POWER PLANT ENGINEERING 3

92 PSP eBook | Power Plant Engineering 3

• Process 1 to 2 is isentropic compression
• Process 2 to 3 is reversible constant volume heating
• Process 3 to 4 is isentropic expansion
• Process 4 to 1 is reversible volume cooling

Isentropic Compression (1-2) Constant Volume Heat Addition (2-3)

Isentropic Expansion (3-4) Constant Volume Heat Rejection (4-1)

Figure 5.2 Otto Cycle Processes

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Consider 1 kg of air: = Cv (T3 − T2 )
• Heat supplied at constant volume = Cv (T4 − T1 )
• Heat rejected at constant volume =
• Work Done = Heat supplied – Heat rejected

• Compression ratio, rc (r) = Cv (T3 − T2 )- Cv (T4 − T1 )

c = v1
v2

• Expansion ratio, re (r) = c = v4
v3

• For isentropic compression process 1 -2,

2 = [ 12] −1 = [ ] −1
1

• For isentropic expansion process 3 -4,

  −1
 T3  v4 
=  v3  =   −1
T4

• Thermal efficiency  =1− ( 1

) −1

( ) ( )•
Mean effective pressure (Pm) = P1   −1 − 1  p − 1

( −1)( −1)

• Compression ratio, r = v1 = v4 = Total cylinder volume / Clearance Volume
v2 v3

• Clearance Volume = Clearance Volume + Swept Volume
Clearance Volume

= VC + VS
VC

= Swept Volume = Vs
 −1

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• Swept Volume = d 2  l
4

• The relations between pressure and temperature or pressure and volume may be

 −1

T1  P1  1
T2  P2 
obtained from the usual isentropic process. =  

T3 VV43  −1 P2 VV12  P3 VV43  P2V2 = P3V3
T4  P1  P4  T2 T3
=  =  = 

5.2 Constant Pressure or Diesel Cycle

• This cycle was introduced by Dr.R. Diesel in 1897.
• It differs from Otto cycle in that heat is supplied at constant pressure instead of at

constant volume.
• Figure (a) and (b) shows the P-v and T-s diagrams of this cycle respectively.

Figure 5.3 Diesel Cycle P-v and T-s Diagram

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Figure 5.4 Diesel Cycle Processes
• This cycle comprises of the following operations:

• 1 – 2 Adiabatic compression
• 2 – 3 Addition of heat at constant pressure
• 3 – 4 Adiabatic expansion
• 4 – 1 Rejection of heat at constant volume
• Point 1 represents that the cylinder is full of air
• Let P1, V1 and T1 be the corresponding pressure, volume, and absolute temperature.
• The piston then compresses the air adiabatically (i.e, PV  = constant) till the values
become P2, V2 and T2 respectively (at the end of the stroke) at point 2.
• Heat is then added from a hot body at constant pressure.
• During this addition of heat let volume increase from V2 to V3 and temperature T2 to T3
corresponding to point 3.
• This point (3) is called the point of cut-off.
• The air then expends adiabatically to the conditions P4, V4 and T4 respectively
corresponding to point 4.
• Finally, the air rejects the heat to the cold body at constant volume till the point 1 where
it returns to its original state.

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Consider 1 kg of air: = C p (T3 − T2 )
• Heat supplied at constant pressure = Cv (T4 − T1 )
• Heat rejected at constant volume

• Work done = Heat supplied – Heat rejected
• Diesel efficiency
= C p (T3 − T2 )- Cv (T4 − T1 )

=  = 1− (T4 −−TT12)) =1− Cv (T4 − T1 )
 (T3 C p (T3 − T2 )

• Let compression ratio, = rc = v1
v2

• Cut off ratio =  = v3
v2

• Now during adiabatic compression 1 - 2,

  −1
 T2  v1 
=  v2  = r  −1
T1

• During constant pressure 2 – 3,

T3 = v3 = 
T2 v2

• During adiabatic expansion 3 -4,

T3  v4  −1  r  −1  v4 v1 v1  v2 r
T4  v3      v3 v3 v2 v3
=   = = = =  


• By inserting values of T2, T3 and T4 into equation:

• Diesel efficiency  = 1− 1   −1
  
 (r ) −1  −1 

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Example 5.1

An engine operates on an air-standard Otto cycle. The pressure and temperature at the
beginning of the compression stroke are 110 kPa and 40°C, respectively. The temperature at
the end of the compression stroke is 500°C. If the heat added during the heat addition process
is 1500 kJ/kg, calculate the maximum pressure that the walls must withstand. Assume
temperature independent properties for air.
Solution:

Solving for P2:

Solving for P3:

Example 5.2

An engine operates on the air-standard diesel cycle with a compression ratio of 18. The air
pressure and temperature at the beginning of the compression are 0.12 MPa and 43°C. If the
maximum temperature is 1992 K and the heat addition is 1274 kJ/kg, find the maximum
pressure for which the cylinder must be designed. Also find T2.
Solution:

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or

Example 5.3

The combustion process in an air-standard diesel cycle adds 800 kJ/kg to the medium. The
minimum temperature and pressure are 20°C and 0.1 MPa while the maximum temperature
is 1000°C. Find the thermal efficiency.
Solution:

PSP eBook | Power Plant Engineering 3 99

Tutorial Air Standard Cycle

Question 1
What is the highest thermal efficiency possible for a heat engine operating between 800ºC
and 15ºC?

Question 2:
In the Carnot cycle operating between 307ºC and 17ºC the maximum and minimum pressure
are 62.4 bar and 1.04 bar. Calculate the thermal efficiency and the work ratio. Assume air to
be the working fluid.

Question 3
In an air standard Otto cycle the maximum and minimum temperatures are 1400ºC and 15ºC.
The heat supplied per kg of air is 800 kJ. Calculate the compression ratio and the thermal
efficiency. Also calculate the ratio of maximum to minimum pressures in the cycle.

Question 4
A four―cylinder petrol engine has a swept volume of 2000 cm³ and the clearance volume in
each cylinder is 60 cm³. Calculate the air standard cycle thermal efficiency. If the induction
conditions are 1 bar and 24ºC, and the maximum cycle temperature is 1400ºC, calculate the
mean effective pressure based on the air standard cycle.

Question 5
Calculate the thermal efficiency and the mean effective pressure of an air standard diesel
cycle with a compression ratio of 15/1, and maximum and minimum cycle temperatures of
1650ºC and 15ºC respectively. The maximum cycle pressure is 45 bar.

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Question 6
An air standard cycle consists of the following processes:
Isentropic compression from 15ºC, 1.01 bar through a compression ratio of 5/1
Heat addition at constant volume of 2600 kJkg
Isentropic expansion to the initial volume
Heat rejection at constant volume
Sketch the cycle on p―v and T―s diagrams, and calculate its ideal efficiency, mean effective
pressure, and peak pressure.

Question 7
An ideal Otto cycle has a compression ratio of 9. At the beginning of compression, air is at 100
kPa, 27ºC. The pressure is doubled during the constant-volume heat addition process.
Determine:
i. The thermal efficiency
ii. The network output
iii. The MEP.
What-if scenario: How would the answers change if the compression ratio were increased to
10?

Question 8
An ideal Otto cycle has a compression ratio of 8. At the beginning of compression process, air
is at 95 kPa and 25ºC and 750 kJkg of heat is transferred to air during the constant―volume
heat―addition process. Considering the variation of specific heats with temperature,
determine:
i. The pressure and temperature at the end of the heat―addition process
ii. The Network output
iii. The thermal efficiency
iv. The mean effective pressure for the cycle

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Question 9
An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The minimum
and maximum temperatures in the cycle are 300 K and 1340 K. Accounting for the variation
of specifics heat with temperature, determine:
i. The amount heat transferred to the air during the heat―addition process
ii. The thermal efficiency
iii. The thermal efficiency of a Carnot cycle operating between the same temperature

limits.

Question 10
The compression ratio of an air―standard Otto cycle is 9.5. Prior to the isentropic
compression process, the air is at 100 kPa, 35ºC, and 600 cm³. The temperature at the end of
isentropic expansion process is 800 K. Using the specific heat values at room temperature,
determine:
i. The highest pressure and temperature in the cycle
ii. The amount of heat transferred in ( in kJ)
iii. The thermal efficiency
iv. The mean effective pressure

Question 11
An air―standard Diesel cycle has a compression ratio of 16 and cutoff ratio of 2. At the
beginning of the compression process, air is at 95 kPa and 27ºC. Accounting for the variation
of specific heats with temperature, determine:
i. The temperature after the heat―addition process
ii. The thermal efficiency
iii. The mean effective pressure for the cycle

CHAPTER 6102 PSP eBook | Power Plant Engineering 3

HEAT TRANSFER

Objectives:
At the end of this chapter, students should be able to:

• Define the mode of Heat transfer
• Perform calculation related to Fourier’s Law of Conduction and Newton's Law

of Cooling

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6.0 Introduction to Heat

Heat may transfer across the boundaries of a system, either to or from the system. It occurs
only when there is a temperature difference between the system and surroundings. Heat
transfer changes the internal energy of the system. Heat is transferred by conduction,
convection and radiation, which may occur separately or in combination.

6.1 Fourier's Law of Conduction

Fourier's law is an empirical law based on observation. It states that the rate of heat flow,
dQ dt , through a homogenous solid is directly proportional to the area, A, of the section at
right angles to the direction of heat flow, and to the temperature difference along the path
of heat flow, dT dx
Consider the transfer of heat through a slab of material.

Figure 6.1 Rate of heat flow

104 PSP eBook | Power Plant Engineering 3

Q  A dt
dx

Q = kA dt
dx

Q.dx = kA.dt

 x Q.dx = t2 kA.dt

0 t1

Qx = −A t2 k.dt
t1

Qx = −Ak t2 .dt
t1

Q = − kA (t2 − t1 )
x

Q = kA (t1 − t2 )
x

Example 6.1

The inner surface of a plane brick wall is at 40C and outer surface is at 20C. Calculate the
rate of heat transfer per m² of surface area of the wall, which is 250 mm thick. The thermal
conductivity of the brick is 0.52 W m.K .

Solution

Q = kA (t1 − t2 )
x

Q =q= k (t1 − t2 )
A x

q = 0.52 (40 − 20) W
250 10-3 m2

q = 2.08  20 W
m2

q = 41.6 W
m2

PSP eBook | Power Plant Engineering 3 105

6.2 Newton's Law of Cooling

At contact surfaces between a fluid and a solid wall, there is always a thin layer of fluid
through which the heat is transferred by conduction. Whenever there is an appreciable
movement of the fluid, conduction heat transfer in fluid may be neglected compared with
convection heat transfer. The heat transfer from the solid surface to the fluid can be described
by Newton's law of cooling. It states that the heat transfer, from a solid surface of area A, at
a temperature Tw, to a fluid of temperature T, is: dQ dT

Q = hA(tw − t)

Figure 6.2 Newton's Law of Cooling
Where:  is the heat transfer coefficient.

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The heat transfer in these films is by conduction only by considering unit surface area.

From fluid A to the wall,

q = kA (tA − t1)
δA

From the wall to fluid B,

q = kB (t2 − tB )
δB

Also

From fluid A to the wall,

q = hA (tA − t1)

From the wall to fluid B,

q = hB (t2 − )tB1

In general, h = k , where  is the thickness of the stagnant film of fluid on the surface.
δ

q = hA (tA − t1) = k (t1 − t2 ) = hB (t2 − tB )
x

q = (tA − t1) : qx = (t1 − t2 ) : q = (t2 − tB )
hA k hB

(tA − t1) + (t1 − t2 ) + (t2 − tB ) = q + qx + q
hA k hB

(tA − tB ) = q 1 + x + 1 
hA k hB

q =  (tA − tB ) 
x
1 k 1
hA + + hB

q = U(tA − tB )

Q = UA(tA − tB )

Where : 1 = 1 + x + 1
U hA k hB

PSP eBook | Power Plant Engineering 3 107

Example 6.2

The mild steel tank of wall thickness 10 mm contains water at 90C. Calculate the rate of heat

loss per m² of tank surface area when the atmospheric temperature is 15C. The thermal
conductivity of mild steel is 50 W m.K , and the heat transfer coefficients for the inside and

outside of the tank are 2800 W and 11 W , respectively. Also calculate the
m2.K m2.K

temperature of the outside surface of the tank

Solution:

 Resistance of water film, R1 = 1
h1A

R1 = 1 1 K W = 3.57110-4 K W
2800


 Resistance of insulating firebrick, R2 = x2
k2A

R2 = x2
k2A

R2 = 10 10-3 KW = 2.0 10−4 KW
50 1

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 Resistance of air gap (fluid film), R2 = 1
hAA

R3 = 1 KW = 0.091 KW
111

RT = 1 = 0.000357 + 0.0002 + 0.0909 K W
U

RT = 1 = 0.0915 K W
U

q= Q = 90 −15 W
A 0.0915 m2

q = Q = 820 W m2
A

Rate of heat loss per m² of surface area = 0.82 kW.

6.3 The composite wall and the electrical analogy

Figure 6.3 The Composite Wall and The Electrical Analogy

PSP eBook | Power Plant Engineering 3 109

Compare Ohm’s law to the thermal resistance analogous.

V = IR or I= V
R

Thermal resistance, R = x
kA

Where Q is analogous to I, and (tA−tB) is analogous to V

Compare Ohm’s law with equation Q = hA(tw − t)

Thermal resistance of a fluid film, R = x
hA

Therefore.

RA = 1, R1 = x1 , R2 = x2 , etc; Rn = xn and RB = 1
hA A k1A k2A knA hBA

RT = RA + R1 + R2 + ....... + Rn + RB

RT = 1 + x1 + x2 + ....... + xn + 1
hA A k1A k2A knA hBA

R T= 1 + x1
hAA kA + hBA

For any number of layers of material,

R T= 1 + x1
hAA kA + hBA

Use the electrical analogy for overall heat transfer we have,

Q = tA − tB
RT

Using equation overall heat transfer coefficient, for any number of layers of walls we have,

1 1 x1
k + hB
U = hA +

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Example 6.3

A furnace wall consists of 125 mm wide refractory brick and 125 wide insulating firebricks
separated ay an air gap. The outside wall is covered with a 12 mm thickness of plaster. The
inner surface of the wall is at 1100C and the room temperature is 25C. Calculate the rate
at which heat is lost per m² of wall surface. The heat transfer coefficient from the outside
wall surface to the air in the room is 17 Wm2.K , and the resistance to the heat flow of the
air gap is 0.16 K W . The thermal conductivity of refractory brick, insulating firebrick, and
plaster are 1.6, 0.3, and 0.14 W m.K , respectively. Also calculate each interface
temperature, and the temperature of the outside surface of the wall.

Solution

Consider 1 m² of surface area, i.e A = 1 m²

PSP eBook | Power Plant Engineering 3 111

Q = U(tA − tB ) or Q = 1 (tA − tB )
RT

By using equation to find: RT = 1 +  x + 1
hAA kA hB A

 Resistance of refractory brick, R1 = x1
k1A

R1 = 125 10-3 K W = 0.0781K W
1.6

 Resistance of air gap (fluid film), R2 = 1
hAA

R2 = 0.16 K W

 Resistance of insulating firebrick, R3 = x3
k3A

R3 = 125 10-3 K W = 0.417 K W
0.3

 Resistance of plaster, R4 = x4
k4A

R4 = 12 10-3 K W = 0.0857 K W
0.14

 Resistance of air film on outside surface, R5 = 1
hAA

R5 = 1 K W = 0.0588 K W
17

RT = 0.0781 + 0.16 + 0.417 + 0.0857 + 0.0588 K W

RT = 0.8 K W

Rate of heat loss per m² of surface area,

Q = U(t A − tB ) or Q = tA − tB
R

Q = 1100 − 25 W
0.8

Q = 1344 W @ 1.344 kW

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To find t1, t2, t3 and t4:

Q = 1100 − t1
R1

1344 = 1100 − t1
0.0781

t1 = 1100 − (1344)(0.0781)C

t1 = 995 C

Q = t1 − t2
R2

1344 = 995 − t2
0.16

t2 = 995 − (1344)(0.16)C

t2 = 780 C

Q = t2 − t3
R3

1344 = 780 − t3
0.417

t3 = 780 − (1344)(0.417)C

t3 = 220 C

To find t4, two methods can be used, Q = t3 − t4 or Q = t4 − 25
R4 R5

Q = t3 − t4 Q = t4 − 25
R4 R5

1344 = 104 − t4 1344 = t4 − 25
0.0857 0.0588

t3 = 780 − (1344)(0.0857)C t3 = 25 + (1344)(0.0588)C

t4 = 104 C t4 = 104 C

PSP eBook | Power Plant Engineering 3 113

6.4 Heat flow through a cylinder

Q = −kA dt = −k(2π 1) dt

dx dr

Q dr = −2πk.dt
r

 r2 Q dr = t2 − 2πk.dt
rr1 t1

 Q r2 dr = −2πk t2 dt
rr1 t1

   Q r2 t2
ln r = 2k t t1
r1

Q(ln r2 − ln r1) = −2k(t2 − t1)

Q ln r2 = 2πk (t2 − t1)
r1

Q = 2πk(t2 − t1)

ln r2
r1

Figure 6.4 Heat flow through a cylinder

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From equation, Q = kA (t1 − t2 ), substitute a mean area, Am and x = r2- r1
x

Q = kAm (t1 − t2 )
r2 − r1

Q = 2πk(t1 − t2 )

ln r2
r1

kAm (t1 − t2 ) = 2πk(t1 − t2 )
r2 − r1
ln r2
r1

k 2πrm (t1 − t2 ) = 2πk(t1 − t2 )
r2 − r1
ln r2
r1

rm = r2 − r1
ln r2
r1

In the case of a composite cylinder (e.g. a metal pipe with several layers of lagging),

R = x and Am = 2π(r2 − r1)
kA m
ln r2
r1

Where x is the thickness of the layer, and Am is the logarithmic mean area for the layer.

The resistance of the fluid film on the inside and outside surface,

Ro = 1 and RI = 1
ho A o hiAi

Where:
ho = heat transfer coefficient for the outside surface.
Ao = the outside surface area, 2ro
hi = heat transfer coefficient for the inside surface.
Ai = the outside surface area, 2rI

PSP eBook | Power Plant Engineering 3 115

From equation,

Q = 2πk(t1 − t2 )

ln r2
r1

• The heat transfer rate depends on the ratio of the radii, r2 r1 not to the difference,
r2 − r1.

• The smaller the ratio r2 r1 then the higher is the heat flow for the same temperature
difference.

• In many practical problems the ratio r2 r1 , tends towards unity since the wall thickness
or lagging thickness is usually small compared with mean radius.

Therefore.

Arithmetic mean radius = r2 + r1
2

Example 6.4

Water at 80oC flow through at 50 mm bore steel pipe of 6 mm thickness and the atmospheric

temperature is 15oC. The internal thermo conductivity of steel is 48 W m.K and the inside

and outside heat transfer coefficient are 2800 and 17 W respectively. Neglecting
m2.K

radiation, calculate the rate of heat flow per unit length of pipe.

Solution

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Q = U(tA − tB ) or Q = 1 (tA − tB )
RT

 Resistance of hot water film, R1 = 1
h1A

R1 = 1 r L K W
h12π 

R1 = 1 K W
2800  2π  25 10−3 1

R1 = 2.274 10-3 K W

 Resistance of steel, R = x
kA m

Am = 2π(31− 25)10−3 1 m2

ln 31
25

Am = 0.0377 m2
0.2151

A m = 0.1753 m2

R2 = 6 10-3 KW
48  0.1753

R 2 = 0.0505 K W

 Resistance of air gap (fluid film), R3 = 1
h3 A

R3 = 1 K W
h12π  r  L

R3 = 1 1K W
17  2π  3110−3

R3 = 0.302 K W

RT = 2.274x10-3 + 0.0505 + 0.302 K W

RT = 0.3527 K W

Q = tA − tB
RT

Q = 80 −15 Wm
0.3527

Q = 184.29 W m @ 0.18429 kW m

PSP eBook | Power Plant Engineering 3 117

Example 6.5

A steel pipe of 100 mm bore and 7 mm wall thickness, carrying steam at 260oC, is insulated
with 40 mm of a moulded high-temperature diatomaceous earth covering. This covering is in
turn insulated with 60 mm of asbestos felt. If the atmospheric temperature is 15oC calculate
the rate at which heat is lost by the steam per meter length of pipe. The heat transfer
coefficients for the inside and outside surfaces are 550 and 15 Wm2.K , respectively and the
thermal conductivities of the steel, diatomaceous earth and the asbestos felt are 50, 0.09 and
0.07 W m.K respectively. Also calculate the temperature of the outside surface.

Solution

Asbestos felt 157 mm
97 mm

Diatomaceous TSTEAM 57 mm
earth 260ºC
50 mm
Steel pipe TATM 15ºC

118 PSP eBook | Power Plant Engineering 3

Q = U(tA − tB ) or Q = 1 (tA − tB )
RT

 Resistance of hot water film, R1 = 1
h1A

R1 = 1 L K W
h12π  r

R1 = 550  2π 1 K W
 50 10−3 1

R1 = 5.79 10-4 K W

 Resistance of steel, R2 = x2
k2Am

Am = 2π(57 − 50)10−3 1 m2

ln 57
50

Am = 0.0440 m2
0.1310

A m = 0.3358 m2

R2 = 7 10-3 KW
50  0.3358

R 2 = 4.169 10-4 K W

 Resistance of diatomaceous earth, R3 = x3
k3Am

Am = 2π(97 − 57)10−3 1 m2

ln 97
57

Am = 0.2513 m2
0.5317

A m = 0.4726 m2

R3 = 40 10-3 KW
0.09  0.4726

R3 = 0.9404 K W

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 Resistance of asbestos felt, R4 = x4
k4Am

Am = 2π(157 − 97)10−3 1 m2

ln 157
97

Am = 0.3770 m2
0.4815

A m = 0.7829 m2

R4 = 60 10-3 K W
0.07  0.7829

R 4 = 1.095 K W

 Resistance of air film on outside surface, R5 = 1
h5 A

R5 = h5 1
 2π r L

R5 = 1 K W
15  2π 157 10-3 1

R5 = 0.0675 K W

RT = 5.79x10-4 + 4.169x10-4 +0.9404 +1.095+0.0675 K W

RT = 2.1987 K W

Q = tA − tB
RT

Q = 260 −15 W m
2.1087

Q = 116 W m @ 0.116 kW m

Rate of heat loss per meter length of pipe = 116 W

Temperature of outside surface,

Q = t SURFACE − tB
R5

116 = tSURFACE − 15
0.0675

tSURFACE = 116  0.0675 + 15 C

tSURFACE = 22.8 C

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Summary:

Q = kA (t1 − t2 ) Q = q = k (t1 − t2 )
x A x

R T= 1 + x1 1 = RTA or U = 1
hAA kA + hBA U RTA

R = x (t1 − t2 ) Am = 2 (r2 − r1)
kAm r2
ln r1

Q = U(tA − tB ) or Q 1 (t A − tB )
= RT

Table 6.1 Thermal Conductivities

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Figure 6.5 Temperature Distribution

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6.5 Heat Exchangers

• One of the most important processes in engineering is the heat exchanger between
flowing fluids.

• In heat exchangers the temperature of each fluid changes as it passes through the
exchanger, and hence the temperature of the dividing wall between the fluids also
changes along the length of the exchanger.

• Example in practice in which flowing fluids exchange heat are:
I. air intercoolers, condenser, and boiler in steam plant
II. condenser and evaporators in refrigeration units

• There are three main types of heat exchangers

6.5.1 Recuperative

The most important type in which the flowing fluids exchanging heat is on either side of a
dividing wall.

6.5.2 Regenerator

This type of heat exchanger is which the hot and cold fluids pass alternately through a space
containing a matrix of material that provides alternately a sink and a source for heat flow.

6.5.3 Evaporative

The third type is the evaporative type in which a liquid is cooled evaporative and continuously
in the same space as the coolant.

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6.6 The Basic Types of Heat Exchangers
6.6.1 Parallel Flow Heat Exchangers

The fluid being heated, and the fluid being cooled flow in the same parallel direction as shown.

Figure 6.6 Heat Exchanger fluid flow
The heat is converted to the wall of the tube, then conducted to the other side and then
converted to the other fluid.
If the tube wall is thin, the heat transfer rate is

A is the surface are of the tubes.
The diagram shows typically how the temperature of the two fluids varies with the path
length. Fluid ‘B’ gets hotter and fluid ‘A’ gets cooler, so the temperature difference is greatest
at inlet.

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6.7 Counter Flow

The fluid flow in opposite but parallel direction as shown.

Figure 6.7 Heat Exchanger fluid flow
The temperature of the fluid being heated can be raised to near the inlet temperature of the
fluid being cooled.
This is important for exchangers that heat up air for combustion such as the exhaust gas heat
exchangers on gas turbines.
For a given heat transfer the surface area is less. The heat transfer formula is the same.

PSP eBook | Power Plant Engineering 3 125

6.8 Cross Flow

The fluids flow at right angles to each other as shown.

Figure 6.8 Heat Exchanger fluid flow

This design is often the result of the plant layout and is common in industrial boilers where
the hot gasses flow in a path normal to the water and steam (Super heaters, recuperates
and economizers).
The tubes are often finned to improve the efficiency.
Sooty deposit on the outside is more likely to be dislodge in this configuration. Large steam
condensers also use this style.

6.9 Parallel Flow and Counter Flow Recuperators

Consider the simple case of a fluid flowing through a pipe and exchanging heat with a second
fluid flowing through an annulus surrounding the pipe.
When the fluid flow in the same direction along the pipe the system is known as parallel-flow,
and when the fluids flow in opposite direction to each other the system is known as counter-
flow.
Let the mean inlet and outlet temperature of fluid A be TA1 and TA2
Let the mean outlet and outlet temperature of fluid B be TB1 and TB2
Let the mass flow rates of the fluid A and fluid B be mA and mB

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Let the specific heats of fluid A and B be cA and cB

Consider any section X-X where fluid A is at TA and fluid B is at TB.
The temperature difference at this section is (TA -TB)=T, and a small amount of heat, dQ, is
transferred across an element of length dl

Using equation. . = UA(t A − tB ), we have,

Q

.

d Q = DdlUt --------------------- (1)

Where D is the mean diameter of the tube

Fluid A increases in temperature by dtA along element dl, and fluid B increases in
temperature by dtB along element dl.
In the case of parallel flow. Temperature TA decreases with the length, while temperature TB
increase with the length. The heat given up by fluid A must equal the heat received by fluid

B for parallel flow

.. .

d Q = − mAcA dtA = mBcB dtB

.. .

d Q = − mAcA dtA = mBcB dtB

Where: = Heat transferred (W or kW)
dQ =
mA = mass flow rate of fluid A (kg/s)
mB =
CA = mass flow rate of fluid B (kg/s)
CB
specific heat of fluid A(J/kgK)

specific heat of fluid B(J/kgK)

Integrating equation (1) above.

. = DUl(1 − 2 )

Q In 1  .................(2)
2

PSP eBook | Power Plant Engineering 3 127

Substituting equation (2) above,

.

Dl = A

(1 − 2 ) = LMTD
In 1 
2

Therefore, equation (2) can be written as:

.

Q = U.A.LMTD...................................(3)

Q = Heat transferred (W or kW)
D =
U = Mean diameter of the tube (m)
l =
A = Overall heat transfer coefficient (W/m2K)
LMTD =
Length of tube (m)

Mean surface area of the tube (m2)

Logarithmic mean temperature difference (K or C)

Tutorial Heat Transfer

Question 1
As its name implies, heat transfer by conduction occurs when heat is conducted through a
material. Let’s consider the simple conductive heat transfer problem shown in Figure a.

Figure a Heat Flow Through A 3-Centimeter Thick Copper Plate

128 PSP eBook | Power Plant Engineering 3

Question 2
A steel tank with 10 mm of thickness contains water at 900C. Calculate the heat losses rate
per m2 of tank surface area when the atmosphere temperature is 150C. The steel
conductivity is 50 W/m K and the conductivity for the inner and outer tank is 2800 and 11
W/m2K respectively. Calculate the outer surface temperature for the tank.

Question 3
The inner surface of the bricks wall temperature is 400C and the outer surface
temperature is 200C. Calculate the heat transfer rate per surface area, m2, with the
thickness of 250 mm. The bricks conductivity is 0.52 W/m. K.

Question 4
Water at 80oC flow through at 50mm bore steel pipe of 6mm thickness and the
atmospheric temperature is 15oC. The internal thermos conductivity of steel is 48 W/m K
and the inside and outside heat transfer coefficient are 2800 and 17 W/m2K respectively.
Neglecting radiation, calculate the rate of heat per unit length of pipe.

Question 5
Exhaust gases from a boiler at a temperature of 425°C is to be used to preheat the feed
water via a heat exchanger. The gas exits the heat exchanger at 120°C. The feed water
enters the heat exchanger at 30°C. The overall heat transfer coefficient for the heat
exchanger is 150 W/m²K. The specific heats of exhaust gases and water may be taken as
1.13 kJ/kg K and 4.19 kJ/kg K respectively. The gas flowrate is 0.3 kg/s and for the water
is 0.6 kg/s. Calculate the surface area required for:

i. Parallel flow heat exchanger
ii. Counter flow heat exchanger

PSP eBook | Power Plant Engineering 3 129

Question 6
An exhaust pipe is 75 mm diameter, and it is cooled by surrounding it with a water jacket.
The exhaust gas enters at 350C and the water enters at 10C. The surface heat transfer
coefficients for the gas and water are 300 and 1500 W/m2K respectively. The wall is thin,
so the temperature drop due to conduction is negligible. The gasses have a mean specific
heat capacity Cp of 1130 J/kgK and they must be cooled to 100C. The specific heat
capacity of the water is 4190 J/kg K. The flow rate of the gas and water is 200 and 1400
kg/h respectively. Calculate the required length of pipe:

i. Parallel flow
ii. Cross flow

130 PSP eBook | Power Plant Engineering 3

Summary

POWER PLANT ENGINEERING 3 provides knowledge and understanding of theory, concepts,
and application of the power generation in steam power plant, gas turbine power plant, air
standard cycle, diesel power plant, refrigeration system and heat transfer. Emphasis of the
course is to determine cycle efficiency and the effect of additional equipment to the power
generation related to thermodynamic processes. Steam Power Plant covers the process in
Rankine cycle, reheat cycle and regenerative cycle. Students are required to draw T-s
diagram and calculate the heat and work, cycle efficiency and specific steam consumption.
Explanation will be done to determine the boiler efficiency, boiler equivalent evaporation
and the factors that effect on boiler efficiency. Gas Turbine provides understanding the
operation of gas turbine power plant, method to increase efficiency and process on Brayton
cycle with intercooling, reheating regeneration and also combine cycle power plant.
Students are required to draw T-s diagram, P-v diagram and calculate the heat supplied,
gross work, network output, isentropic efficiency, work ratio and thermal efficiency for each
cycle. Air Standard Cycle provides understanding of processes involved in the internal
combustion engine and the heat engine. It requires drawing of Otto and Diesel cycle and
calculation on heat efficiency mean effective pressure, compression ratio for the respective
cycle. Diesel power plant describes application of diesel power plant and processes involved
in the internal combustion engine. This topic also explain the various parameter related to
the performance of internal combustion engine such as indicated power, indicated mean
effective pressure, brake power, friction power, engine torque, specific fuel consumption,
mechanical efficiency, thermal efficiency, compression ratio and volumetric efficiency.
Students are required to calculate the respective parameters and to complete the heat
balance sheet. This topic explains operation of refrigeration system. Refrigeration includes
the operation of refrigerated vapour compression cycle which requires students to calculate
the coefficient of performance, refrigerating effect, heat loss, mass flow rate, refrigerant
temperature, input power and work done. The student needs to draw T-s and p-h diagram,
and Finally Heat Transfer states the concepts of heat transfer between bodies. Three types
of heat transfer according to the Fourier’s law of conduction and Newton’s law of cooling.
The student needs to draw block diagram, elaborate on the general equations and calculate
heat transfer as well as wall temperature.



viii PSP eBook | Power Plant Engineering 3

References

Thomas C. E, Kao Chen and Swanekamp, Robert C. (2005).Standard Handbook Of
Power Plant Engineering. 2nd Edition. Tata McGraw-Hill. New Delhi. ISBN:
0070194351

Rajput, R.K (2010). Thermal Engineering in S.I Units. 8th Edition. Laxmi Publication,
New Delhi

Eastop,T.D and McConkey. (2009) Applied Thermodynamics for Engineering
Technologists. 5th Edition. Pearson Education Ltd. New Delhi.

Nag P.K. (2005). Power Plant Engineering.2nd Edition. Tata McGraw-Hill. New Delhi
Rajput, R.K. (2006). A Textbook of Power System Engineering.1st Edition.
Laxmi Publication, New Delhi.

Roger G.F.C and Y.R. Mayhew (2005). Engineering Thermodynamics, Work and Heat
Transfer. S.I Units, 3rd Edition, Longman

Sharma,P,C (2018). A Text book of Power Plant Engineering. S.K Kataria & Sons.
ISBN-13: 978-93-5014-384-1

Yunus A. Cengel, Micheal A. Boles. (2015). Thermodynamics: An Engineering
Approach, 8th Edition. McGraw-Hill Education. New York. ISBN-13: 978-
0073398174

Richardson, Duncan C (2014). Plant Equipment and Maintenance Engineering
Handbook. McGraw Hill Professional. ISBN 9780071809900

Afzairizal, M (2018). Pengenalan kepada Teknologi Kejuruteraan Loji. Pelangi
Engineering Series. Pelangi Professional Publishing Sdn Bhd Selangor.

PSP eBook | Power Plant Engineering 3 ix

Nag P.K (2017). Power Plant Engineering. 4th Kindle Edition.
Kumar, D.S (2017), Refrigeration and Air Conditioning. S.K Kataria & Sons.

ISBN:978-93-5014-634-7

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