regen, act 5 2  regen q = h − h regen, max 4 2

For ideal gases using the cold-air-standard assumption with constant specific heats, the regenerator effectiveness becomes

T − T  5 2 regen T − T 4 2

Using the closed cycle analysis and treating the heat addition and heat rejection as steady-flow processes, the regenerative cycle thermal efficiency is

qout th, regen = 1 − q in h − h = 1 − 6 1 h3 − h5

Notice that the heat transfer occurring within the regenerator is not included in the efficiency calculation because this energy is not a heat transfer across the cycle boundary.

Assuming an ideal regenerator regen = 1 and constant specific heats, the thermal efficiency becomes (take the time to show this on your own)

T F P2 I ( k −1)/ k  = 1 − 1 th, regen T G P J 3 H 1 K T1 = 1 − ( k −1)/ k (rp ) T3

When does the efficiency of the air-standard Brayton cycle equal the efficiency of the air-standard regenerative Brayton cycle? If we set th, Brayton = th, regen then

th, Brayton = th, regen

1 T1 1 − = 1− ( k −1)/ k ( k −1)/k (rp ) (rp ) T3 F T3 I k /[2( k −1)] r =

p G T J H 1 K

Recall that this is the pressure ratio that maximizes the net work for the simple Brayton cycle and makes T4 = T2. What happens if the regenerative Brayton cycle operates at a pressure ratio larger than this value?

For fixed T3 and T1, pressure ratios greater than this value cause T4 to be less than T2, and the regenerator is not effective.

What happens to the net work when a regenerator is added?

What happens to the heat supplied when a regenerator is added?

The following shows a plot of the regenerative Brayton cycle efficiency as a function of the pressure ratio and minimum to maximum temperature ratio, T1/T3.

Example 8-3: Regenerative Brayton Cycle

Air enters the compressor of a regenerative gas-turbine engine at 100 kPa and 300 K and is compressed to 800 kPa. The regenerator has an effectiveness of 65 percent, and the air enters the turbine at 1200 K. For a compressor efficiency of 75 percent and a turbine efficiency of 86 percent, determine (a) The heat transfer in the regenerator. (b) The back work ratio. (c) The cycle thermal efficiency.

Compare the results for the above cycle with the ones listed below that have the same common data as required. (a) The actual cycle with no regeneration,  = 0. (b) The actual cycle with ideal regeneration,  = 1.0. (b) The ideal cycle with regeneration,  = 0.65. (d) The ideal cycle with no regeneration,  = 0. (e) The ideal cycle with ideal regeneration,  = 1.0.

We assume air is an ideal gas with constant specific heats, that is, we use the cold-air-standard assumption.

The cycle schematic is the same as above and the T-s diagram showing the effects of compressor and turbine efficiencies is below. T-s Diagram for Gas Turbine with Regeneration

800 kPa

3

100 kPa

T 5 4a 2a 2 4s s 6

1

s

Summary of Results

Cycle type Actual Actual Actual Ideal Ideal Ideal

regen 0.00 0.65 1.00 0.00 0.65 1.00

comp 0.75 0.75 0.75 1.00 1.00 1.00

turb 0.86 0.86 0.86 1.00 1.00 1.00

qin kJ/kg 578.3 504.4 464.6 659.9 582.2 540.2

wcomp kJ/kg 326.2 326.2 326.2 244.6 244.6 244.6

wturb kJ/kg 464.6 464.6 464.6 540.2 540.2 540.2

wcomp/wturb 0.70 0.70 0.70 0.453 0.453 0.453

th 24.0% 27.5% 29.8% 44.8% 50.8% 54.7%

Compressor analysis

The isentropic temperature at compressor exit is

( k −1)/ k T P I 2 s = F G P2 T1 H 1 KJ F P2 I T = T ( k −1)/ k

2 s 1 G P J H 1 K 800kPa = 300K( )(1.4−1) /1.4 = 543.4K 100kPa

To find the actual temperature at compressor exit, T2a, we apply the compressor efficiency

w h − h T − T  = isen,comp = 2 s 1  2 s 1 comp h T wact ,comp 2a − h1 2a − T1 1 T2a = T1 + (T2 s − T1 ) comp 1

= 300K + (543.4 − 300) K 0.75 = 624.6K

Since the compressor is adiabatic and has steady-flow

wcomp = h2a − h1 = Cp (T2a − T1 ) kJ kJ = 1.005 (624.6 − 300)K = 326.2

kg  K kg

Turbine analysis

The conservation of energy for the turbine, process 3-4, yields for constant specific heats (let’s take a minute for you to get the following result)

W˙ = m˙ (h − h ) turb 3 4a W˙ = m˙ C (T − T ) turb p 3 4a W˙ turb w = = Cp (T3 − T4a ) turb m˙

Since P3 = P2 and P4 = P1, we can find the isentropic temperature at the turbine exit.

( k −1)/ k T P I 4 s = F G P4 T3 H 3 KJ F P4 I T = T ( k −1)/ k

4 s 3 G P J H 3 K 100kPa = 1200K( )(1.4−1)/1.4 = 662.5K 800kPa

To find the actual temperature at turbine exit, T4a, we apply the turbine efficiency.

wact ,turb h3 − h4a T3 − T4a = =   turb h − h T − T wisen,turb 3 4 s 3 4 s

T4a = T3 − turb (T3 − T4 s ) = 1200K − 0.86(1200 − 662.5) K

= 737.7K  T2a

The turbine work becomes

wturb = h3 − h4a = Cp (T3 − T4a ) kJ = 1.005 (1200 − 737.7) K kg  K kJ = 464.6 kg

The back work ratio is defined as

w w BWR = in = comp wout wturb kJ 326.2 kg = kJ = 0.70 464.6 kg

Regenerator analysis

To find T5, we apply the regenerator effectiveness.

T T   5 − 2a regen − T T4a 2a

T = T +  regen (T4a − 5 2a T2a ) = 624.6K + 0.65(737.7 − 624.6) K = 698.1K

To find the heat transferred from the turbine exhaust gas to the compressor exit gas, apply the steady-flow conservation of energy to the compressor gas side of the regenerator.

˙ m˙ 2a h2a + Qregen = m˙ 5h5

m˙ 2a = m˙ 5 = m˙ Q˙ q = regen = h − h

regen 5 2a m˙

= Cp (T5 − T2a ) kJ = 1.005 (698.1− 624.6) K kg  K kJ = 73.9 kg

Using qregen, we can determine the turbine exhaust gas temperature at the regenerator exit.

˙ m˙ 4a h4a = Qregen + m˙ 6h6

m˙ 4a = m˙ 6 = m˙ ˙ Qregen qregen = = h4a − h = Cp (T4a − T6 ) m˙ 6 kJ 73.9 q kg T = T − regen = 737.7K − 6 4a C kJ p 1.005 kg  K = 664.2K

Heat supplied to cycle

Apply the steady-flow conservation of energy to the heat exchanger for process 5-3. We obtain a result similar to that for the simple Brayton cycle.

qin = h3 − h5 = Cp (T3 − T5 ) kJ = 1.005 (1200 − 698.1) K kg  K kJ = 504.4 kg

Cycle thermal efficiency

The net work done by the cycle is

wnet = wturb − wcomp kJ kJ = (464.6 − 326.2) = 138.4 kg kg

The cycle efficiency becomes

w  = net th, Brayton qin kJ 138.4 kg = kJ = 0.274 or 27.4% 504.4 kg

You are encouraged to complete the calculations for the other values found in the summary table.

Other Ways to Improve Brayton Cycle Performance

Intercooling and reheating are two important ways to improve the performance of the Brayton cycle with regeneration.

Intercooling

When using multistage compression, cooling the working fluid between the stages will reduce the amount of compressor work required. The compressor work is reduced because cooling the working fluid reduces the average

specific volume of the fluid and thus reduces the amount of work on the fluid to achieve the given pressure rise.

To determine the intermediate pressure at which intercooling should take place to minimize the compressor work, we follow the approach shown in Chapter 6.

For the adiabatic, steady-flow compression process, the work input to the compressor per unit mass is

4 3 2 0 4 wcomp = v dP = v dP + v dP + v dP  1  3 1 2

For the isentropic compression process

k k w = − Pv ) + (P v (P v − Pv ) comp k-1 2 2 1 1 k-1 4 4 3 3 k = R(T kR

− T ) + (T − T ) k-1 2 1 k-1 4 3 k = R T (T / T −1)  / T −1) + T (T  k-1 1 2 1 3 4 3   k  P (k −1)/ k  (k −1)/ k    2    P4   = R T  1  P −1 + T3   −1 k-1   1    P3       

Notice that the fraction kR/(k-1) = C p.

( k −1)/ k ( k −1)/ k L T 2 − 1 + T 4 I O wcomp = C 1 GF F P I JI 3 GF F P I − 1 p M G J G J J P P P NM HH 1 K K HH 3 K KPQ

Can you obtain this relation another way? Hint: apply the first law to processes 1-4.

For two-stage compression, let’s assume that intercooling takes place at constant pressure and the gases can be cooled to the inlet temperature for the compressor, such that P3 = P2 and T3 = T1.

The total work supplied to the compressor becomes

( k −1)/ k ( k −1)/ k L 2 4 I O w = CpT1 F P − 1I + F P comp MG GF IJ J G GF IJ − 1J P P H P M HH 1K K H 2 K KP N 2 ( k −1)/ k + 4 ( k −1)/ k Q = CpT1 L − 2 O GMF P JI GF P JI P P P MNH 1K H 2 K PQ

To find the unknown pressure P2 that gives the minimum work input for fixed compressor inlet conditions T1, P1, and exit pressure P4, we set

dwcomp ( P2 ) = 0 dP2 This yields

P2 = P1 P4 or, the pressure ratios across the two compressors are equal.

P P P 2 = 4 = 4 P1 P2 P3

Intercooling is almost always used with regeneration. During intercooling the compressor exit temperature is reduced; therefore, more heat must be

supplied in the heat addition process. Regeneration can make up part of the required heat transfer.

To supply only compressed air, using intercooling requires less work input. The next time you go to a home supply store where air compressors are sold, check the larger air compressors to see if intercooling is used. For the larger air compressors, the compressors are made of two piston-cylinder chambers. The intercooling heat exchanger may be only a pipe with a few attached fins that connects the large piston-cylinder chamber with the smaller piston- cylinder chamber.

Reheating When using multistage expansion through two or more turbines, reheating between stages will increase the net work done (it also increases the required heat input).The regenerative Brayton cycle with reheating is shown above.

The optimum intermediate pressure for reheating is the one that maximizes the turbine work. Following the development given above for intercooling and assuming reheating to the high-pressure turbine inlet temperature in a constant pressure steady-flow process, we can show the optimum reheat pressure to be

P7 = P6 P9

or the pressure ratios across the two turbines are equal.

The refrigeration cycle

The Reversed Carnot Cycle Unlike the Carnot heat engine, the Carnot refrigeration cycle undergoes a process with opposite direction. We see from the model, heat QL is absorbed from the low−temperature reservoir (TL=constant) and heat QH is rejected to a high−temperature reservoir (TH=constant). In this case a work input in the amount st of Wrev is required to achieve this process. And we know from the 1 law of thermodynamics, the required work can be determined in Wrev=− QH − QL. Here QH<0 and QL>0.

The reversed Carnot cycle also consists of two isentropic and two isothermal processes. The process undergoes in direction 3−2−1−4−3 p 1 QH

2 4 3 QL

0 v

Process 3−2: Reversible Adiabatic Compression This process is isentropic. The engine is perfect insulated so that no heat is lost and absorbed. Gas is compressed slowly until the temperature rises from TL to TH.

Process 2−1: Reversible Isothermal Compression (TH=constant) During this process, heat is rejected. Gas is compressed reversibly at the constant temperature TH.

Process 1−4: Reversible Adiabatic Expansion This process is isentropic. The engine is perfect insulated so that no heat is lost and absorbed. Gas expands slowly until the temperature drops from TH to TL.

Process 4−3: Reversible Isothermal Expansion (TL=constant) During this process, heat is absorbed. Gas is compressed reversibly at the constant temperature TH.

The coefficient of performance (COP) of any refrigerator or heat pump, reversible or irreversible, can be calculated with the general expression: • For a Carnot refrigerator: QL QL COP = = Wrev —QK — R QL Where Q <0 and Q >0 • For a Carnot heat pump: H L —QK —QK COP = = KP W rev —QK — QL Where QH<0 and QL>0 T QH 1 TH 2

TL 3 4 QL

0 s From the T−s−diagram, we obtain: Qab=QH = TH· 6S2(1 Qzu=QL = TL· 6S4(3 6S2(1= − 6S4(3

Therefore • For a Carnot refrigerator: QL TL · 6S4‹3 TL 1 = = = = COPR TK —QK — —TK · 6S2‹1 — TL · TK — — 1 QL 6S4‹3 TL TL • For a Carnot heat pump: —QK —TK · 6S2‹1 TK 1 = = = = COPK TL —QK — —TK · 6S2‹1 — TL · TK — 1 — P QL 6S4‹3 TL TK

According to coefficient of performance (COP), we can also draw the following conclusions: 1. The coefficient of performance (COP)of a reversed Carnot cycle only depends on the highest and lowest temperature. 2. Normally TH>TL, so that means COPR>1 and COPHP>1 3. Both coefficients of performance (COP) have a relationship: COPHP = COPR +1 4. If TL decreases, both COPR and COPHP decrease.

As mentioned above, the reversed Carnot cycle is a reversible process. Hence, if a real refrigerator has the coefficient of performance of COP, then: • COPR < COPR,rev: irreversible refrigerator • COPR = COPR,rev: reversible refrigerator • COPR > COPR,rev: unrealistic refrigerator

Ideal reverse Brayton cycle

Fig. 9.2(a). Schematic of a closed reverse Brayton cycle

This is an important cycle frequently employed in gas cycle refrigeration systems. This may be thought of as a modification of reversed Carnot cycle, as the two isothermal processes of Carnot cycle are replaced by two isobaric heat transfer processes. This cycle is also called as Joule or Bell-Coleman cycle. Figure 9.2(a) and (b) shows the schematic of a closed, reverse Brayton cycle and also the cycle on T-s

diagram. As shown in the figure, the ideal cycle consists of the following four processes:

Process 1-2: Reversible, adiabatic compression in a compressor Process 2-3: Reversible, isobaric heat rejection in a heat exchanger Process 3-4: Reversible, adiabatic expansion in a turbine Process 4-1: Reversible, isobaric heat absorption in a heat exchanger

Fig. 9.2(b). Reverse Brayton cycle in T-s plane

Process 1-2: Gas at low pressure is compressed isentropically from state 1 to state 2. Applying steady flow energy equation and neglecting changes in kinetic and potential energy, we can write:

. . W1−2 = m(h 2 − h1 ) = m cp (T2 − T1 ) s2 = s1 (9.9) −1   P2  −1

= T r  and T = T   1 p 2 1 P 1  where rp = (P2/P1) = pressure ratio

Process 2-3: Hot and high pressure gas flows through a heat exchanger and rejects heat sensibly and isobarically to a heat sink. The enthalpy and temperature of the gas drop during the process due to heat exchange, no work transfer takes place and the entropy of the gas decreases. Again applying steady flow energy equation and second T ds equation: . . Q2−3 = m(h2 − h3 ) = m cp (T2 − T3 ) T2 ln s2 − s3 = cp (9.10) T3 P2 = P3

Process 3-4: High pressure gas from the heat exchanger flows through a turbine, undergoes isentropic expansion and delivers net work output. The temperature of the gas drops during the process from T3 to T4. From steady flow energy equation:

. . W3−4 = m(h 3 − h 4 ) = m cp (T3 − T4 ) s3 = s4 (9.11) −1   P3  −1

= T r  and T = T   4 p 3 4 P 4  where rp = (P3/P4) = pressure ratio

Process 4-1: Cold and low pressure gas from turbine flows through the low temperature heat exchanger and extracts heat sensibly and isobarically from a heat source, providing a useful refrigeration effect. The enthalpy and temperature of the gas rise during the process due to heat exchange, no work transfer takes place and the entropy of the gas increases. Again applying steady flow energy equation and second T ds equation: . . Q4−1 = m(h1 − h 4 ) = m cp (T1 − T4 ) T4 ln s4 − s1 =cp (9.12) T1 P4 = P1

From the above equations, it can be easily shown that:

 T2   T3  T  =T  (9.13)  1   4 

Applying 1st law of thermodynamics to the entire cycle:

 q =(q 4−1 − q 2−3 ) =  w =(w 3−4 − w1−2 ) = − w net (9.14)

The COP of the reverse Brayton cycle is given by:

q  (T − T )  COP = 4−1 =  T(T ) − Tl ) −4 (T −  (9.15) w net  2 1 3 4  using the relation between temperatures and pressures, the COP can also be written as:   −1 −1  (Tl − T4 )   T4   (Tl − T4 )  (9.16)

COP = (T − T ) − (T − T )  =  T − T  = −1  =(rp  − 1) 4  2 1 3 4   3   (T1 − T4 )(rp  − 1)  



From the above expression for COP, the following observations can be made:

a) For fixed heat rejection temperature (T3) and fixed refrigeration temperature (T1), the COP of reverse Brayton cycle is always lower than the COP of reverse Carnot cycle (Fig. 9.3), that is  T4   T1  COPBrayton =  T −   COP =   3 T4  Carnot  T 3 − T 1 

Fig. 9.3. Comparison of reverse Carnot and reverse Brayton cycle in T-s plane

b) COP of Brayton cycle approaches COP of Carnot cycle as T1 approaches T4 (thin cycle), however, the specific refrigeration effect [cp(T1-T4)] also reduces simultaneously. c) COP of reverse Brayton cycle decreases as the pressure ratio rp increases

Actual reverse Brayton cycle:

The actual reverse Brayton cycle differs from the ideal cycle due to:

i. Non-isentropic compression and expansion processes ii. Pressure drops in cold and hot heat exchangers

Fig. 9.4. Comparison of ideal and actual Brayton cycles T-s plane

Figure 9.4 shows the ideal and actual cycles on T-s diagram. Due to these irreversibilities, the compressor work input increases and turbine work output reduces. The actual work transfer rates of compressor and turbine are then given by:

W = 1−2,isen W1−2,act  (9.17) c,isen

W3−4,act = t,isen W3−4,isen (9.18)

where c,isen and t,isen are the isentropic efficiencies of compressor and turbine, respectively. In the absence of pressure drops, these are defined as:

(h 2 − h1 ) (T2 − T1 ) (9.20)  c,isen = = (h 2' − h1 ) (T2' − T1 )

(h3 − h4' ) (T3 − T4' )  = = (9.21) t ,isen (h − h ) (T − T ) 3 4 3 4

The actual net work input, wnet,act is given by:

(9.22) Wnet,act = W1−2,act − W3−4,act thus the net work input increases due to increase in compressor work input and reduction in turbine work output. The refrigeration effect also reduces due to the irreversibilities. As a result, the COP of actual reverse Brayton cycles will be considerably lower than the ideal cycles. Design of efficient compressors and turbines plays a major role in improving the COP of the system.

In practice, reverse Brayton cycles can be open or closed. In open systems, cold air at the exit of the turbine flows into a room or cabin (cold space), and air to the compressor is taken from the cold space. In such a case, the low side pressure will be atmospheric. In closed systems, the same gas (air) flows through the cycle in a closed manner. In such cases it is possible to have low side pressures greater than atmospheric. These systems are known as dense air systems. Dense air systems are advantageous as it is possible to reduce the volume of air handled by the compressor and turbine at high pressures. Efficiency will also be high due to smaller pressure ratios. It is also possible to use gases other than air (e.g. helium) in closed systems.

Simple Vapour Compression Refrigeration System

A vapour compression refrigeration system is an improved type of air refrigeration system in which a suitable working substance, termed as refrigerant is used. It condensed and evaporates at temperatures and pressures close to the atmospheric conditions. The refrigerants usually used for this purpose are ammonia, carbon dioxide and sulphur dioxide.

Advantage and disadvantages of vapour compression and air refrigeration system:

• Advantage: 1. It has smaller size for given capacity of refrigeration. 2. It has less running cost. 3. It can be employed over a large range of temperatures 4. The coefficient of performance is quite high.

• Disadvantages: 1. The initial cost is high 2. The prevention of leakage of refrigerant is the major problem in vapour compression system.

Comparison between gas cycles and vapor cycles

Thermodynamic cycles can be categorized into gas cycles and vapour cycles. In a typical gas cycle, the working fluid (a gas) does not undergo phase change, consequently the operating cycle will be away from the vapour dome. In gas cycles, heat rejection and refrigeration take place as the gas undergoes sensible cooling and heating. In a vapour cycle the working fluid undergoes phase change and refrigeration effect is due to the vaporization of refrigerant liquid. If the refrigerant is a pure substance then its temperature remains constant during the phase change processes.. Hence, the required mass flow rates for a given refrigeration capacity will be much smaller compared to a gas cycle. Vapour cycles can be subdivided into vapour compression systems, vapour absorption systems, vapour jet systems etc. Among these the vapour compression refrigeration systems are predominant.

Mechanism of simple vapour compression refrigeration system:

Compression refrigeration cycles take advantage of the fact that highly compressed fluids at a certain temperature tend to get colder when they are allowed to expand. If the pressure change is high enough, then the compressed gas will be hotter than our source of cooling (outside air, for instance) and the expanded gas will be cooler than our desired cold temperature. In this case, fluid is used to cool a low temperature environment and reject the heat to a high temperature environment. Vapour compression refrigeration cycles have two advantages. First, a large amount of thermal energy is required to change a liquid to a vapor, and therefore a lot of heat can be removed from the air-conditioned space. Second, the isothermal nature of the vaporization allows extraction of heat without raising the

temperature of the working fluid to the temperature of whatever is being cooled. This means that the heat transfer rate remains high, because the closer the working fluid temperature approaches that of the surroundings, the lower the rate of heat transfer. The refrigeration cycle is shown in Figure below and can be broken down into the following stages:

1 – 2 Low-pressure liquid refrigerant In the evaporator absorbs heat from its surroundings, usually air, water or some other process liquid. During this process it changes its state from a liquid to a gas, and at the evaporator exit is slightly superheated.

2 – 3 The superheated vapour Enters the compressor where its pressure is raised. The temperature will also increase, because a proportion of the energy put into the compression process is transferred to the refrigerant.

3 – 4 The high pressure superheated gas Passes from the compressor into the condenser. The initial part of the cooling process (3-3a) superheats the gas before it is then turned back into liquid (3a-3b). The cooling for this process is usually achieved by using air or water. A further reduction in temperature happens in the pipe work and liquid receiver (3b - 4), so that the refrigerant liquid is sub-cooled as it enters the expansion device.

4 - 1 The high-pressure sub-cooled liquid Passes through the expansion device, which both reduces its pressure and controls the flow into the evaporator.

Figure 1. Schematic representation of the vapour compression refrigeration cycle

Figure 2. Schematic representation of the refrigeration cycle including pressure changes

Figure3 .Layout of simple vapor compression refrigeration machine

Simple Vapor Compression Refrigeration Cycle

It is shown on T-S below at point 1, let T1, P1, and s1 be the properties of vapour refrigerant . the four processes of the cycle are as follows:

Figure 5: T-s diagram of simple vapor compression refrigeration cycle

Figure 6:P-h diagram for simple vapor compression refrigeration cycle

1. Compression process:

The vapour refrigerant at low pressure p1 and temperature T1 is compressed isentropically to dry saturated vapour as shown by the vertical line 1- 2 on T-s diagram and by the curve 1-2 on p-h diagram. The pressure and temperature rises from 1 to 2. The work done during isentropic compression is given by:

W=h2-h1 ...... 1

2. Condensing process The high pressure and temperature vapour refrigerant from the compressor is passed through the condenser where it is completely condensed at

constant pressure p2 and temperature T2. The vapour refrigerant is changed into liquid refrigerant. The refrigerant while passing through the condenser, gives its latent heat to the surrounding condensing medium.

3. Expansion process:

The liquid refrigerant at pressure p3=p2 expanded by throttling process

through the expansion valve to a low pressure p4=p1 and temperature T4=T1. Some of the liquid refrigerant evaporates as it passes through expansion valve, but the greater portion is vaporized in the evaporator. During the throttling process no heat is absorbed or rejected by the liquid refrigerant.

4. Vaporizing process:

The liquid vapour mixture of the refrigerant at pressure p4=p1 and

temperature T4=T1 is evaporated and changed into vapour refrigerant at constant pressure and temperature. During evaporation, the liquid vapour refrigerant absorbs its latent heat of vaporization from medium (air, water or brine) which is to be cooled.

The heat absorbed or extracted by the liquid vapour refrigerant during evaporation is given by:

RE=h1-h4=h1-hf3 ...... 2

Where hf3 is sensible heat at T3(enthalpy of liquid refrigerant leaving the condenser). The coefficient of performance is

Types of refrigerant used in vapour compression systems A variety of refrigerants are used in vapor compression systems. The required cooling temperature largely determines the choice of fluid. Commonly used refrigerants are in the family of chlorinated fluorocarbons (CFCs, also called Freons): R-11, R-12, R-21, R-22 and R-502. The properties of these refrigerants are summarized in Table 1 and the performance of these refrigerants is given in Table 2 below. Table 1. Properties of commonly used refrigerants

* At -10 oC ** At Standard Atmospheric Pressure (101.325 kPa)

Table 2. Performance of commonly used refrigerants

* At -15 oC Evaporator Temperature, and 30 oC Condenser Temperature VAPOUR ABSORPTION REFRIGERATION SYSTEMS

Introduction

The vapour absorption refrigeration system is one of the oldest methods of producing refrigerating effect. The principle of vapour absorption was first discovered by Michael Faraday in 1824 while performing a set of experiments to liquefy certain gases. The first vapour absorption refrigeration machine was developed by a French scientist, Ferdinand Carre, in 1860. This system may be used in both the domestic and large industrial refrigerating plants. The refrigerant, commonly used in a vapour absorption system, is ammonia.

The vapour absorption system uses heat energy, instead of mechanical energy as in vapour compression systems, in order to change the conditions of the refrigerant required for the operation of the refrigeration cycle. We have discussed in the previous chapters that the function of a compressor, in a vapour compression system, is to withdraw the vapour refrigerant from the evaporator. It then raises its temperature and pressure higher than the cooling agent in the condenser so that the higher pressure vapors can reject heat in the condenser.

The liquid refrigerant the condenser is now ready to the evaporator conditions again.

In the vapour absorption system, the compressor is replaced by an absorber, a pump, a generator and a pressure reducing value. These components in vapour absorption system perform the same function as that of a compressor in vapour compression system in this system, the vapour refrigerant from the evaporator is drawn into an absorber where it is absorbed by the weak solution of the refrigerant forming a strong solution. This strong solution is pumped to the generator where it is heated by some external source. During the heating process, the vapour refrigerant is driven off by the solution and enters into the condenser where it is liquefied. The liquid refrigerant then flows into the evaporator and thus the cycle is completed. Simple vapour absorption system

The simple vapour absorption system, as shown in fig.1, consists of an absorber, a pump, a generator and a pressure reducing value to replace the compressor of vapour compression system. The other components of the system are condenser, receiver, expansion value and evaporator as in the vapour compression system.

Figure 1 Simple vapour absorption system

In this system, the low pressure ammonia vapour leaving the evaporator entres the absorber where it is absorbed by the cold water in the absorber. The water has the ability to absorb very large quantities of ammonia vapour and the solution, thus formed, is known as aqua-ammonia. The absorption of ammonia vapour in water lowers the pressure in the absorber which in turn draws more ammonia vapour from the evaporator and thus raises the temperature of solution. Some form of cooling arrangement (usually water cooling) is employed in the absorber to remove the heat of solution evolved there. This is necessary in order to increase the absorption capacity of water, because at higher temperature water absorbs less ammonia vapour. The strong solution thus formed in the absorber is pumped to the generator by the liquid pump. The pump increase the pressure of the solution up to 10 bar.

The strong solution of ammonia in the generator is heated by some external source such as gas or steam. During the heating process, the ammonia vapour is driven off the solution at high pressure leaving behind the hot weak ammonia solution in the generator. This weak ammonia solution flows back to the

absorber at low pressure after passing through the pressure reducing value. The high pressure ammonia vapour from the generator is condensed in the condenser to a high pressure liquid ammonia. This liquid ammonia is passed to the expansion value through the receiver and then to the evaporator. This completes the simple vapour absorption cycle.

Practical vapour absorption system

The simple absorption system as discussed in the previous article is not very economical. In order to make the system more practical, it is fitted with an analyser, a rectifier and two heat exchangers as shown in fig. 2. These accessories help to performance and working of the plant, as discussed below:

1. Analyser. When ammonia is vaporized in the generator, some water is also vaporized and will flow into the condenser, along with the ammonia vapours in the simple system. If these unwanted water particles are not removed before entering into the condenser, they will enter into the expansion value where they freeze and choke the pipeline. In order to remove these unwanted particles flowing to the condenser, an analyser is used. The analyser may be built as an intergral part of the generator or made as a separate piece of equipment. It consists of a series of trays mounted above the generator. The strong solution from the absorber and the aqua from the rectifier are introduced at the top of the analyser and flow downward over the trays and into the generator. In this way, considerable liquid surface area is exposed to the vapour rising from the generator. The vapour is cooled and most of the water vapour condenses, so that mainly ammonia vapour (approximately 99%) leaves the top of the analyser. Since the aqua is heated by the vapour, less external heat is required in the generator. 2. Rectifier. In case the water vapours are not completely removed in the analyser, a closed type vapour cooler called rectifier (also known as dehydrator) is used. It is generally water cooled and may be of the double pipe, shell and coil or shell and tube type. Its function is to cool further the ammonia vapours leaving the analyser so that remaining water vapours are condensed. Thus, only dry or anhydrous ammonia vapours flow to the condenser. The condensate from the rectifier is returned to the top of the analyser by a drip return pipe.

Figure 2 Practical vapour absorption systems

3. Heat exchangers. The heat exchanger provided between the pump and the generator is used to cool the weak hot solution returning from the generator to the absorber. The heat removed from the weak solution raises the temperature of the strong solution leaving the pump and going to analyser and generator. This operation reduces the heat supplied to the generator and the amount of cooling required for the absorber. Thus the economy of the plant increases. The heat exchanger provided between the condenser and the evaporator may also be called liquid sub-cooler. In this heat exchanger, the liquid refrigerant leaving the condenser is sub-cooled by the low temperature ammonia vapour from the evaporator as shown in fig. 2. this sub- cooled liquid is now passed to the expansion value and then to the evaporator. In this system, the net refrigerating effect is the heat absorbed by the refrigerant in the evaporator. The total energy supplied to the system is the sum of work done by the pump and the heat supplied in the generator. Therefore, the coefficient of performance of the system is given by Heat absorbed in evaporator C.O.P =

Work done by pump + heat supplied in generator

Thermodynamic requirements of refrigerant-absorbent mixture

The two main thermodynamic requirements of the refrigerant-absorbent mixture are as follows:

1. Solubility requirement. The refrigerant should have more than Raoult’s law solubility in the adsorbent so that a strong solution, highly rich in the refrigerant, is formed in the absorber by the absorption of the refrigerant vapour. 2. Boiling points requirement. There should be a large difference in the normal boiling points of the two substances, at least 200˚C, so that the absorbent exerts negligible vapour pressure at the generator temperature. Thus, almost absorbent-free refrigerant is boiled off from the generator and the absorbent alone returns to the absorber. In addition, the refrigerant-absorbent mixture should possess the following desirable characteristics: (a) It should have low viscosity to minimize pump work. (b) It should have low freezing point. (c) It should have good chemical reactions of all kinds such as decomposition; polymerization, corrosion etc. are to be avoided.

Properties of ideal refrigerant-absorbent combination

The ideal refrigerant-absorbent combination should possess the following qualities:

1. The refrigerant should have high affinity for the absorber at low temperature and less affinity at high temperature.

2. The combination should have high degree of negative deviation from Raoult’s law.

3. The mixture should have low specific heat and low viscosity.

4. The mixture (solution) should be non-corrosive.

5. The mixture should have a small heat.

6. The mixture should have low freezing point.

7. There should be a large difference in normal boiling points of the refrigerants and the absorbent.

(a) Ammonia-water combination; and (b) lithium-bromide water combination.

The ammonia-water absorption system finds a significant place in large tonnage industrial applications. Comparison of refrigerant-liquid absorbent combination (say NH3-water) with refrigerant-solid absorbent combination (say NH3-CaCl2)

The solid absorbent system has the following two main advantages:

8. The amount of refrigerant cycled in relation to the amount of absorbent is larger for the solid absorbent than for the liquid absorbent. Therefore, in solid absorbent system, the thermal capacity of the salt contributes little, and hence heat required in the generator is less as compared to that for water in the aqua-ammonia system.

9. The solid absorbent system is extremely robust, not only in mechanical sence, but also with respect to adverse operating conditions. The performance is remarkably insensitive to changes in both condensing and evaporating temperatures. However the solid-absorption cycle has one major drawback. The heat of reaction is large compared with those found for liquid absorbent, approximately twice the latent heat of vaporization.

However the C.O.P. for solid absorption cycle is still higher than that of liquid absorption cycle, particularly when the difference between the condensing temperature and the evaporating temperature is large.

The solid-absorbent combination is ideally suited for intermittent operation on solar energy.

Advantages of vapour absorption refrigeration system over vapour compression refrigeration system.

Following are the advantages of vapour absorption system over vapour compression system.

1. In the vapour absorption system, the only moving part of the entire system is a pump which has a small motor. Thus, the operation on this system is essentially quiet and is

subjected to little wear.

2. The vapour absorption system uses heat energy to change the condition of the refrigerant from the evaporator. The vapour compression system uses mechanical energy to change the condition of the refrigerant from the evaporator. 3. The vapour absorption systems are usually designed to use steam, either at high pressure or low pressure. The exhaust steam from furnaces and solar energy may also be used. Thus this systems can be used where the electric power is difficult to obtain or is very expensive. 4. The vapour absorption systems can operate at reduced evaporator pressure and temperature by increasing the steam pressure to the generator, with little decrease in capacity. But the capacity of vapour compression system drops rapidly with lowered evaporator pressure. 5. The load variations do not affect the performance of a vapour absorption system. The load variations are met by controlling the quantity of aqua circulated and the quantity of steam supplied to the generator. 6. In the vapour absorption systems, the liquid refrigerant leaving the evaporator has no bad effect on the system except that of reducing the refrigerating effect. In the vapour compression system, it is essential to superheat the vapour refrigerant leaving the vapour so that no liquid may enter the compressor. 7. The vapour absorption systems can be built in capacities well above 1000 tonnes of refrigeration each, which is the large size for single compressor units. 8. The space requirements and automatic control requirements favor the absorption system more and more as the desired evaporator temperature drops.

Coefficient of performance of an ideal vapour absorption refrigeration system

(a) The heat (QG) is given to the refrigerant in the generator.

(b) The heat (QC) is discharged to the atmosphere or cooling water from the condenser and absorber.

(c) The heat (QE) is absorbed by the refrigerant in the evaporator, and

(d) The heat (QP) is added to the refrigerant due to pump work.

Neglecting the heat due to pump work (Qp), we have according to First Law of Thermodynamics,

Qc = QG + QE

Let TG = Temperature at which heat (QG) is given to the generator,

TC = Temperature at which heat (QC) is discharged to

atmosphere or cooling water from the condenser and

absorber, and

TE = Temperature at which heat (QE) is absorbed in the evaporator.

Since the vapour absorption system can be considered as a perfectly reversible system, therefore the initial entropy of the system must be equal to the entropy of the system after the change in its condition.

Q Q Q G + E = C ... (ii) T T T G E C

Q + Q = G E T C

Q Q Q Q = G − G = E − E TG TC TC TE

G  T − T  E  T − T  Q TC TG = Q TE TC  G C   C E 

Q = Q TE −TC   TG TC  G E     T T  T − T  C E   C G 

E  T − T   T T  = Q TC TE TG − TC  C E   G C 

 TC − TE   T−G T  = QE  T  T  E   G C 

Maximum coefficient of performance of the system is given by

Q E QE (C.O.P) = = max Q  T − T  T  G C E G QE  T T T  E  G − C 

    TE  TG − TC  = T − T T  C E  G 

It may be noted that,

T E is the C.O.P. of a Carnot refrigerator working T − T 1. The expression C E between

the temperature limits of TE and TC

TG − TC 2. The expression is the efficiency of a Carnot engine working between

TG

the temperature limits of TG and TC

Thus an ideal vapour absorption refrigeration system may be regarded as a combination of a Carnot engine and a Carnot refrigerator to produce the desired refrigeration effects as shown in Fig.3.

Figure 3 Representation of vapour absorption

refrigeration system The maximum C.O.P. may be written

as

(C.O.P.)max = (C.O.P)carnot carnot

In case the heat is discharged at different temperatures in condenser and absorber, then

(C.O.P.)  TE  TG − TA 

=  max T − T   T   C E   G 

Domestic Electrolux (ammonia hydrogen) refrigerator

The domestic absorption type refrigerator was invented by two Swedish engineers, Carl munters and baltzer von platan, in 1925 while they were studying for their undergraduate course of royal institute of technology in Stockholm. The idea was first developed by the ‘Electrolux company’ of Luton, England. This type of refrigerator is also called three-fluid absorption system. The main purpose of this system is to eliminate the pump so that in the absence of moving parts, the machine becomes noiseless. The three fluids used in this system are ammonia is used as a refrigerant because it possesses not of the desirable properties. It is toxic, but due to absence of moving parts, there is very little chance for the leakage and the total amount of refrigerant used is small. The hydrogen, being the lightest gas, is used to increase the rate of evaporation of the liquid ammonia passing through the evaporator. The hydrogen is also non-corrosive and insoluble in water. This is used in the low-pressure side of the system. The water is used as a solvent because it has the ability to absorb ammonia readily. The strong ammonia solution from the absorber through heat exchanger is heated in the generator by applying heat from an external source, usually a gas burner. During this heating process, ammonia vapours are removed from the solution and passed to the condenser. A rectifier or a water separator fitted before the condenser removes water vapour carried with the ammonia vapours, so they dry ammonia vapours are supplied to the condenser. These water vapours, if not removed, will enter into the evaporator causing freezing and choking of the machine.

Figure 4 Domestic Electrolux type refrigerator

The hot weak solution left behind in the generator flows to the absorber through the heat exchanger. This hot weak solution while passing through the exchanger is cooled. The heat removed by the weak solution is utilized in raising the temperature of strong solution passing through the heat exchanger. In this way, the absorption is accelerated and the improvement in the performance of a plant is achieved.

The ammonia vapours in the condenser are condensed by using external cooling souce. The liquid refrigerant leaving the condenser flows under gravity to the evaporator where it meets the hydrogen gas. The hydrogen gas which is being fed to the evaporator permits the liquid ammonia to evaporation at a low pressure and temperature according to Dalton’s principle. During the process of evaporation the ammonia absorbs latent heat from the refrigerated space and thus produces cooling effect.

The mixture of ammonia vapour and hydrogen is passed to the absorber where ammonia absorbed in water while the hydrogen rises to the top and flow a back to the evaporator. This completes the cycle. The coefficient of performance of this refrigerator is given by:

Heat absorbed in the evaporator C.O.P = Heat supplied in the generator

Lithium Bromide Absorption Refrigeration System

The lithium bromide absorption refrigeration system uses a solution of lithium bromide in water. In this system, the water is being used as a refrigerant whereas lithium bromide, which is a highly hydroscopic salt, as an absorbent. The lithium bromide solution has a strong affinity for water vapour because of its very low vapour pressure. Since lithium bromide solution is corrosive, therefore, inhibitors should be added in order to protect the metal parts of the system against corrosion. Lithium chromate is often used as a corrosion inhibitor. This system is very popular for air-conditioning in which low refrigeration temperatures (not below 0˚C) are required.

Fig.5 shows a lithium bromide vapour absorption system. In this system, the absorber and the evaporator are place in one shell which operates at the same low

pressure of the systems. The generator and condenser are placed in another shell which operates at the same high pressure of the system. The principle of operations of this system is discussed below:

Figure 5 Lithium-Bromide absorption refrigeration system