UNIT I GAS POWER CYCLES
1. Define: Thermodynamic cycles.
Ans: Thermodynamic cycle is defined as the series of processes performed on the system, so
that the system attains to its original state.
2. Define the term compression ratio.
Ans: Compression ratio is the ratio between total cylinder volume to clearance volume. It is
denoted by the letter ‘r’
3. What is the range of compression ratio for SI and diesel engine?
Ans: For petrol of SI engine 6 to 8 : For diesel engine 12 to 18.
4. Write the expression for efficiency of the otto cycle?
Ans:
1
Efficiency n =1 - ---------
(r)r-1
5. What is meant by cutoff ratio?
Ans: Cutoff ratio is defined as the ratio of volume after the heat addition to before the heat
addition. It is denoted by the letter ‘p’
6. What are the assumptions made for air standard cycle.
Ans:
1. Air is the working substance.
2. Throughout the cycle, air behaves as a perfect gas and obeys all the gas laws.
3. No chemical reaction takes place in the cylinder
4. Both expansion and compression are strictly isentropic
5. The values of specific heats of the air remain constant throughout the cycle.
7. What is the difference between otto and Diesel cycle.
Otto Cycle Diesel Cycle
1. Otto cycle consist of two adiabatic and
two constant volume process.
1. It consists of two adiabatic, one constant
volume and one constant pressure
processes.
2. Compression ratio is equal to expansion
ratio
2. Compression ratio is greater than
expansion ratio.
3. Heat addition takes place at constant
volume.
3. Heat addition takes place at constant
pressure
4. Compression ratio is less. It is varies
from 6 to 8.
4. Compression ratio is more. It varies
from 12 to 18.
8. Define: Mean effective pressure of an I.C. engine.
Ans: Mean effective pressure is defined as the constant pressure acting on the piston during the
working stroke. It is also defined as the ratio of work done to the stroke volume or piston
displacement volume.
9. Define: Specific fuel consumption.
Ans: SFC is defined as the amount of fuel consumed per brake power hour of work.
10. What is meant by calorific value of a fuel.
Ans: Calorific value of a fuel is defined as the amount of heat liberated by the compete
combustion of unit quantity of a fuel.
11. Give the expression for efficiency of the Dual cycle.
Ans:
1 Kpr - 1
Efficiency n = 1 - ------- -----------------
(r)r-1 (K-1) + yK(p-1)
where,
r – Compression ratio
k – pressure or Expassion ratio
p – cut off ratio and
y – adiabatic index
11. What are the factors influencing of the Dual cycle?
Ans: 1. Compression ratio 2.cut off ratio 3. pressure ratio and 4. heat supplied at constant
volume and constant pressure.
12. Give the expression for efficiency of the Brayton cycle.
Ans:
1
Efficiency n = 1 - ---------- where Rp – pressure ratio.
(Rp)y-1
13. What is a Gas turbine? How do you classify.
Ans: Gas turbine is an axial flow rotary turbine in which working medium is gas.
Classification of gas turbine.
1. According to the cycle of operation
a) open cycle b) closed cycle and c) semi – closed cycle.
2. According to the process
a) constant volume and b) constant pressure process.
14. What is meant by open cycle gas turbine?
Ans: In open cycle gas turbine, the exhaust gas form turbine is exhausted to the atmosphere
and fresh air is taken in compressor for every cycle.
15. Differentiate open and closed cycle gas turbines.
Open cycle gas turbine Closed cycle gas turbine
1. Working substance is exhausted to the
atmosphere after one cycle.
1. The same working substance is
recirculated again and again.
2. Pre-cooler is not required 2. Pre-cooler is required to cool the exhaust
gas to the original temperature.
3. High quality fuels are used 3. Low quality fuels are used
4. For the same power developed size and
weight of the plant is small
4. Size and weight are bigger.
16. What is the function of intercooler in gas turbines? Where it is placed?
Ans: The intercooler is placed between L.P. and H.P. compressors. It is used to cool the gas
coming form L.P. compressor to its original temperature.
17. Why re-heater is necessary in gas turbine? What are its effects?
Ans: The expansion process is very often performed in two sperate turbine stages. The re-
heater is placed between the H.P. and L.P. turbines to increase the enthalpy of the exhaust gas
coming from H.P. turbine.
Effects:
1. Turbine output is increased for the same compression ratio
2. Thermal efficiency is less.
18. What is the function of regenerator in gas turbine?
Ans: The main function of heat regenerator is to exchange the heat from exhaust gas to the
compressed air for preheating before combustion chamber. It increases fuel economy and
increase thermal efficiency.
1. Derive the air standard efficiency of Otto cycle with neat sketch
2. Problem
3.Derive the air standard efficiency of diesel cycle with neat sketch.
4. PROBLEM
5 PROBLEM
6. PROBLEM
7 PROBLEM
8 PROBLEM
UNIT II REFRIGERATION CYCLES AND RECIPROCATING CYCLES 1. What is meant by single acting compressor? In single acting compressor, the suction, compression and delivery of air take place on one side of the piston. 2. What is meant by double acting compressor? In double acting reciprocating compressor, the suction compressin and delivery of air take place on both side of the piston. 3. What is meant by single stage compressor? In single stage compressor, the compression of air from the initial pressure to the final pressure is carried out in one cylinder only. 4. Define clearance ratio Clearance ratio is defined as the ratio of clearance volume to swept volume (or) stroke volume. Vc Vc – clearance volume C = -------- Vs – swept volume Vs 5. What is compression ratio? Compression ratio is defined as the ratio between total volume and clearance volume. Total volume Compression ratio = ------------------- Clearance Volume 6. What are the factors that effect the volumetric efficiency of a reciprocating compressor? 1) Clearance volume 2) Compression ratio. 7. What is the difference between complete (or) perfect inter cooling and incomplelte (or) imperfect inter cooling. Perfect Inter cooling When the temperature of air leaving the intercooler (T3) is equal to the original atmospheric air temperature (T1), then the inter cooling is known as perfect inter cooling. Imperfect Inter cooling When the temperature of air leaving the inter cooler (T3) is more than original atmospheric air temperature (T1), then the inter cooling is known as Imperfect inter cooling. 8. Power requirement of a refrigerator is _________ Ans: Inversely proportional to cop 9. In SI Units, one ton of refrigeration is equal to __________ Ans: 210KJ/min 10. The capacity of a domestic refrigerator is in the range of __________ Ans: 0.1 to 0.3 tonnes.
11. COP of a refrigerator working on a reversed carnot code is _________ Ans: T2 ----------- T1 – T2 12. The vapour compression refrigerator employs the __________cycle Ans: Reversed carnot 13. In vapour compression cycle the condition of refrigerant is dry saturated vapour ________ Ans: Before entering the compressor. 14. Name four important properties of a good refrigerant Ans: 1. Low boiling point 2. High critical temperature & pressure 3. Low sp.heat of liquid 4. Non – flammable and non explosive. 15. Name some of the equipments used in air conditioning system Ans: 1. Filter 2. Cooling coil 3. Heating coil 4. Compressor 5. Condeser 6. Evaporator 16. Name any four commonly used refrigerants Ans; 1. Ammonia (NH3) 2. Carbon di oxide (CO2) 3. Sulphur di oxide (SO2) 4. Freon – 12. 17. What are the factors to be considered in air conditioning a room? Ans: 1. Temperature of air 2. Humidity of air 3. Purity of air 4. Motion of air.
1. PROBLEM
2. PROBLEM
3.PROBLEM
4.What are the difference between vapour compression and absorbtion system?
5.Explain with neat sketch vapour absorbtion
6 Problem
7 Problem
RECIPROCATING AIR COMPRESSOR 1. Explain with neat sketch the Operation of an Air Compressor
Operation:
To understand the operation of an air compressor, let us assume the cycle and indicator diagram for a
simple single stage reciprocating air compressor, as shown below. (Click the image to enlarge.)
The simple reciprocating air compressor has a piston which reciprocates inside the cylinder wall and cylinder head. The piston is attached to the crankshaft with the help of a connecting rod and thus the rotation of the crankshaft causes the piston to move up and down inside the cylinder. The crankshaft is mounted on the crank case. The cylinder head contains valve pockets where the suction and delivery valve are fixed.
These suction and delivery valves are of simple pressure differential types. They open and close, due to the pressure difference on either side of the valve plates.
1. When the compressor stops or idles for some time, it is always assumed that there is some residual compressed air left in the cylinder space. This residual air expands when the piston moves down. The pressure drops in the cylinder space at a particular point as the piston moves down, where the pressure inside the cylinder becomes lesser than the atmospheric pressure. Thus this difference in pressure makes the suction or inlet valve open.
2. This opening of inlet valve allows fresh air to be drawn inside the cylinder space as the piston still continues to move in the downward direction. The inlet valve will remain open till there is pressure difference between the atmosphere and inside of the cylinder space. As the pressure difference starts to reduce, the inlet valve starts slowly closing.
The inlet valve closes completely when there is no pressure difference and then the piston reaches bottom dead center (BDC), and it starts to travel in an upward direction. At this position, both the inlet and delivery valve remains closed. Thus as the piston moves up, the pressure starts to build inside the cylinder space.
3. The delivery valve starts to open when there is a pressure difference between the cylinder space and air receiver. Let us assume the air receiver is at a pressure of 7 bar. The delivery valve will not open until the pressure inside the cylinder space is slightly above 7 bar. As the piston moves in upward direction, the pressure increases and at some point the pressure grows beyond 7 bar making the delivery valve open. Thus the compressed air is delivered into the air receiver.
4. As the piston reaches top, the pressure starts to fall and the delivery valve starts to close. The residual compressed air remaining in the space again starts to expand as the piston moves down continuing the next cycle.
2. Explain with neat sketch theoretical P-V diagram of air compressor:
Referring to the diagram, the theoretical air compressor P-V diagram can be understood.
4-1:
The air compressor draws in air from the atmosphere. The atmospheric pressure is P1. The initial volume when the piston is at top is zero (Assuming there is no bumping clearance). Thus as the piston moves from top to bottom, a volume of air V1 is drawn into the compressor. The temperature of air is T1.
1-2: As the piston moves up, the air is compressed polytropically (PV^n= C). The pressure of air increases from P1 to P2. The volume decreases from V1 to V2. The temperature increases from T1 to T2.
2-3:
The compressed air at the pressure P2, volume V2 and temperature T2 is delivered out of the compressor to the air receiver.
3. Explain with neat sketch the Practical P-V diagram of air compressor:
In my last article, we have seen the P-V diagram and understood the operation of an air compressor. But in practice, the diagram is not so perfect. The figure shown here represents the actual practical P-V diagram of an air compressor. The points, 1234 represents the theoretical diagram. But there are some shaded portions above and below the work done area. It is necessary to give certain explanation for these additional areas which add up to the work done by the compressor.
Referring to the diagram, at point 4, when the clearance air has reduced to the atmospheric pressure, the inlet valve will not open immediately. The pressure drops lower than the atmospheric pressure and the inertia of the valves are overcome by the pressure difference. Thus the valve is forced open by the atmospheric air and it rushes into the cylinder chamber. There is a “valve bounce” and the pressure does not remain constant inside the cylinder. The pressure slightly increases and then decreases after which reaches somewhat steady intake of air. This negative pressure difference is called as the “Intake Depression”.
The same occurs at the point 2, where the delivery valve delays to open. The compressed air pressure inside the cylinder of compressor reaches a pressure slightly more than the air receiver pressure. The delivery valve then opens causing a slight decrease in pressure but always above the air receiver pressure enabling the compressor to deliver air. Then it reaches a stable point after some “Valve Bounce” and then reaches point 3 where the delivery valve closes to continue its cycle. Thus in practice, the compressor requires more power to compensate for the additional work done due to the depressions in delayed opening of the valves.
4. What are the Limitations of a Single-Stage Air Compressor?
Refer to the enclosed diagram, the single stage air-compressor is compressing from pressure P1 to Pressure P2, completing the cycle 1234, where 3-4 is the clearance air expansion. Also V1-V4 is the effective swept volume or the effective volume where the fresh air from atmosphere is sucked in. The mass of air flowing through the compressor is controlled by this effective swept volume V1-V4.
If any restriction is placed on the delivery of the air compressor, for example: the discharge valve throttled, then the delivery pressure of the air compressor increases. From the diagram, let us say the new delivery pressure is P5. Then the operating cycle will be 1567, where 6-7 is the clearance expansion of air and the effective swept volume is V1-V7. Thus it is evident that the effective swept volume (V1-V4) is more than (V1- V7). Thus when the delivery pressure of the single-stage air compressor is increased, the effective swept volume is reduced.
If the delivery pressure is further increased (assuming the compressor is so strong to work), the delivery pressure reaches P8, and the compression follows the curve 1-8, where there will be no delivery of compressed air. Thus when the delivery pressure of a single-stage compressor is increased, the mass flow rate also increases.
Since the delivery pressure increases, the associated temperature also increases. Thus the temperature of the air after compression is so high as to cause mechanical problems and the amount of heat is actually theenergy loss.
If a single-stage machine is required to deliver a high-pressure compressed air, then it requires
1. Heavy moving/working components to compress air to such a high pressure,
2. There might be some balancing problems due to heavy moving parts,
3. The power requirement for such heavy parts movement is too high,
4. There will high torque fluctuations,
5. To compensate for the torque fluctuations, a heavy flywheel is required.
6. Better cooling arrangements are required, and
7. Lubricating oil which does not get vaporized at such high temperatures.
A multi-stage compressor is one in which there are several cylinders of different diameters. The intake of air in the first stage gets compressed and then it is passed over a cooler to achieve a temperature very close to ambient air. This cooled air is passed to the intermediate stage where it is again getting compressed and heated. This air is again passed over a cooler to achieve a temperature as close to ambient as possible. Then this compressed air is passed to the final or the third stage of the air compressor where it is compressed to the required pressure and delivered to the air receiver after cooling sufficiently in an after-cooler.
Advantages of Multi-stage compression: 1. The work done in compressing the air is reduced, thus power can be saved
2. Prevents mechanical problems as the air temperature is controlled
3. The suction and delivery valves remain in cleaner condition as the temperature and vaporization of lubricating oil is less
4. The machine is smaller and better balanced 5. Effects from moisture can be handled better, by draining at each stage
6. Compression approaches near isothermal
7. Compression ratio at each stage is lower when compared to a single-stage
machine
8. Light moving parts usually made of aluminum, thus less cost and better maintenance
Refer to the diagram of a multi-stage compressor, where it is evident that the work done by the compressor is less when compared to a single-stage machine for same delivery pressure.
Practical Understanding of Multi-stage Compressor
We know that PV^n = C, where n is the polytropic compression index.
If we want to compress air from atmospheric pressure to a pressure of 30 bar, and say the ambient temperature is 27 degree Celsius:
The compression index n = 1.35 and the compression ratio for single-stage compressor would be 30:1.
Also we know that T1/T2 = (P1/P2) ^ ((n-1)/n).
Thus when calculated using the above expression, T2= 450 degree Celsius. Thus it is evident that the delivery temperature of compressed air is 450 degree Celsius.
Do you think of any harm to compressor will happen at this temperature? The main issue is the lubricating oil mist and associated explosion. The lubricating oil at this temperature will ignite and cause very severe problem. Thus if multi-staging is used, the delivery air temperature is controlled very close to the ambient air and there is no possibility of lubricating oil and associated problems.
Why Cylinder Diameter reduces as Pressure Increases Now let us understand a very practical issue.
Have you ever noticed in a multi-stage compressor, the diameter of cylinder liners of each stage is different? To be more elaborate, the first stage cylinder diameter is biggest and the intermediate stage is the second biggest and the third or final stage is the smallest of all. Do you know the reason for this?
As the pressure of compression increases, the cylinder diameter decreases. The mass flow rate or the amount of air passing through each stage is same. Thus as the pressure increases with the same mass flow rate, the volume occupied by air must reduce. This is accomplished by reducing the diameter and thus the volume of the cylinder.
UNIT- III CONDUCTION
1. What is thermal diffusivity?
It is the ration of thermal conductivity to the thermal capacity
2. Define overall heat transfer coefficient
The overall heat transfer U gives the heat transmitted per unit area per unit time per
degree temperature difference between the bulk fluids on each side to the metal or solid.
3. Define the Fourier Law of heat conduction
The rate of flow of heat in a simple homogeneous solid is directly proposional to the
area of the section at right angle to the direction of heat flow and change of temperature with
respect to the length of the path of the heat flow.
4. What are the mechanisms in heat transfer through solids?
Latice Vibration
Transport of Free Electrons
5. Define thermal conductivity of metals
Thermal conductivity is defined as the amount of energy conducted through a body of
unit area and unit thickness in unit time when the difference in unit temperature between the
faces causing the heat flow in unit temperature difference.
6. What is thermal contact resistance?
Due to the reduced area and presence of voids, a large resistance to heat flow occurs at
the interface. This resistance is known as thermal contact resistance.
7. What ate the purpose of thermal insulation?
Prevent the heat flow from the system to surroundings
Prevent the heat flow from the surroundings to system
8. Name few common types of fin
Uniform straight fin, Tapered straight fin, Annular fin, Pin fins
9. Define fin efficiency
The efficiency of fin is defined as the ratio of the actual hat transferred by the fin to the
maximum heat transferable by fin, if entire fin area were at a base temperature.
10. What are the conditions that make the fin effective?
Thermal conductivity should be large
Convective heat transfer coefficient should be small
Thickness of the fin should be small
11. Define thermal diffusivity
Thermal diffusivity = Thermal conductivity/Thermal capacity
What are the various mode of heat transfer?
Conduction, Convection, Radiation
12. What are the factors influence the thermal conductivity?
Material structure, Moisture content, Density, Pressure and Temperature, Homogeneous
mixture of components
13. Define critical thickness of insulation
The thickness up to which the heat flow increases and after which heat flow decreases is
termed as critical thickness.
14. Define thermal insulation
A material, which retards the flow of heat with reasonable effectiveness, is known as
insulation.
15. Name few applications of finned surfaces
Radiator of automobiles, Air-cooled engine cylinder heads, Condenser and evaporator in
refrigeration and AC systems, Electric motor bodies and Electronic chip heat sinks
16. What are the assumptions made in the analysis of heat flow through fins?
Steady state heat conduction
No heat generation within the fin
Uniform convective heat transfer coefficient
One dimensional heat conduction
Homogeneous properties of materials
17. Define fin effectiveness
Fine effectiveness is the ratio of the fin heat transfer rate to the heat transfer rate that
would exist without a fin.
18. What are the conditions that make the fin effective?
Thermal conductivity should be large
Convective heat transfer coefficient should be small
Thickness of the fin should be small
PART –B
1. Consider a 4-m-high, 6-m-wide, and 0.3-m-thick brick Wall whose thermal conductivity
is k =0.8 W/m · °C. On a Certain day, the temperatures of the inner and the outer surfaces Of
the wall are measured to be 14°C and 6°C, respectively. Determine the rate of heat loss
through the wall on that day.
Assumptions 1 Heat transfer through the wall is steady since the surface temperatures remain
constant at the specified values. 2 Heat transfer is one-dimensional since any significant
temperature gradients will exist in the direction from the indoors to the outdoors. 3. Thermal
conductivity is constant.
Properties The thermal conductivity is given to be k = 0.8 W/m°C.
Analysis The surface area of the wall and the rate of heat loss through the wall are
A ( (4 6 24 m) m) m2
( . )( )
.Q kA
T T
L
1 2 08
14 6
0 3 W / m. C)(24 m
C
m
2 512 W
2. Consider a 4-m-high, 6-m-wide, and 0.3-m-thick brick wall whose thermal conductivity is
k =0.8 W/m · °C. On a certain day, the temperatures of the inner and the outer surfaces of the
wall are measured to be 14°C and 6°C, respectively. Determine the rate of heat loss through
the wall on that day. assuming the space between the two glass layers is evacuated.
Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor
temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since
any significant temperature gradients will exist in the direction from the indoors to the
outdoors. 3 Thermal conductivities of the glass and air are constant. 4 Heat transfer by
radiation is negligible.
Properties The thermal conductivity of the glass and air are given to be kglass = 0.78 W/m°C
and kair = 0.026 W/m°C.
Analysis The area of the window and the individual resistances are
A ( . ( .12 2 2 4 m) m) m2
6C 14C
L=0.3 m
Q
Wall
Air
C/W 2539.0
0167.01923.0)0016.0(20417.02
C/W 0167.0)m 4.2(C).W/m 25(
11
C/W 1923.0)m 4.2(C)W/m. 026.0(
m 012.0
C/W 0016.0)m 4.2(C)W/m. 78.0(
m 003.0
C/W 0417.0)m 4.2(C).W/m 10(
11
2,211,
o
2o22
2,o
22
22
21
1glass31
221
1,i
convconvtotal
conv
air
conv
RRRRR
AhRR
Ak
LRR
Ak
LRRR
AhRR
The steady rate of heat transfer through window glass then becomes
W114
C/W2539.0
C)]5(24[21
totalR
TTQ
The inner surface temperature of the window glass can be determined from
C19.2
=C/W)W)(0.0417 114(C24o
1,111,
11conv
conv
RQTTR
TTQ
3. A cylindrical resistor element on a circuit board dissipates 0.15 W of power in an
environment at 40°C. The resistor is 1.2 cm long, and has a diameter of 0.3 cm. Assuming
heat to be transferred uniformly from all surfaces, determine (a) the amount of heat this
resistor dissipates during a 24-h period, (b) the heat flux on the surface of the resistor, in
W/m2, and
(c) the surface temperature of the resistor for a combined convection and radiation heat
transfer coefficient of 9 W/m2 · °C.
Assumptions 1 Steady operating conditions exist. 2 Heat is transferred uniformly from all
surfaces of the resistor.
Analysis (a) The amount of heat this resistor dissipates during a 24-hour period is
Wh3.6 h)W)(24 15.0(tQQ
(b) The heat flux on the surface of the resistor is
222
m 000127.0m) m)(0.012 003.0(4
m) 003.0(2
42
DL
DAs
2 W/m1179
2m 000127.0
W 15.0
sA
(c) The surface temperature of the resistor can be determined from
C171
)m 7C)(0.00012.W/m (1179
W 15.0)(
22s
ssshA
QTTTThAQ
4. Heat is to be conducted along a circuit board that has a copper layer on one side. The
circuit board is 15 cm long and 15 cm wide, and the thicknesses of the copper and epoxy
layers are 0.1 mm and 1.2 mm, respectively. Disregarding heat transfer from side surfaces,
determine the percentages of heat conduction along the copper (k =386 W/m · °C) and epoxy
(k = 0.26 W/m · °C) layers. Also determine the effective thermal Conductivity of the board.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since
heat transfer from the side surfaces is disregarded 3 Thermal conductivities are constant.
Properties The thermal conductivities are given to be k = 386 W/m°C for copper and 0.26
W/m°C for epoxy layers.
Resistor
0.15 W
Q
Analysis We take the length in the direction of heat transfer to be L and the width of the
board to be w. Then heat conduction along this two-layer board can be expressed as
L
Twktkt
L
TkA
L
TkAQQQ
epoxycopper
epoxycopper
epoxycopper )()(
Heat conduction along an “equivalent” board of thickness t = tcopper + tepoxy and thermal
conductivity keff can be expressed as
L
Twttk
L
TkAQ
)( epoxycoppereff
board
Setting the two relations above equal to each other and solving for the effective conductivity
gives
k t t kt kt kkt kt
t teff eff( ) ( ) ( )
( ) ( )copper epoxy copper epoxy
copper epoxy
copper epoxy
Note that heat conduction is proportional to kt. Substituting, the fractions of heat conducted
along the copper and epoxy layers as well as the effective thermal conductivity of the board
are determined to be
( ) ( .
( ) ( . .
( ) ( ) ( ) . . .
( )
( )
.
..
( )
( )
.
..
kt
kt
kt kt kt
fkt
kt
fkt
kt
copper
epoxy
total copper epoxy
epoxy
epoxy
total
copper
copper
total
386 0 0386
0 26 0 000312
0 0386 0 000312 0 038912
0 000312
0 0389120 008
0 0386
0 0389120 992
W / m. C)(0.0001 m) W/ C
W / m. C)(0.0012 m) W/ C
W/ C
0.8%
99.2%
and keff
( . . . )
( . . )
386 0 0001 0 26 0 0012
0 0001 0 0012
W/ C
m29.9 W / m. C
5. A1-mm-thick copper plate (k = 386 W/m · °C) is sandwiched between two 5-mm-thick
epoxy boards (k = 0.26 W/m · °C) that are 15 cm 20 cm in size. If the thermal contact
conductance on both sides of the copper plate is estimated to be 6000 W/m · °C, determine
the error involved in the total thermal resistance of the plate if the thermal contact
conductance are ignored.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since
the plate is large. 3 Thermal conductivities are constant.
Properties The thermal conductivities are given to be k = 386 W/m°C for copper plates and
k = 0.26 W/m°C for epoxy boards. The contact conductance at the interface of copper-epoxy
layers is given to be hc = 6000 W/m2C.
Analysis The thermal resistances of different layers for unit
surface area of 1 m2 are
C/W 00017.0)m C)(1.W/m 6000(
1122
ccontact
cAhR
Epoxy
Q
Copper
Ts tepoxy tcopper
RL
kAplate 2
m
(386 W / m. C)(1 m ) C / W
0 001
2 6 10 6..
RL
kAepoxy 2
m
(0.26 W / m. C)(1 m ) C / W
0 0050 01923
..
The total thermal resistance is
C/W 03914.001923.02106.200017.02
22
6
epoxyplatecontacttotal
RRRR
Then the percent error involved in the total thermal resistance of the plate if the thermal
contact resistances are ignored is determined to be
%87.0
10003914.0
00017.02100
2Error%
total
contact
R
R
which is negligible.
Epox
y
5 mm
Epox
y
5 mm
Copp
er
plate
Q
Rcontact
Rcontact
Repoxy
T1 T2 Repoxy
Rplate
6. A 4-m-high and 6-m-wide wall consists of a long 18-cm 30-cm cross section of horizontal
bricks (k =0.72 W/m · °C) separated by 3-cm-thick plaster layers (k = 0.22 W/m · °C). There
are also 2-cm-thick plaster layers on each side of the wall, and a 2-cm-thick rigid foam (k
=0.026 W/m · °C) on the inner side of the wall. The indoor and the outdoor temperatures are
22°C and =4°C, and the convection heat transfer coefficients on the inner and the outer sides
are h1 = 10 W/m2 · °C and h2 =20 W/m2 · °C, respectively. Assuming one-dimensional heat
transfer and disregarding radiation, determine the rate of heat transfer through the wall.
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat
transfer through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Heat
transfer by radiation is disregarded.
Properties The thermal conductivities are given to be k = 0.72 W/m°C for bricks, k = 0.22
W/m°C for plaster layers, and k = 0.026 W/m°C for the rigid foam.
Analysis We consider 1 m deep and 0.33 m high portion of wall which is representative of
the entire wall. The thermal resistance network and individual resistances are
R Rh A
R RL
kA
R R RL
kA
R R RL
h A
R R
i conv
foam
plasterside
plastercenter o
,( ( . )
.
.
( . ( . ).
.
( . ( . ).
.
( . ( . ).
11
1
6
5
1 1
10 0 33 10 303
0 02
0 026 0 33 12 33
0 02
0 22 0 30 10 303
018
0 22 0 015 154 55
W / m . C) m C / W
m
W / m. C) m C / W
m
W / m. C) m C / W
m
W / m. C) m C / W
2 2
2
2 2
3 2
4 brick
conv
midmid
total i mid o
L
kA
R Rh A
R R R RR
R R R R R R
018
0 72 0 30 10833
1 1
20 0 33 10152
1 1 1 1 1
54 55
1
0833
1
54 55081
2 0 303 2 33 2 0 303 081 0152
4 201
22
3 4 5
1 2
.
( . ( . ).
( ( . ).
. . ..
. . ( . ) . .
.
,
m
W / m. C) m C / W
W / m. C) m C / W
C / W
C / W
2
o 2
The steady rate of heat transfer through the wall per 0 33. m2 is
[( ( )]
..Q
T T
Rtotal
1 2 22 4
4 201619
C
C / W W
Then steady rate of heat transfer through the entire wall becomes
( .( )
.Qtotal
619
4 6
0 33 W)
m
m
2
2450 W
7. A50-m-long section of a steam pipe whose outer diameter is 10 cm passes through an open
space at 15°C. The average temperature of the outer surface of the pipe is measured to be
150°C. If the combined heat transfer coefficient on the outer surface of the pipe is 20 W/m2 ·
R2
R3
R4
R5
R6
R7
T2
R1
Ri
T1
°C, determine (a) the rate of heat loss from the steam pipe, (c) the thickness of fiberglass
insulation (k =0.035 W/m · °C) needed in order to save 90 percent of the heat lost. Assume
the pipe
Temperature to remain constant at 150°C.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2
Heat transfer is one-dimensional since there is thermal symmetry about the center line and no
variation in the axial direction. 3 Thermal conductivity is constant. 4 The thermal contact
resistance at the interface is negligible. 5 The pipe temperature remains constant at about 150
C with or without insulation. 6 The combined heat transfer coefficient on the outer surface
remains constant even after the pipe is insulated.
Properties The thermal conductivity of fiberglass insulation is given to be k = 0.035 W/m°C.
Analysis (a) The rate of heat loss from the steam pipe is
2m 71.15m) 50(m) 1.0( DLAo
W42,412=C)15150)(m 71.15(C).W/m 20()( 22 airsobare TTAhQ
(c) In order to save 90% of the heat loss and thus to reduce it to 0.142,412 = 4241 W, the
thickness of insulation needed is determined from
kL
rr
Ah
TT
RR
TTQ
oo
airs
insulationo
airsinsulated
2
)/ln(1 12
Substituting and solving for r2, we get
m 0692.0
)m 50(C)W/m. 035.0(2
)05.0/ln(
)]m 50(2(C)[.W/m 20(
1
C)15150(W 4241 2
2
22
r
r
r Then the thickness of insulation becomes
cm 1.92 592.612 rrtinsulation
8. Consider a 2-m-high electric hot water heater that has a diameter of 40 cm and maintains
the hot water at 55°C. The tank is located in a small room whose average temperature i 27°C,
and the heat transfer coefficients on the inner and outer surfaces of the heater are 50 and 12
W/m2 · °C, respectively. The tank is placed in another 46-cm-diameter sheet metal tank of
negligible thickness, and the space between the two tanks is filled with foam insulation (k
=0.03 W/m · °C). The thermal resistances of the water tank and the outer thin sheet metal
shell are very small and can be neglected. The price of electricity is $0.08/kWh, and the home
owner pays $280 a year for water heating. Determine the fraction of the hot water energy cost
of this household that is due to the heat loss from the tank. Hot water tank insulation kits
consisting of 3-cm-thick fiberglass insulation (k = 0.035 W/m · °C) large enough to wrap the
entire tank are available in the market for about $30. If such an insulation is installed on this
water tank by the home owner
himself, how long will it take for this additional insulation to pay for itself?s
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2
Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no
variation in the axial direction. 3 Thermal properties are constant. 4 The thermal contact
resistance at the interface is negligible. 5 Heat transfer coefficient accounts for the radiation
effects, if any.
Ro Ta
ir
Rinsulatio
n Ts
Properties The thermal conductivity of plastic cover is given to be k = 0.15 W/m°C.
Analysis In steady operation, the rate of heat transfer from the wire is equal to the heat
generated within the wire,
W80)A 10)(V 8( VIWQ e
The total thermal resistance is
C/W 4051.00735.03316.0
C/W 0735.0)m 10(C) W/m.15.0(2
)1/2ln(
2
/ln(
C/W 3316.0m)] m)(10 (0.004C)[. W/m24(
11
plasticconvtotal
12plastic
2conv
RRR
kL
rrR
AhR
oo
Then the interface temperature becomes
C62.4
)C/W 4051.0)( W80(C30total1total
21 RQTTR
TTQ
The critical radius of plastic insulation is
mm 25.6m 00625.0C.W/m 24
CW/m. 15.02
h
krcr
Doubling the thickness of the plastic cover will increase the outer radius of the wire to 3 mm,
which is less than the critical radius of insulation. Therefore, doubling the thickness of plastic
cover will increase the rate of heat loss and decrease the interface temperature.
9. Obtain a relation for the fin efficiency for a fin of constant cross-sectional area Ac,
perimeter p, length L, and thermal conductivity k exposed to convection to a medium at T
with a
heat transfer coefficient h. Assume the fins are sufficiently long so that the temperature of the
fin at the tip is nearly T. Take the temperature of the fin at the base to be Tb and neglect heat
transfer from the fin tips. Simplify the relation for (a) a circular fin of diameter D and (b)
rectangular fins of thickness t..
Assumptions 1 The fins are sufficiently long so that the temperature of the fin at the tip is
nearly T . 2 Heat transfer from the fin tips is negligible.
Analysis Taking the temperature of the fin at the base to be Tb and using the heat transfer
relation for a long fin, fin efficiency for long fins can be expressed as
fin
Actual heat transfer rate from the fin
Ideal heat transfer rate from the fin
if the entire fin were at base temperature
hpkA T T
hA T T
hpkA
hpL L
kA
ph
c b
fin b
c c( )
( )
1
This relation can be simplified for a circular fin of diameter D and rectangular fin of
thickness t and width w to be
fin,circular
fin,rectangular
1 1 4 1
2
1 1
2
1
2
1
2
2
L
kA
ph L
k D
D h L
kD
h
L
kA
ph L
k wt
w t h L
k wt
wh L
kt
h
c
c
( / )
( )
( )
( )
( )
Rconv
T2
Rplastic
T1
h, T
D
p= D
Ac = D2/4
Tb
10. A hot plate is to be cooled by attaching aluminum pin fins on one side. The rate of heat
transfer from the 1 m by 1 m section of the plate and the effectiveness of the fins are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 The temperature along the fins varies in
one direction only (normal to the plate). 3 Heat transfer from the fin tips is negligible. 4 The
heat transfer coefficient is constant and uniform over the entire fin surface. 5 The thermal
properties of the fins are constant. 6 The heat transfer coefficient accounts for the effect of
radiation from the fins.
Properties The thermal conductivity of the aluminium plate and fins is given to be k = 237
W/m°C.
Analysis Noting that the cross-sectional areas of the fins are constant, the efficiency of the
circular fins can be determined to be
ahp
kA
h D
k D
h
kDc
2 4
4 4 35
237 0 00251537
/
( )
( )( . ).
W / m . C
W / m. C m m
2-1
fin
-1
-1
m m
m m
tanh tanh( . . )
. ..
aL
aL
1537 0 03
1537 0 030 935
The number of fins, finned and unfinned surface areas, and heat transfer rates from those
areas are
n 1
0 006 0 00627 777
m
m) m)
2
( . ( .,
W2107
C)30100)(m 86.0)(C. W/m35()(
W300,15
C)30100)(m 68.6)(C. W/m35(935.0
)(
m 86.04
)0025.0(277771
4277771
m 68.64
)0025.0()03.0)(0025.0(27777
427777
22unfinnedunfinned
22
finfinmaxfin,finfinned
222
unfinned
222
fin
TThAQ
TThAQQ
DA
DDLA
b
b
Then the total heat transfer from the finned plate becomes
kW 17.4 W1074.12107300,15 4unfinnedfinnedfintotal, QQQ
The rate of heat transfer if there were no fin attached to the plate would be
A
Q hA T Tb
no fin2
no fin no fin2 2
m m m
W / m C m C W
( )( )
( ) ( . )( )( )
1 1 1
35 1 100 30 2450
Then the fin effectiveness becomes
7.102450
400,17
fin no
finfin
Q
Q
3 cm
0.6 cm D=0.25 cm
UNIT- IV CONVECTION
1. What is dimensional analysis?
Dimensional analysis is a matehematical technique which make use of the study of
the dimensions for solving seferal engineering problems.
2. Define free convection
If the fluid motion is produced due to change in density resulting from temperature
gradient, the mode is said to be natural convection.
3. What is forced convection?
If the fluid motion is artificially created by means of external force like a blower or
fan that type of heat transfer is know as forced convection.
4. What are the advantages of dimensional analysis?
It express the functional relationship between the variables in dimensional terms
It enable getting up a theoretical solution in a simplified dimensional form
5.What are the factors changes the thickness of boundary layer?
Turbulence in the ambient flow
Surface roughness, Pressure gradient
Viscosity of the fluid and temperature difference between the surface and fluid
6. What is meant by Newtonian and Non-Newtonian flow?
The fluid, which obeys Newton’s law of viscosity, is called Newtonian flow and other
type of fluid is called Non-Newtonian flow.
7. Define hydrodynamic and thermal boundary layer.
In hydrodynamic boundary layer the velocity of flow is less than 99% of free stream
velocity.
In thermal boundary layer, temperature of the fluid is less than 99% of free stream
temperature.
8. What are the limitations of dimensional analysis?
The complete information is not provided by this analysis
No information about the internal mechanism of physical phenomenon
Des not give any clue regarding the selection of variables.
9.Define lower critical Reynolds number
It defines the limit below which all turbulence, no matter how severe, entering the
flow from any source will eventually be damped out by viscous action.
10.Define Nusselt Number.
It is the ratio of heat transfer rate, Q to the rate at which heat would be conducted
within the fluid under a temperature gradient.
11. Define Grashoff number
It is related with natural convection heat transfer. It is defined as the ratio of the
product of inertia force and buoyancy force to the square of viscous force.
12. What is boundary layer?
A layer adjacent to the boundary is called boundary layer.
13. What are the factors that change the boundary layer from laminar to turbulent?
Turbulence in ambient flow, surface roughness, pressure gradient, plate curvature and
temperature difference between fluid and boundary.
14. Define skin friction coefficient.
It is defined as the ration of shear stress at the plate to the dynamic head caused by
free stream velocity.
15. Define thermal boundary layer.
In a boundary layer wherein the temperature variation exists is called the thermal
boundary layer.
16. Define free convection.
Free or natural convection is the process of heat transfer which occurs due to
movement of the fluid particles by density changes associated with temperature differential in
a fluid.
17. Give few examples of free convection
The cooling of transmission lines, electrical transformers and rectifiers
The heating of rooms by use of radiators
The heat transfer from hot pipes and ovens surrounded by cooler air
Cooling of reactor core.
PART –B
1. Hot engine oil flows over a flat plate. The temperature and velocity of the oil are 30C&3
m/s respectively. The temperature of the plate is 30C. compute the total drag force and the
rate of heat transfer per unit width of the plate.
Assumptions 1 Steady operating condition exists. 2 The critical Reynolds number is Recr =
5105. 3 Radiation effects are negligible.
Properties The properties of engine oil at the film temperature of (Ts + T)/2 = (80+30)/2
=55C = 328 K are (Table A-13)
1505PrC W/m.141.0
/sm 10123kg/m 867 263
k
Analysis Noting that L = 6 m, the Reynolds number at the end of the plate is
Re(
.L
L
V
3
123 10146 10
6
5 m / s)(6 m)
m / s2
which is less than the critical Reynolds number. Thus we have
laminar flow over the entire plate. The average friction
coefficient and the drag force per unit width are determined
from
N 3.81
2
m/s) )(3 kg/m867()m 16)(00347.0(
2
00347.0)1046.1(328.1Re328.1
232
2
5.055.0
VsfD
Lf
ACF
C
Similarly, the average Nusselt number and the heat transfer
coefficient are determined using the laminar flow relations for
a flat plate,
Nu
hL
k
hk
LNu
L
0 664 0 664 146 10 1505 2908
0141
62908 68 3
0 5 1 3 5 0 5 1 3. Re Pr . ( . ) ( )
.( ) .
. / . /
W / m. C
m W / m . C2
The rate of heat transfer is determined from Newton's law of cooling
3. Wind is blowing parallel to the wall of a house. The temperature and velocity of the air is
5oC& 55 km/hr. Calculate the rate of heat loss from the wall.
Assumptions 1 Steady operating condition exists. 2 The critical Reynolds number is Recr =
5105. 3 Radiation effects are negligible. 4 Air is an ideal gas with constant properties.
Properties The properties of air at 1 atm and the film
temperature of (Ts + T)/2 = (12+5)/2 = 8.5C
7340.0Pr
/sm 10413.1
C W/m.02428.0
25-
k
Analysis Air flows parallel to the 10 m side:
Ts = 30C
Oil
V = 3 m/s
T = 30C
L = 6 m
Ts = 12C
Air
V = 55 km/h
T = 5C
L
The Reynolds number in this case is
7
2510081.1
/sm 10413.1
m) m/s](10)3600/100055[(Re
LL
V
which is greater than the critical Reynolds number. Thus we have combined laminar and
turbulent flow. Using the proper relation for Nusselt number, heat transfer coefficient and
then heat transfer rate are determined to be
C. W/m43.32)10336.1(m 10
C W/m.02428.0
10336.1)7340.0](871)10081.1(037.0[Pr)871Re037.0(
24
43/18.073/18.0
NuL
kh
k
hLNu L
kW 9.08
W 9081C5)-)(12m C)(40.W/m 43.32()(
m 40=m) m)(10 4(
22
2
ss
s
TThAQ
wLA
If the wind velocity is doubled:
7
2510163.2
/sm 10413.1
m) m/s](10)3600/1000110[(Re
LL
V
which is greater than the critical Reynolds number. Thus we have combined laminar and
turbulent flow. Using the proper relation for Nusselt number, the average heat transfer
coefficient and the heat transfer rate are determined to be
C. W/m88.57)10384.2(m 10
C W/m.02428.0
10384.2)7340.0](871)10163.2(037.0[Pr)871Re037.0(
24
43/18.073/18.0
NuL
kh
k
hLNu L
kW 16.21
W 206,16C5)-)(12m C)(40.W/m 88.57()(
m 40=m) m)(4 10(
22
2
ss
s
TThAQ
wLA
4. Air is flowing over the steam pipe having steam temperature of 90 oC .The velocity
and temperature of the air are 70Cand 50 km/ hr respectively. Calculate rate of heat
loss by the air on the steam pipe
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is
an ideal gas with constant properties.
Properties The properties of air at 1 atm and the film temperature of (Ts + T)/2 = (90+7)/2
= 48.5C are (Table A-15)
7232.0Pr
/sm 10784.1
CW/m. 02724.0
25-
k
Analysis The Reynolds number is
Air
V = 50 km/h
T = 7C
Pipe
D = 8 cm Ts = 90C
4
2510228.6
/sm 10784.1
m) (0.08]s/h) 0m/km)/(360 1000(km/h) (50[Re
DV
The Nusselt number corresponding to this Reynolds number is
1.159
000,282
10228.61
7232.0/4.01
)7232.0()10228.6(62.03.0
000,282
Re1
Pr/4.01
PrRe62.03.0
5/48/5
4
4/13/2
3/15.04
5/48/5
4/13/2
3/15.0
k
hDNu
The heat transfer coefficient and the heat transfer rate become
C. W/m17.54)1.159(m 08.0
C W/m.02724.0 2
NuD
kh
length) m(per =C7)-)(90m C)(0.2513.W/m 17.54()(
m 0.2513=m) m)(1 08.0(
22
2
W1130
TThAQ
DLA
ssconv
s
5. The components of an electronic system located in a horizontal square duct (20cm×20 cm)
is cooled by air flowing over the duct. The velocity and temperature of the air are 200 m/min
& 300C.Determine the total power rating of the electronic device.
Assumptions 1 Steady operating condition exists. 2 Radiation effects are negligible. 3 Air is
an ideal gas with constant properties. 4 The local atmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the film
temperature of (Ts + T)/2 = (65+30)/2 = 47.5C are
7235.0Pr
/sm 10774.1
CW/m. 02717.0
25-
k
Analysis The Reynolds number is
4
2510758.3
/sm 10774.1
m) (0.2m/s (200/60)Re
DV
Using the relation for a square duct from Table 7-1, the Nusselt number is determined to be
2.112)7235.0()10758.3(102.0PrRe102.0 3/1675.043/1675.0 k
hDNu
The heat transfer coefficient is
C. W/m24.15)2.112(m 2.0
C W/m.02717.0 2
NuD
kh
6. Water at 15ºC is to be heated to 65ºC by passing it over a bundle of 4-m-long 1-cm-
diameter resistance heater rods maintained at 90ºC. Water approaches the heater rod bundle
in normal direction at a mean velocity of 0.8 m/s. The rods arc arranged in-line with
longitudinal and transverse pitches of SL = 4 cm and ST = 3 cm. Determine the number of tube
rows NL in the flow direction needed to achieve the indicated temperature rise.
Assumptions 1 steady operating condition exists. 2 The surface temperature of the rods is
constant.
Properties The properties of water at the mean temperature of (15C +65C)/2=40C
Air
30C
200 m/min
20 cm
65C
k = 0.631 W/m-K = 992.1 kg/m3
Cp =4.179 kJ/kg-K Pr = 4.32
= 0.65310-3
kg/m-s Prs = Pr@ Ts = 1.96
Also, the density of water at the inlet temperature of 15C (for use in the mass flow rate
calculation at the inlet) is i =999.1 kg/m3.
Analysis It is given that D = 0.01 m, SL = 0.04 m and ST = 0.03 m, and V = 0.8 m/s.
Then the maximum velocity and the Reynolds number
based on the maximum velocity become
m/s 20.1m/s) 8.0(01.003.0
03.0max
VV
DS
S
T
T
232,18skg/m 10653.0
m) m/s)(0.01 20.1)(kg/m 1.992(Re
3
3max
DD
V
The average Nusselt number is determined using
the proper relation from Table 7-2 to be
3.269)96.1/32.4()32.4()232,18(27.0
)Pr(Pr/PrRe27.0Nu
25.036.063.0
25.036.063.0
sDD
Assuming that NL > 16, the average Nusselt number and heat transfer coefficient for all the
tubes in the tube bank become
3.269NuNu , DND L
C W/m994,16
m 0.01
C) W/m631.0(3.269 2,
D
kNuh LND
Consider one-row of tubes in the transpose direction (normal to flow), and thus take NT =1.
Then the heat transfer surface area becomes
LLtubes NNDLNA 1257.0 m) m)(4 01.0()1( Then the log mean temperature difference, and the expression for the rate of heat transfer
become
C51.45)]6590/()1590ln[(
)6590()1590(
)]/()ln[(
)()(ln
esis
esis
TTTT
TTTTT
LLs NNThAQ 220,97)C51.45()C)(0.1257W/m 994,16( 2ln
The mass flow rate of water through a cross-section corresponding to NT =1 and the rate of
heat transfer are
kg/s91.95m/s) )(0.8m 0.03)(4 kg/m1.999( 23 VAm c
W 10004.2C)1565(J/kg.C) 9 kg/s)(41791.95()( 7 iep TTCmQ
Substituting this result into the heat transfer expression above we find th e number of tube
rows
206 LLs NNThAQ 220,97W 10004.2 7ln
7. Cooling water available at 10°C is used to condense steam at 30°C in the condenser of a
power plant at a rate of 0.15 kg/s by circulating the cooling water through a bank of 5-m-long
1.2-cm-internal-diameter thin copper tubes. Water enters the tubes at a mean velocity of 4
m/s, and leaves at a temperature of 24°C. The tubes are nearly isothermal at 30°C.Determine
SL
ST
V=0.8 m/s
Ti=15C
Ts=90C
D
the average heat transfer coefficient between the water and the tubes, and the number of tubes
needed to achieve the indicated heat transfer rate in the condenser.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is
constant. 3 The thermal resistance of the pipe is negligible.
Properties The properties of water at the average temperature of (10+24)/2=17C are
CJ/kg. 5.4184
kg/m 7.998 3
pC
Also, the heat of vaporization of water at 30C is kJ/kg 2431fgh
.
Analysis The mass flow rate of water and the surface area are
kg/s 0.4518=m/s) (44
m) (0.012)kg/m 7.998(
4
23
2
mmc
DAm VV
The rate of heat transfer for one tube is
W 468,26)C1024)(CJ/kg. 5.4184)( kg/s4518.0()( iep TTCmQ
The logarithmic mean temperature difference and the surface area are
C63.11
1030
2430ln
1024
ln
ln
is
es
ie
TT
TT
TTT
2m 0.1885=m) m)(5 012.0( DLAs
The average heat transfer coefficient is determined from
C.kW/m 12.1
2
W 1000
kW1
)C63.11)(m 1885.0(
W 468,262
lnln
TA
QhThAQ
ss
The total rate of heat transfer is determined from
kW65.364 kJ/kg)2431)( kg/s15.0( fgcondtotal hmQ
Then the number of tubes becomes
13.8
W468,26
W650,364
Q
QN total
tube
9. Water is to be heated from 10°C to 80°C as it flows through a 2-cm-internal-diameter, 7-
m-long tube. The tube is equipped with an electric resistance heater, which provides uniform
heating throughout the surface of the tube. The outer surface of the heater is well insulated, so
that in steady operation all the heat generated in the heater is transferred to the water in the
Steam, 30C
L = 5 m
D = 1.2 cm
Water
10C
4 m/s
24C
tube. If the system is to provide hot water at a rate of 8 L/min, determine the power rating of
the resistance heater. Also, estimate the inner surface temperature of the pipe at the exit.
Assumptions 1 Steady flow conditions exist. 2 The surface heat flux is uniform. 3 The inner
surfaces of the tube are smooth.
Properties The properties of water at the average
temperature of (80+10) / 2 = 45C are (Table A-9)
91.3Pr
CJ/kg. 4180
/sm 10602.0/
C W/m.637.0
kg/m 1.990
26-
3
pC
k
Analysis The power rating of the resistance heater is
kg/s 132.0kg/min 921.7)/minm 008.0)(kg/m 1.990( 33 Vm
W38,627 C)1080)(CJ/kg. 4180)(kg/s 132.0()( iep TTCmQ
The velocity of water and the Reynolds number are
Vmc
V
A
(8 / )
( . /.
10 60
0 02 40 4244
3 m / s
m) m / s
3
2
101,14/sm 10602.0
m) m/s)(0.02 (0.4244Re
26
hm DV
which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this
case are roughly
m 20.0m) 02.0(1010 DLL th
which is much shorter than the total length of the duct. Therefore, we can assume fully
developed turbulent flow in the entire duct, and determine the Nusselt number from
79.82)91.3()101,14(023.0PrRe023.0 4.08.04.08.0 k
hDNu h
Heat transfer coefficient is
C. W/m2637)79.82(m 02.0
C W/m.637.0 2
NuD
kh
h
Then the inner surface temperature of the pipe at the exit becomes
C113.3
es
s
eess
T
T
TThAQ
,
2
,
C)80)](m 7)(m 02.0()[C.W/m 2637(W 627,38
)(
Water
10C
3 m/s
80C
(Resistance heater)
L
D = 2 cm
10. Water is boiling in a 12-cm-deep pan with an outer diameter of 25 cm that is placed on
top of a stove. The ambient air and the surrounding surfaces are at a temperature of 25°C, and
the emissivity of the outer surface of the pan is 0.95. Assuming the entire pan to be at an
average temperature of 98°C, determine the rate of heat loss from the cylindrical side surface
of the pan to the surroundings by (a) natural convection and (b) radiation. (c) If water is
boiling at a rate of 2 kg/h at 100°C, determine the ratio of the heat lost from the side surfaces
of the pan to that by evaporation of water. The heat of vaporization of water at 100°C is 2257
kJ/kg.
Assumptions 1 Steady operating conditions exist. 2 Air is an
ideal gas with constant properties. 3 The local atmospheric
pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature
of (Ts+T)/2 = (98+25)/2 = 61.5C are (Table A-15)
1-
25
K 00299.0K)2735.61(
11
7198.0Pr
/sm 10910.1
CW/m. 02819.0
fT
k
Analysis (a) The characteristic length in this case is the height of the pan, m. 12.0 LLc
Then,
6
225
3-12
2
3
10299.7)7198.0()/sm 10910.1(
)m 12.0)(K 2598)(K 00299.0)(m/s 81.9(Pr
)(
LTTgRa s
We can treat this vertical cylinder as a vertical plate since
4/14/164/1
35 and thus 0.25< 07443.0
)7198.0/10299.7(
)12.0(3535
Gr
LD
Gr
L
Therefore,
60.28
7198.0
492.01
)10299.7(387.0825.0
Pr
492.01
Ra387.0825.0
2
27/816/9
6/16
2
27/816/9
6/1
Nu
Vapor
2 kg/h
Water
100C
Pan
Ts = 98C = 0.1
Air
T = 25C
2
2
m 09425.0)m 12.0)(m 25.0(
C.W/m 720.6)60.28(m 12.0
CW/m. 02819.0
DLA
NuL
kh
s
and
W46.2 C)2598)(m 09425.0)(C.W/m 720.6()( 22TThAQ ss
(b) The radiation heat loss from the pan is
W5.9
444282
44
)K 27325()K 27398().KW/m 1067.5)(m 09425.0)(10.0(
)( surrssrad TTAQ
(c) The heat loss by the evaporation of water is
W 1254 kW254.1) kJ/kg2257)( kg/s3600/2( fghmQ
Then the ratio of the heat lost from the side surfaces of the pan to that by the evaporation of
water then becomes
4.2%
042.01254
9.52.46f
11. Consider a 1.2-m-high and 2-m-wide glass window with a thickness of 6 mm, thermal
conductivity k =0.78 W/m °C, and emissivity 0.9. The room and the walls that face the
window are maintained at 25°C, and the average temperature of the inner surface of the
window is measured to be 5°C. If the temperature of the outdoors is 5°C, determine (a) the
convection heat transfer coefficient on the inner surface of the window, (b) the rate of total
heat transfer through the window, and (c) the combined natural convection and radiation heat
transfer coefficient on the outer surface of the window. Is it reasonable to neglect the thermal
resistance of the glass in this case?
Assumptions 1 Steady operating conditions exist. 2 Air is an
ideal gas with constant properties. 3 The local atmospheric
pressure is 1 atm.
Properties The properties of air at 1 atm and the film
temperature of (Ts+T)/2 = (5+25)/2 = 15C are (Table A-15)
Q
Outdoors
-5C
Glass
Ts = 5C
= 0.9
L = 1.2 m
Room
T = 25C
1-
25
K 003472.0K)27315(
11
7323.0Pr
/sm 10471.1
CW/m. 02476.0
fT
k
Analysis (a) The characteristic length in this case is the height of the window, m. 2.1 LLc
Then,
9
225
3-12
2
3
10986.3)7323.0()/sm 10471.1(
)m 2.1)(K 525)(K 34720.0)(m/s 81.9(Pr
)(
0
cs LTTgRa
7.189
7323.0
492.01
)10986.3(387.0825.0
Pr
492.01
Ra387.0825.0
2
27/816/9
6/19
2
27/816/9
6/1
Nu
2m 4.2m) m)(2 2.1(
)7.189(m 2.1
CW/m. 02476.0
sA
NuL
kh C. W/m3.915 2
(b) The sum of the natural convection and radiation heat transfer from the room to the
window is
W 9.187C)525)(m 4.2)(C.W/m 915.3()( 22convection ss TThAQ
W3.234])K 2735()K 27325)[(.KW/m 1067.5)(m 4.2)(9.0(
)(
444282
44radiation
ssurrs TTAQ
W422.2 3.2349.187radiationconvectiontotal QQQ
(c) The outer surface temperature of the window can be determined from
C65.3)m 4.2)(CW/m. 78.0(
)m 006.0)(W 346(C5)(
2
total,,,,total
sisososis
s
kA
tQTTTT
t
kAQ
Then the combined natural convection and radiation heat transfer coefficient on the outer
window surface becomes
C. W/m20.35 2
C)]5(65.3)[m 4.2(
W 346
)( or
)(
2,,
totalcombined
,,combinedtotal
ooss
ooss
TTA
Qh
TTAhQ
Note that T QR and thus the thermal resistance R of a layer is proportional to the
temperature drop across that layer. Therefore, the fraction of thermal resistance of the glass is
equal to the ratio of the temperature drop across the glass to the overall temperature
difference,
4.5%)(or 045.0)5(25
65.35
total
glass
total
glass
TR
T
R
R
which is low. Thus it is reasonable to neglect the thermal resistance of the glass.
UNIT – V RADIATION
PART- A
1. What are the factors influencing the rate of emission of radiation?
The temperature of surface
The nature of surface
The wave length and frequency of radiation
2. Define emissive power
Emissive power is defined as the total amount of radiation emitted by a body per unit time
and unit area.
3. What is meant by gray body?
It is the body absorbs a definite percentage of incident radiation, irrespective of their
wave length, the body is known as gray body.
4. Define shape factor
The shape factor is defined as the fraction of the radiation energy that is diffused from
one surface element and strikes the other surface directly with no intervening reflections.
5. What is radiation shield?
The radiation shield reduces the radiation heat transfer by effective increasing of surface
resistance without actually removing any heat from the overall system.
6. What are the properties of black body?
It absorbs all the incident radiation falling on it and does not transmit or reflect
regardless of the wave length.
It emits maximum amount of thermal radiation at all wave length at any specified
temperature
It is a diffuse emitter
7. Define emissivity
It is defined as the ability of the surface of a body to radiate heat.
8. What are the assumption made to calculate radiation exchange between surfaces
All surfaces are considered to be either black body or gray body.
9. Radiation and reflection process are assumed to be definite
The absorptivity of a surface is taken equal to its emissivity and independent of
temperature of the source of the incident radiation.
10. Define irradiation and radiosity
Irradiation is defined as the total radiation incident upon a surface per unit time per unit
area
Radiosity is used to indicate the total radiation leaving a surface per unit time per unit
area.
PART- B
1. Determine the view factors F13 and F23 between the rectangular surfaces
Assumptions The surfaces are diffuse emitters and reflectors.
Analysis From Fig.
24.0
5.02
11
5.02
1
31
3
F
W
L
W
L
and
29.0
12
2
5.02
1
)21(321
3
F
W
LL
W
L
We note that A1 = A3. Then the reciprocity and superposition rules gives
0.24 3113313131A FFFAF
05.024.029.0 32323231)21(3 FFFFF
Finally, 0.05 322332 FFAA
2. Consider a 4-m × 4-m × 4-m cubical furnace whose floor and ceiling are black and whose
side surfaces are reradiating. The floor and the ceiling of the furnace are maintained at
temperatures of 550 K and 1100 K, respectively. Determine the net rate of radiation heat
transfer between the floor and the ceiling of the furnace.
Assumptions 1 Steady operating condition exist 2 The surfaces are opaque, diffuse, and gray.
3 Convection heat transfer is not considered.
Properties The emissivities of all surfaces are = 1 since they are black or reradiating.
Analysis We consider the ceiling to be surface 1, the floor to be surface 2 and the side
surfaces to be surface 3. The furnace can be considered to be three-surface enclosure with a
radiation network shown in the figure. We assume that steady-state conditions exist. Since
the side surfaces are reradiating, there is no heat transfer through them, and the entire heat
lost by the ceiling must be gained by the floor. The view factor from the ceiling to the floor of
the furnace is F12 0 2 . . Then the rate of heat loss from the ceiling can be determined from
1
231312
211
11
RRR
EEQ bb
where
W = 2 m
(2) L2 = 1 m
L1 = 1 m
L3 = 1 m A3 (3)
A2
A1 (1)
T2 = 550 K
2 = 1
T1 = 1100 K
1 = 1
Reradiating
side surfacess
a = 4 m
244284
22
24428411
W/m 5188)K 550)(K.W/m 1067.5(
W/m 015,83)K 1100)(K.W/m 1067.5(
TE
TE
b
b
and
A A1 224 16 ( ) m m2
RA F
R RA F
121 12
13 231 13
1 1
16 0 20 3125
1 1
16 080 078125
( )( . ).
( )( . ).
m m
m m
2
-2
2
-2
Substituting,
kW 747
W1047.7
)m 078125.0(2
1
m 3125.0
1
W/m)5188015,83( 5
1
2-2-
2
12Q
3. A thin aluminum sheet with an emissivity of 0.15 on both sides is placed between two very
large parallel plates, which are maintained at uniform temperatures T1 =900 K and
T2 =650 K and have emissivities 0.5 and 2 ,0.8, respectively. Determine the net rate of
radiation heat transfer between the two plates per unit surface area of the plates and Compare
the result with that without the shield.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and
gray. 3 Convection heat transfer is not considered.
Properties The emissivities of surfaces are given to be 1 = 0.5, 2 = 0.8, and 3 = 0.15.
Analysis The net rate of radiation heat transfer with a thin aluminum shield per unit area of
the plates is
2 W/m1857
115.0
1
15.0
11
8.0
1
5.0
1
])K 650()K 900)[(K W/m1067.5(
111
111
)(
44428
2,31,321
42
41
shield one,12
TTQ
The net rate of radiation heat transfer between the plates in the case of no shield is
244428
21
42
41
shield ,12 W/m035,12
18.0
1
5.0
1
])K 650()K 900)[(K W/m1067.5(
111
)(
TTQ no
Then the ratio of radiation heat transfer for the two cases becomes
,
,
,
Q
Q
12
12
1857
12 035
one shield
no shield
W
W
1
6
T2 = 650 K 2 = 0.8
T1 = 900 K 1 = 0.5
Radiation
shield
3 = 0.15
4. Two very large parallel plates are maintained at uniform temperatures of T1=1000 K and
T2 = 800 K and have emissivities of 1 ,2 and 0.2, respectively. It is desired to reduce the net
rate of radiation heat transfer between the two plates to one-fifth by placing thin aluminum
sheets with anemissivity of 0.15 on both sides between the plates. Determine the number of
sheets that need to be inserted.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and
gray. 3 Convection heat transfer is not considered.
Properties The emissivities of surfaces are given to be 1 = 0.2, 2 = 0.2, and 3 = 0.15.
Analysis The net rate of radiation heat transfer between the plates in the case of no shield is
2
44428
21
42
41
shield ,12
W/m3720
12.0
1
2.0
1
])K 800()K 1000)[(K W/m1067.5(
111
)(
TTQ no
The number of sheets that need to be inserted in order to reduce the net rate of heat transfer
between the two plates to onefifth can be determined from
3
92.2
115.0
1
15.0
11
2.0
1
2.0
1
])K 800()K 1000)[(K W/m1067.5( ) W/m(3720
5
1
111
111
)(
shield
shield
444282
2,31,3shield
21
42
41
shields,12
N
N
N
TTQ
5. A2-m-internal-diameter double-walled spherical tank is used to store iced water at 0°C.
Each wall is 0.5 cm thick, and the 1.5-cm-thick air space between the two walls of the tank is
evacuated in order to minimize heat transfer. The surfaces surrounding the evacuated space
are polished so that each surface has an emissivity of 0.15. The temperature of the outer wall
T2 = 800 K
2 = 0.2
T1 = 1000 K
1 = 0.2
Radiation shields
3 = 0.15
of the tank is measured to be 20°C. Assuming the inner wall of the steel tank to be at 0°C,
determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice
at 0°C that melts during a 24-h period.
Assumptions 1 Steady operating conditions exist 2 The surfaces are opaque, diffuse, and
gray.
Properties The emissivities of both surfaces are given to be 1 = 2 = 0.15.
Analysis (a) Assuming the conduction resistance s of the walls to be negligible, the rate of
heat transfer to the iced water in the tank is determined to be
m m2A D1 12 22 01 12 69 ( . ) .
W107.4
2
444282
2
2
1
2
2
1
41
421
12
04.2
01.2
15.0
15.01
15.0
1
])K 2730()K 27320)[(K W/m1067.5)(m 69.12(
11
)(
D
D
TTAQ
(b) The amount of heat transfer during a 24-hour period is
kJ 9275)s 360024)(kJ/s 1074.0( tQQ
The amount of ice that melts during this period then becomes
kg 27.8kJ/kg 7.333
kJ 9275
ifif
h
QmmhQ
D2 = 2.04 m
T2 = 20C
2 = 0.15
D1 = 2.01 m
T1 = 0C
1 = 0.15
Vacuum
Iced
water
0C