Thermo Theory

THERMODYNAMICS

RECIPROCATING ENGINES

TDC - Top Dead Centre
BDC - Bottom Dead Centre
* The distance between TDC & BDC is the largest distance that piston can travel in one direction & is called
Stroke.
* Diameter of piston is calle Bore.
* The minimum volume formed in cycle when piston is at TDC is called clearance volume.
* The volume displaced by cycle between TDC & BDC is called swept volume stroke volume or displace-
ment volume.

* Compression ratio rk = Vmax = VBDC
Vmin VTDC

* Mean Effective Pressor (MEP) : It is the pressure that if acted alone on the piston during entire power
stroke would produce the same amount of net work as that produced druing actual cycle.

Wnet = mep x A L
= mep x Vswept
= mep (Vmax - Vmin)

mep = Wnet
Vmax - Vmin

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Otto Cycle

Two reversible adiabatic & Two reversible isochores

η =1- 1
otto (r) γ−1

V1
r = compression ratio =

V2

work output = P1 V1 (rP - 1) (rγ−1 − 1)
γ −1 k

For maximum work output

T2 = T4 = √ T1T3 , Wmax = Cv (√T1- √T3)2
Note :
* The η of air standard otto cycle is thus a function of compression ratio only. The higher rk, higher is η.It
is independent of temperature levels at which cycle operates.

* If rk increased beyond certain limit, premature ignition of fuel called autoignition produces engine knock or
detonation.

Diesel Cycle

Two reversible adiabatics, one rev. isobar and one reversible isochor

η = 1 - 1 (ργ − 1) r = compression ratio = V1
diesel (r) γ−1 γ(ρ − 1) V2

ρ = cut - off ratio = V3
V2

Note :η of Diesel cycle is less than that of Otto cycle for same compression ratio.

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THERMODYNAMICS

Dual Cycle

η= 1 rprcγ − 1
1-
(rk) γ−1 rp − 1+γr (r − 1)
pc

Brayton Cycle

W -W
Tc

η Brayton = 1 - 1
r γ−1/γ

p

rp = Pressure ratio = P2
P1

η = 1 − T min
T

max

Two reversible isobars & Two reversible adiaabatics

T max γ/(γ −1)
Tmin
(r ) =
p max

(rp)opt = √ (rp)max

(Wnet)max = Cp ( √ Tmax - √Tmin)2

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Lenoir Cycle

η =1−γ r 1/γ − 1
p

rp - 1

Note : Efficiency of Lenoir cycle depends on pressure ratio and specific heat ratio.
Atkinson Cycle

η Atkinson = 1 - γ re - rk
r γ-r γ

ek

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THERMODYNAMICS

Comparison between Brayton and Otto Cycle

* For the same rk and work capacity the Brayton cycle handles a large range of volume and a smaller
range of pressure and temperature
* A reciprocating engine cannot efficiently handle a large volume flow of low pressure gas for which

π
the engine size d2 l becomes large and the friction losses also becomes large so the otto cycle

4

is more suitable in the reciprocating engine field.
* An I.C. engine is exposed to the highest temperature (after the combustion of fuel) only for a short
while and it gets time to cool in the other processes of the cycle. On the other hand a gas turbine plant is a
steady flow device, is always exposed to the highest temperature, thus the material is to be protected.

Comparison between Otto, Diesel and Dual cycle

1. For same compresson ratio and heat rejected (rk & Q2)
η > η Dual > η Diesel
Otto

2. For same max. pressure and temperature and heat rejection

η > η Dual > η Otto
Diesel
3. For same compression ratio and heat supplied (rk & Q1) - η Otto > η Dual > η Diesel
4. For constant maximum pressure and heat input η Diesel > η Dual > η Otto

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Comparison between Brayton cycle and Rankine Cycle

* The working fluid undergoes phase change in Rankine cycle but not in Brayton cycle.
* The ratio of turbine work to compressor work is much more in Rankine cycle than that in Brayton
cycle.

Effect of Regeneration on Brayton Cycle

∈, Effectiveness = t3 - t2
t5 - t2

* Efficiency is increased but work output remains same.
* As the pressure ratio if increased, the η steadily increases.

η= 1 - T1 rP γ−1
T4 γ

Note : Above certain pressure ratio (p2/p1) the addition of regenerater causes a loss in cycle η when
compared to original brayton cycle.

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THERMODYNAMICS

Effect of Intercooling & Reheating on Brayton Cycle :
1) Intercooling :

* 1 - 2 - 5 - 6 - cycle without intercooling
1-2-3-4-5-6 - cyle with intercooling.

* Cycle 2-3-4-2 is added
so, more work but also more heat supplied

* η reduces with intercooling & since Tm1 is lower & Tm2 is higher.
ii) Reheat

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* The work capacity increase but heat supplied also increase.
* On cycle 4-5-6-4 rp is lower than in basic cycle 1-2-3-4 so its η is lower

∴ η of cycle decreases with use of reheat.

* If T6 > T4 so regeneration is employed, there is more energy that can be recovered from turbine
exhaust gases so if regeneration is employed in conjuction with reheat, there may be net gain in η

without regen eration

η= 1- 1 γ−1 rp decrease η decrease
rp γ

with regen eration

η= 1- T1 rP γ−1
T4 γ

Note : If in one cycle several stages of intercooling & several stages of reheat are employed. When

number of such stages is large, the cycle reduces to Ericsson cycle, with 2 reversible isobars & 2 revers-
ible isotherms. With ideal regenaration the cycle η becomes equal to carnot η .

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THERMODYNAMICS

Ideal Regenerative cycle with Intercooling & Reheat

If assumes P3 = P2 = P7 = P8 = √P1P4 = √P6P9
Aslo T1 = T3 & T6 = T8

We get

1 - T1 P4 γ−1
T6 P1
η cycle = 2γ

*********************

“Be intent on the perfection of the present day”

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MECHANICAL ENGINEERING

9 RANKINE CYCLE

STEAM CYCLES
* A steam power plant continuously converts the energy stored in fossil flues into shaft work and ulti-
mately into electricity. The working substance is water which is some times in the liquid phase and sometimes
in the vapour phase.
* The fluid is undergoing a cyclic process, there will be no net change in its internal energy over the cycle
(∫ dE = 0) and consequently the net energy transferred to the unit mass of the fluid as heat during the cycle
must equal the net energy transfer as work from the fluid.

∑Q net = ∑W net
cycle cycle

Q1 - Q2 = WT - WP

where,

Q1 = heat transferred to the working fluid kJ/kg
Q2 = heat rejected from the working fluid kJ/kg
W = work transferred from the working fluid kJ/kg

T

WP = work transferred into the working fluid kJ/kg

η =cycle Wnet WT - WP =1- Q2
= Q1

Q1 Q1

RANKINE CYCLE

* This cycle contains four processes :
• For steam boiler: reversible constant pressure heating process of water
• For turbine: reversible adiatatic expansion of steam.
• For condenser: reversible constant pressure heat rejection
• For pump: reversible adiabatic compression
* When all these four processes are ideal the cycle is an ideal cycle, called a rankine cycle.

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* Now,
For 1 kg of fluid, the steady flow energy equation to each processes:
For boiler Q = h - h

1 14

For turbine WT = h1 - h2
For condenser Q2 = h2 - h3
For pump WP = h4 - h3
Efficiency of Rankine cycle

η = Wnet = (h1-h2) - (h4-h3)
Q1 h1 - h4

* The pump work is small compared to the turbine work and is often negleted.
* Steam rate: The capacity of a steam plant is often expressed in terms of steam rate or specific steam
consumption. It is defined as the rate of steam flow (kg/s) required to produce unit shaft output (1kW).

Steam rate = 1 kg/kWs
Wnet

* Heat rate: The cycle efficiency is sometimes expressed alternatively as heat rate which is the rate of
heat input (kJ/s) required to produce unit shaft output (1kW)

Heat rate (H.R.) = Q1 = 1 kJ
WT-WP η kWs.

Note : In steam power plants, the pump handles liquid, specific volume is many times langer. Therefore, the
work output of the turbine is much larger than the work input to the pump.

Internal Irreversibility

* Internal irreversibility of Rankine cycle is caused by fluid fricton, throttling and mixing. Though the
assumption of adiabatic flow in them is still valid, due to fluid friction the expansion and compression pro-

cesses are not reversible and entropy of the fluid in both increases.
• The isentropic Effciency (ηT) of the turbine is

ηΤ = h1 - h2
h1 - h2s

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Piping Losses : States 1’ and 1 represents the states of the steam leaving the boiler and entering the turbine
respectively, 1’- 1” represents the frictional losses, and 1” - 1 shows the constatnt pressure heat loss to the
surroundings.

Turbine Losses : For the reversible adiabatic expansion, the path will be 1 - 2s. For an ordinary real tubine
the heat loss is small, and WT is h1 - h2 , with Q2 , equal to zero.

η = Wr = h1 - h2
h1 - h2s
r h1 - h2s

Pump Losses :

The liquid leaving the pump must be at higher pressure than turbine inlet as there pressure drop in boiler.
Pipes, valves etc. and reaches at P1 from P4 as shown above.
• The isentropic Efficiency of the Pump (ηp) is

η = h4s - h3
p h4 - h3

• the actual pump work would be

h - h v (p - p )
4s 3 34 3
W= η = η
pp

Thus turbine produces less work and the pump absorbs more work.

Mean Temperature of Heat Addition

* In the Rankine cycle, heat is added reversibly at a constant pressure but at infinite temperatures. Tm1 is
the mean temperature of heat addition then

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THERMODYNAMICS

Heat added is
Q1 = h1 - h4 = Tm1 (S1 - S4)

∴ Tm1 = h1 - h4
S1- S4

Heat rejected

Q2 = h2 - h3 = T2 (S1 - S4)

η = 1 - Q 1 - T2
Rankine 2 Tm1

Q1

* Lower is the condenser pressure, the higher will be the Efficiency of the Rankine cycle. Since it is fixed

due to ambient conditions η = f (Tm1) only.
Rankine

* The higher the mean temperature of heat addition, the higher will be the cycle efficiency.

Note :
a) If mean temperature of heat addition is increased, efficiency is increased.
b) The quality of steam at turbine exhaust is increased as the expansion line shifts to right. Hence perfor-
mance of turbine is improved.

* The maximum temperature of steam that can be used is fixed from metallurgical considerations.
* As the operating steam pressure at which heat is added in the boiler increases, the mean temperature of
heat addition increases. But when the turbine inlet pressure increases the ideal expansion line of steam shifts
to the left and the moisture content of steam in the later steages of the turbine is high and strike the blade with
high velocity and erode their edges, as a result of which the life of the blades decreases.
* At turbine exhaust quality of steam should not fall below 88%.

Reheat Cycle :
* Reheating is done to utilize higher boiler pressure while maintaining better quality of steam at turbine
exhaust.
* In reheating, the expansion of steam from intial state 1 to condenser pressure is carried out in two or
more steps.

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* Initially the steam is expanded from state 1 to 2s in high pressure (HP) turbine, then reheated from 2s
to 3 in a reheater and then expanded from 3 to 4s in low pressure (LP) turbine.

Q1 = h1 - h6s + h3 - h2s
Q2 = h4s - h5
WT = h1 - h2s + h3 - h4s’ Wp = h6s - h5

η = (h1 - h2s + h3 + h4s) - (h6s - h5)
h1 - h6s + h3 - h2s

Note :
* The net work output of the plant increases with reheat, and hence the steam rate decreases. Reheating
also improves the quality at turbine exhaust.
* By increasing the number of reheats, still higher steam pressure could be used, but the mechanical
stresses increases in much higher proportion than the pressure because of prevailing high temperature. In
that way the maximum steam pressure in cycle result in complication and increases capital cost that are not
justified by improvement in the cycle efficiency.
* The optimum reheat pressure for most of the modern power plant is 0.2 to 0.25 of the intial steam
pressure.

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THERMODYNAMICS

* For too low reheat pressure the exhaust steam may even be in the-supersaturated state, which is not
good for the condenser.

Ideal Regenerative Cycle : -
* In regeneraton, the energy is exchanged internally between the expanding fluid in turbine and com-
pressed fluid (after pump work) before heat addition.
* A well known gas cycle that uses regeneration is the stirling cycle comprising two reversible isotherms
and two reversible isochores, ideal stirling cycle has the same efficiency as the carnot cycle.
* In the ideal regenerative cycle the condensate after leaving the pump circulates around the turbine
casing so that heat is transferred from the vapour expanding in the turbine to the condensate circulating
around it. It is assumed that this heat transfer process is reversible.

Q1 = h1 - h5 = T1 (S1 - S5)
Q2 = h2 - h3 = T2 (S2 - S3)
for reversible heat transfer:
∆S univ = ∆S water + ∆S steam = 0
∆S water = - ∆S steam

η = 1 − Q2 = 1 − T2
Q1 T1

Pump work remains same as in Rankine cycle i.e.
wp = h4 - h3

* The efficiency of ideal regenerative cycle is equal to Carnot cycle.
* The net work output of the ideal regenerative cycle is thus less and hence, its steam rate will be more;
although it is more efficient compared to the Rankine cycle. However the cycle is not practicable-because

• reversible heat transfer cannot be realized in finite time.
• heat exchanger in the turbine is mechanically impracticable
• the moisture content of the steam in the turbine is high, which leads to excessive erosion of turbine
blades.

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Regenerative Cycle
* In practical regenrative cycle, steam is bled from the turbine and feed water is heated with it.
* In multistage regenrative cycle, more no. of feedwater pumps are used and heat added at low tempera-
ture (in economizer) is minimized.

WT = 1 (h1 - h2) + (1 - m1) (h2 - h3) + (1- m1- m2) (h3 - h4)
WP = (1 - m1 - m2) (h6 - h5) + (1- m1)

(h8 - h7) + (h10 - h9)
Q1 = 1(h1- h10)
Q2 1 (1 - m1 - m2 ) (h4 - h5)

η= Q1 - Q2 = WT - WP
Q1 Q1

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* The effects of regenerative feedwater heating for the same turbine output may be summerized as below:

• It significantly increases the cycle effciency and reduces the heat rate.
• It increases the steam flow rate. (requiring bigger boiler.)
• It reduces the steam flow to the condenser (needing smaller condenser)
• If there is no change of boiler output, the turbine output drops.

Feed Water Heaters
* These are of two types viz., open heaters and closed heaters. In an open or contact type heater, the
extracted steam is allowed to mix with feed water and both leave the heater at a common temperature.
* In a closed heater, the fluids are kept separate and are not allowed to mix together.
* The condensate (saturated water at the steam extraction pressure). sometimes called the heater drip,
then passes through a trap into the next lower pressure heater.
* The drip from the lowest pressure heater could similarly be trapped to the condenser, but this would be
throwing away energy to the condenser cooling water.
* To avoid this waste a drip pump, pumps the drip directly into feed water stream.
* A terminal temperature difference (TTD) is defined for all closed feed water heaters as.

TTD = saturation temperature of bled steam Exit water temprature.
The value of TTD varies with pressure.
* Too small a value, although good for plant efficiency, would require a larger heater. Too large a value
would reduce the cycle efficiency.
* If the extracted steam upon condensation gets subcooled, a drain cooler may be used.
* The advantages of the open heater are simplicity lower cost, and high heat transfer capacity. The
disadvantage is the necessity of a pump at each heater to handle the larger feedwater stream.
* In most steam power plants, closed heater are favoured but at least one open heater is used, Primarily
for the purpose of feedwater deaeration. The open heater in such a system is called the dearator. Closed
heaters are mostly horizontal.

Carnotizaton of Rankine Cycle
* With infinite number of extraction stages, the irreversible process DE - the heating from condenser to
boiler saturation temprature could thus be made reversible.
* The area of parallelogram CEAG which represent cycle output will be equal to the area of rectangle
AFGE, which represent the output of Carnot cycle.

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* Regenerative feed water heating by turbine extraction is, therefore, also termed as the carnotization of
the Rankine cycle.Aregenerative feed heating cycle with an infinite number of feedwater heaters has thus an
efficiency equal to that of carnot cycle.

* Complete carnotization of Rankine cycle is not possible with a finite number of heaters. If there is one
feedwater heater used, m kg of steam is extracted from the turbine for each kg of steam thermal efficiency of
the cycle is

η = 1 - (h2 - h6) (h3 - h4)
(h2 - h4) (h1 - h6)

* Maximum efficiency is obtained in regeneration, when the total enthalpy rise of feed water from condenser
temperature to boiler temperature is equally distributed in feedwater heaters and economizers.

Binary Vapour Cycles
* No single fluid can meet at all the requirement of ideal working fluid, water, diphenyl ether, aluminium
bromide mercury are few working fluids which give satisfactory performance.
* Among these fluids mercury has the best performance at higher temperature due to its low saturation
pressure.
* To take advantage of beneficial features of mercury at high temperature to avoid its deleterious effect at
low temperature, mercury cycle is used in combination with water vapour cycles.
* Thus in Binray cycle, two cycle with different working fluid are coupled in series and heat rejected by
one is utilized by the other.

Efficiencies in a Steam Power Plant
* The overall efficiency of a power plant is defined as

η =overall Power available at the generator terminals
Rate of energy release by the combustion of fuel

mW x 103
= e , Wf = Fuel burning rate

Wf x C.V.

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* The boiler efficiency is

η =boiler Rate of energy absorption by water to form steam
Rate of energy release by the combustion of fuel

= Ws (h1 - h4) , Wf = Fuel burning rate
Wf x C.V.

Ws = is the steam generation rate
The cycle efficiency is given by

η =cycle h1 - h2
h1 - h4

* The mechanical efficiency of the turbine will be

η =turbine(mech) Brake output of the trubine
Internal output of the turbine

brake output
=

Ws (h1 - h2)

* The generator Efficiency of the electric alternator is

η =generator Electrical output at generator terminals
Brake output of the turbine

η = η η η ηoverall
boiler. cycle. turbine. generator

*******************

“To improve is to change; to be perfect is to change often”

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MECHANICAL ENGINEERING

10 REFRIGERATION CYCLES

REVERSED HEAT ENGINE CYCLE
Areverse heat engine cycle is visualized as an engine operating in the reverse way, i.e. receiving heat from a
low temperature region, discharging heat to a high temperature region, and receiving a net inflow of work.
Under such conditions the cycle is called a heat pump cycle or a refrigeration cycle .

(COP)H.P. = Q1 = Q1
W Q1 - Q2

and for a refrigerator

(COP)ref = Q2 = Q
2

W Q1 - Q2

The working fluid in a refrigeration cycle is called a refrigerant. In the reversed Carnot cycle, the

refrigerant is first compressed reversibly and adiabatically in process 1-2 where the work input per kg of

refrigerant is WC then it is condensed reversibly in process 2-3 where the heat rejection is Q1, the refrigerant
then expands reversibly and adiabatically in process 3-4 where the work output is WE, and finally it absorbs
heat Q2 reversibly by evaporation from the surroundings in process 4-1.

Here Q1 = T1 (s2 - s3), Q2 = T2(s1 - s4)
and Wnet = WC - WE = Q1 - Q2 = (T1 - T2) (s1 - s4)

Where T1 is the temperature of heat rejection and T2 the temperature of heat absorption

(COPref)rev = Q2 = T
2

and Wnet T1 - T2

(COPH.P.)rev = Q2 = T1
Wnet T1 - T2

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THERMODYNAMICS

VAPOUR COMPRESSION REFRIGERATION

In an actual vapour refrigeration cycle, an expansion engine, as shown in is not used.Athrottling valve or a
capillary tube is used for expansion in reducing the pressure from p1 to p2. The basic operations involved in
a vapour compression refrigeration plant.

Fig : Vapour Compression Refrigeration Plant-flow Diagram

Fig : Vapour Compression Refrigeration Cycle-Property Diagram

1. Compression : A reversible adiabatic process 1 - 2 or 1’- 2’either starting with saturated vapour (state
1), called dry compression, or starting with wet vapour (state 1’), called wet compression, (1’ -2’), because
with wet compression there is a danger of the liquid refrigerant being trapped in the heat of the cylinder by
the rising piston which may damage the valve or the cylinder head, and the droplets of liquid refrigerant may
wash away the lubricating oil from the walls of the cylinder, thus accelerating wear.

2. Cooling and Condensing : A reversible constant pressure process, 2-3 first desuperheated and then
condensed, ending with saturated liquid Heat Q1 is transferred out.

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3. Expansion : An adiabatic throttling process 3-4, for which enthalpy remains unchanged. States 3 and 4
are equilibrium points. Process 3-4 is adiabatic (then only h3 = h4 by S.F.E.E.), but not isentropic,

∫ p2 vdp

Tds = he - vdp, or s4 - s3 = -
p1 T

Hence it is irreversible and cannot be shown in property diagrams. States 3 and 4 have simply been joined
by a dotted line.

4. Evaporation :Aconstant pressure reversible process, 4-1, which completes the cycle. The refrigerant is
throttled by the expansion valve to a pressure, the saturation temperature at this pressure being below the
temperature of the surroundings. Heat then flow, by virtue of temperature difference, from the surroundings,
which gets cooled or refrigerated, to the refrigerant, which then evaporates, absorbing the latent heat of
evaporation. The evaporator thus produces the cooling or the refrigerating effect, absorbing heat Q2 from
the surrounding by evaportion.

Fig : Phase Diagram with Constant Property Lines on p-h Plot

Fig : Vapour Compression Cycle on p-h Diagram

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PERFORMANCE AND CAPACITY OFAVAPOUR COMPRESSION PLANT
The simplified diagram of a vapour compressions refrigeration plant.

Compressor

Fig : Vapour Compression Plant

When steady state has been reached, for 1 kg flow of refrigerant through the cycle, the steady flow

energy equations (neglecting K.E. and P.E. changes) may be written for each of the components in the cycle

as given below :

Compressor h1 + Wc = h2
∴ Wc = (h2 - h1) kJ/kg
h2 + Q1 + h3
Compressor Q = (h - h ) kJ/kg
∴
1 23
Expansion valve
h3 = h4
or (hf)p1 = (hf)p2 + x4 (hfg)p2

∴ (hf) p1 - (hf) p2
x4 = (hf) p2

This is the quality of the refrigerant at the inlet to the evaporator (mass fraction of vapour in liquid-vapour
mixture).
Evaporator

h4 = Q2 = h1
∴ Q2 = (h1 - h4) kJ/kg
This is known as the refrigerating effect, the amount of heat removed from the surroundings per unit mass
flow of refrigerant.

COP = Q2 = h1 - h4
WC h2 - h1

If w is the mass flow of refrigerant in kg/s, then the rate of heat removal from the surroundings
= w(h1 - h4) kJ/s = w(h1 -h4) x 3600 kJ/h
One tonne of refrigeration is defined as the rate of heat removal from the surrounding equivalent to the

heat reuired for melting 1 tonne of ice in one day. If the latent heat of fusion of ice is taken as 336 kJ/kg, then
1 tonne is equivalent to heat revoval at the rate of (1000 x 336) / 24 or 14,000 kJ/h

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∴ Capacity of the refrigerating plant

w(h - h ) x 3600
14 tonnes

14,000

The rate of heat removal in the condenser

Q1 = w(h2 - h3) kJ/s
If the condenser is water-cooled , mc the flow-rate of cooling water in kg/s, and (tc2 - tc1) the rise in tempera-
ture of water, then

Q1 = w(h2 - h3) = mccc (tc2 - tc1) kJ/s
The rate of work input to the compressor

Wc = w(h2 - h1) kJ/s

VOLUMETRIC EFFICIENCY

η = Actual volume of gas drawn at evaporator pressure and temperature
vol Piston displacement

∴ Volume of gas handled by the compressor

= w . v1 (m3/s) = π D2L N n x η
4 60 vol

where w is the refrigerant flow rate ,

v1 is the specific volume of the regrigeratnt at the compressor inlet,
D and L are the diameter and stroke of the compressor,

n is the number of cylinders in the compressor, and

N is the r.p.m.

The clearnace volumetric efficiency is given by Eq.

η =1+C-C p2 1/n C= Vc = clearnace volume
vol p1 Vs swept vol. or P.D.

where C is the clearance

For a reciprocarting compressor, a high pressure ratio across a single stage means low volumetric efficiency.
Also, with dry compression the high pressure ratio results in high compressor discharge tempature whcih
may damage the refrigerant. To reduce the work of compression and improve the COP, multistage com-
pression with intercolling may be adopted.

For minimum work, the intercooler pressure pi is the geometric mean of the evaporator and condenser
pressures, p1 and p2, or

Pi = √P1 . P2

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By making an energy balance of the direct contact heat exchanger,
m2h2 + m1h6 = m2h7 + m1h3

∴ m1 = m2 - h7
m2 h3 - h6

The desire refrigerating effect determiners the flow rate in the low pressure loop, m2, as given below

m2 (h1 - h8) = 14000 xP
3600
where P is the capcity, in tonnes of refrigeration

∴ m = 3.89 P kg/s
2 h1 - h8

********************

“Better to do something imprefectly than to do nothing flawlessly”

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11 PSYCHROMETRY

DALTON’S LAW

Pi = ni P
Σ ni

n = No. of moles

AMAGAT’S LAW

Vi = ni V
Σ ni

=> Pi = Vi = ni = γ
PV Σni

γ = mole fraction of gas i in mixture of gases.

* Molecular mass of mixture

M mix = Σ Mi γ
i

* Universal gas constant R = 8.314 kJ/kg mole K

R
characteristic gas constant Ra = Ma

* Psychometric properties :
F+P=C+2

F = Degree of freedom
P = Phases in mixture
C =Components of mixture
For a mixture of 2 gases we need 3 properties to fix its thermodynamic state whereas for pure
substance we need only 2 properties.
* Dry Bulb Temperature (DBT) is the actual temperature ‘t’of the moist air.

* Dry Air Mass = ma
Specific Volume = va

Water Vapour Mass = mw
Specific Volume = Vw

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where, V = Volume
m = Mass
P = Pressure,
T = Temperature

Specific humidity/Humidity ratio

ω = mw = Va
ma Vw

= 0.622 Pv
P-P

v

Pv = Partial pressure of vapour
P = Total pressure of moist air
* Specific humidity is a function of partial pressure

Dew Point Temperature
* The temperature at which the super heated vapour present in a sample of unsaturated.
* At a given partial pressure of water vapour, the saturation temperature is DBT.

Degree of Saturation
* The capacity of air to absorb moisture is called degree of saturation

ω Pv [1 - Ps / P]
µ= ω = Ps [1 - Pv / P]

ST

ω = specific humidity at partial pressure Pv.
ω
S = specific humidity at saturation pressure PS.

* Ralativehumidity

RH/φ = Mass of water vapour in a fixed volume of moist air temp. T
_________________________________________________

Mass of water vapour in a saturated sample of moist air at same
temp. and volume

= mv = Pv V/RT = PV = PS
mvs Ps V/ RT Vs VV

µ=φ 1 - P /P
s

1 - PV /P

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ENTHALPY OF MOISTAIR

h = ha + ωhv
For 1 kg of dry air and w kg of water vapour.

ha = enthalpy of dry air

hv = enthalpy of water vapour

h = C t + (h ) + C (t - t )
v pw d fg td pv d

or = (hfg) 0oc + C pv (t - 0) = (hfg) 0oc + C pvt

h = h + ω (h ) 0oc + ωC pv t
a fg

hfg = Latent heat of vapourization
= (Cpa +ω Cpv)t + ω (hfg) 0oc
Cpt + ω hfg

WET BULB TEMPERATURE (WBT)

* WBT is used to measure humidity by a psychrometer.
The temperature measured by thermometer whose bulb is covered with wet wick is wet bulb tem-

perature while the bare bulb mesures dry bulb temperature.
* Wet Bulb Temperature (WBT) is always less than Dry Bulb Temperature (DBT) except when the
air saturated. That time wdt is equal to DBT.
* Wet bulb depression = DBT - WBT
* WBT is an indirect measure of the dryness of the moist air.

Thermodynamic wet bulb temperature or temperature of adiabatic saturation
* For any state of unsaturated moist air, there exit a temperature at which the air becomes adiabati-
cally saturated by the evaporation of water into air. That temperature is temperature of adiabatic saturation.
* Only in case of saturated moist air, wet bulb temperature and thermodynamic wet bulb temperature
temperature of adiabatic saturation are equal.

• For moist air having three degree of freedom, three of the four properties can be measured
* total pressure or barometer pressure.
* dry bulb temperature.
* dew point temperature.

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* wet bulb temperature.
Thermodynamic WBT is given as

t* = t - hfg (ω* − ω)
Cp

PSYCHROMETRIC CHARTS

Constant enthalpy lines are along the constant wet bulb temperature lines.

PSYCHROMETRY OFAIR CONDITIONING PROCESS

* Mixing process
• Specific humidity of the mixture (w)

m w +m w
a1 1 a2 2
w = ma1 + ma2

• Enthalpy of the mixture (h)

ma1 h1 + ma2 h2
h= m +m

a1 a2

• Temperature of the mixture (t)

ma1 t1 + ma2 t2
t = ma1 + ma2
where for the two moist air stream

ma1, wa2 - Mass of the dry air
w , w - Specific humidity

12

h , h - Specific enthalpy
12

t , t - Temperature (in oC)

12

* Mixing with Condensation : When a cool current mixes with hot current with high relative humidity,

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the moisture may get condensed and relative humidity of mixed stream might be less that without condensa-

tion mass condensed per unit dry air

ω = ω - ω
C 3 4

ω m ω + m ω2 ω
c a1 1 a2 c
= -
ma1 + ma2

h4 = ma1 h1 + ma2 h2 - hf4 ω
c
ma1 + m
2

Basic Air Conditioning Process

Process in diagram Type ofAir-Conditioning
OA Sensible Heating
OB Sensible Cooling
OC Humidification
OD Dehumidificaton
OE Heating and Humidification
OF Cooling and Dehumidification
OG Cooling and Humidification
OH Heating and Dehumidification

* If CMM be the cubic meter per minute supply of air and

Then P = 1.2,C pair = 1.02
air

QS = 0.0204 (CMM) ∆t kW
QL = 50 (CMM) ∆ω kW

Sensible Heating and Cooling
QS = ma (1.005 + 1.88ω) (t1 - t2)
ma - mass flow rate of dry air
ω - specific humidity of moist air

t1 , t2 - initial and final temperature

Latent Heating and Cooling

QL = ma [h1 - h2]
ma - mass flow rate of dry air
h1 , h2 - initial and final specific enthalpy

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Sensible Heat Factor (SHF)

SHF QS
= =

QS + QL

hB -hA = 0.0204∆t

hC - hA 0.0204∆t + 50 ∆ω

• SHF = 1
⇒ Sensible Heating or Cooling

• SHF = 0
⇒ Humidification or Dehumidification

By Pass Factor (x)
* By pass factor is defined as the fraction of air which doesn’t come in contact with or which by
passes the cooling surface.
* Due to bypass factor, the exit state of air is a complex mixture of contacted and uncontacted air.

h1

h2 2 1 ω1
h (1-x)
ω2
s ω

S sω

x

t2 - ts ω - ω h2 - hs
t1 - ts 2 s h1 - hs
X = = ts t2
ω - ω t t1
1 s

* By pass factor represents how much effective the coil is high by pass factor coil implies less effective
and vice-versa.
Apparaturs Dew Point Temperature
* For dehumidification, the surface temperature of coil should be below the dew point temperature of

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the air. This surface temperature of coil isADP.
This point on the ω-t graph, can be found by joining the intial and final condition of air and joining it to
saturation curve which showsADP. Point S in the previous figure isADP.
Driving potential for sensible heat transfer is temperature difference whereas that for latent heat transfer is
partial pressure of sp. humidity difference i.e., (PV - PS) or (ω − ωS)

Heating Load and Cooling Load
* If a conditioned space gains heat from atmosphere, it requires cooling of air to a lower temperature.
This is called cooling load. If conditioned space losses heat to outsibe it is heating load.
* Both these loads can be sensible or latent loads.
* Total heating or total cooling load is sum of sensible and latent loads.

SummerAir-Conditioning System

O

m1- fresh air inducted in the room
m2 - recirculated air
m3 - air rejected from the room = m1

Room SHF = RSH = RSH

RSH + RLH RTH

Minimum quantity of air to be supplied is

RSH
(CMM) =

0.0204 (ti - tADP)

RLH
=

50 (ωi - ωADP)

RTH
=

0.02 (h - h )
i ADP

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* RSH = Room sensible Heat
RLH = Room Latent Heat
RTH = Room total heat

* When ventilation air is used, another load, that of bringing a ventilation air from outside condition to
inside condition, comes on apparatus. This is called ventilation load.

OASH = Outside air sensible heat
OALH = Outside air latent heat
* Air conditioning equipment load
Total sensible TSH = RSH + OASH
Total latent TLH = RLH + OALH
Grand sensible heat factor

TSH TSH
GSHF = =
TSH + TLH GTH

***********************

“What is once well done is done forever”

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