End of Chapter Questions and Solutions

Chapter 1 Solutions

Alternating Current Theory

1. 1. A vector of length 1.0 makes an angle of \( 45^\circ \) with the horizontal plane. What is the vertical component of the vector?

$$ \sin \phi = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{\text{vertical component}}{\text{Vector}} $$

$$ \begin{aligned}\text{vertical component} &= \sin 45^\circ \times \text{Vector} \\ &= 0.707 \times 1.0 = 0.707\end{aligned} $$

1. 2. For the right-hand rule for generators, the thumb points in the direction of the motion of the conductor and the index finger points in the direction of the magnetic field. What is represented by the third finger?

The third finger of the right hand rule for generators points in the direction of the induced current.

1. 3. In North America, most power distribution systems are 60 Hz or 60 cycles per second. What is the definition for one cycle?

The transition from zero volts to a positive maximum, to zero, to a negative maximum and back to zero is called one cycle.

1. 4. What is the rms value of 170 volts?

The RMS value of 170 volts is: \( 170 \times 0.707 = 120.19 \text{ V} \) .

1. 5. What is the most common method of correcting a lagging power factor?

Capacitors are generally used to correct power factor in the industrial workplace.

1. 6. In a circuit containing only resistance and inductance, will current lead or lag voltage?

In a circuit containing only resistance and inductance, current will lag voltage.

1. 7. What is the inductive reactance for coil of 0.05 H connected in to a 50 Hz. supply system?

$$ \begin{aligned}\text{Inductive reactance: } X_L(\text{ohms}) &= 2\pi fL \\ &= 2 \times 3.1416 \times 50 \times 0.05 \\ &= 15.71 \text{ ohms}\end{aligned} $$

1. 8. A wattmeter connected to the 240 volt main supply conductors measures 21.6 kW. An ammeter connected in the mains supply conductors shows 100 amps. What is the power factor?

$$ \begin{aligned}\text{Power factor} = \cosine \phi &= \frac{\text{Watts}}{VA} = \frac{\text{TruePower}}{\text{Apparent}} \\ &= \frac{21600}{240} \times 100 \\ &= 0.9\end{aligned} $$

1. 9. What are the two most common types of 3-phase connections?

The two most common types of 3-phase connections are delta and star.

1. 10. Each phase of a 3-phase delta-connected generator supplies a full-load current of 80 A at a voltage of 220 V and at a power factor of 0.75 lagging. Calculate the following:

1. The line voltage
2. The line current
3. The three-phase kilovoltamperes
4. The three-phase power in kilowatts

1. a) The line voltage is equal to the phase voltage in a delta system.

$$ \begin{aligned}V_L &= V_p \\ &= 220 \text{ V} \\ I_L &= \sqrt{3} I_p\end{aligned} $$

$$ \begin{aligned}\text{b)} \quad &= 1.73 \times 80 \\ &= 138.40 \text{ A}\end{aligned} $$

$$ \begin{aligned}kVA &= \frac{\sqrt{3} V_L I_L}{1000} \\ \text{c)} \quad &= \frac{1.73 \times 220 \times 138.40}{1000} \\ &= 52.68 \text{ kVA}\end{aligned} $$

$$ \begin{aligned} P &= kVA \times \text{Power factor} \\ \text{d)} \quad &= 52.68 \times 0.75 \\ &= 39.51 \text{ kW} \end{aligned} $$

Chapter 2 Solutions

Direct Current Machines

1. Explain the operation of a commutator.

The commutator is a mechanical rectifier. It converts AC power into DC power.

2. How is ripple reduced in a DC generator?

The ripple in the DC waveform can be improved further by adding more armature coils and commutator segments. It can also be improved by adding more poles.

3. What is the term for the action of the armature currents that establish a field that distorts and weakens the stator field?

The action of the armature currents in establishing a field that distorts and weakens the stator field is called armature reaction.

4. What is one method of reducing armature reaction in a DC generator?

Compensating windings or interpoles can be used to reduce armature reaction.

5. What is the most common application in industry for a DC generator with lap windings? Wave windings?

DC generators with lap windings are used high current applications and wave windings are used in high voltage applications.

6. Explain the difference between series wound DC generators and shunt wound generators.

In series generators, the field winding is in series with the armature winding. In shunt generators, the field winding is in parallel with the armature winding.

7. The voltage of a generator changes from 130 volts at a no-load condition to 120 volts at a full-load condition. What is the voltage regulation of this machine.

$$ \begin{aligned}\text{Voltage Regulation} &= \frac{(\text{No-load volts}) - (\text{Full-load volts})}{\text{Full-load volts}} \times 100 \\ &= (130-120)/120 \\ &= 8.3\%\end{aligned} $$

8. Describe back emf for a motor.

An emf is induced in the armature winding of a DC motor as it moves through the magnetic field. This voltage is called back emf or counter emf (cemf). Back emf is counter to, or opposes the applied voltage at the armature brushes. In other words, the back emf tends to restrict current flow in the armature.

1. 9. What is the approximate magnitude of the starting current compared to the full load current?

A motor draws approximately 12 times more much more current during the first few seconds while it is started than it draws during normal full load operation.

1. 10. Describe two methods of limiting armature resistance during the startup period for motors.

Another method is the cemf starter. Cemf starters monitor back emf and adjust the amount of armature resistance as the motor speeds up. Resistance is added to the circuit through the voltage sensitive relays.

1. 11. Explain the difference between dynamic braking and regenerative braking.

The technique of dynamic braking involves using the motor as a generator. To do this, the armature is disconnected from the supply lines and connected across heavy duty resistors. The shunt field of the motor remains connected to the power source for the motor. The motor now behaves like a generator and feeds armature current to the resistors. The motor will rapidly slow down since the armature current will be in a direction to opposed the motion of the armature (Lenz's Law).

Regenerative braking also used the motor as a generator. Take for example a crane when a heavy load is lowered. When the container is lowered, the DC motor will be driven faster than its rated speed. This will causes its back emf to be greater than the applied voltage and result in generator action. This energy is put back into the electrical system. The voltage produced while the load is being lowered is of the opposite polarity to the applied voltage. This acts like an electrical brake.

1. 12. Name three types of losses in a generator or motor.

Friction losses, iron losses and copper losses.

1. 13. A shunt generator has the following particulars at full-load rating: output 15 kilowatts, volts 240, armature circuit resistance 0.5 ohms, field circuit resistance 180 ohms; friction and iron losses combined 700 watts. Calculate the generator efficiency at this load.

$$ \begin{aligned}\text{Generator Efficiency} &= \frac{\text{Output}}{\text{Input}} \\ \text{Output} &= \text{Input} - \text{Losses}\end{aligned} $$

The losses will consist of Copper loss + Iron loss + Friction loss
and the Copper loss = Armature \( I^2R \) + Field Circuit \( I^2R \)

$$ \begin{aligned}\text{Generator Output Current} &= \frac{15\,000 \text{ watts}}{240 \text{ volts}} \\ &= 62.5 \text{ amps}\end{aligned} $$

$$ \begin{aligned}\text{Field Circuit Current} &= \frac{240 \text{ volts}}{180 \text{ ohms}} \\ &= 1.24 \text{ amps}\end{aligned} $$

$$ \begin{aligned}\text{Armature Current} &= 62.50 + 1.33 \\ &= 63.83 \text{ amps}\end{aligned} $$

$$ \begin{aligned}\text{Armature } I^2R &= 63.83^2 \times 0.5 \\ &= 4074.27 \times 0.5 \\ &= 2037.13 \text{ Watts}\end{aligned} $$

$$ \begin{aligned}\text{Field Circuit } I^2R &= 1.33^2 \times 180 \\ &= 1.77 \times 180 \\ &= 318.60 \text{ Watts}\end{aligned} $$

$$ \begin{aligned}\text{Total Copper loss} &= 2037.13 + 318.60 \\ &= 2355.73 \text{ Watts}\end{aligned} $$

$$ \begin{aligned}\text{Total losses at full load} &= 2355.73 + 700 \\ &= 3055.73 \text{ Watts}\end{aligned} $$

$$ \begin{aligned}\text{Generator Efficiency} &= \frac{15 \, 000}{15 \, 000 + 3055.73} \\ &= 0.8308\end{aligned} $$

$$ \text{Generator Efficiency} = \mathbf{83.08\% \text{ (Ans.)}} $$

Chapter 3 Solutions

Alternating Current Generators

1. 1. With the aid of simple sketches, explain the differences between AC and DC generators.

AC generators are often called alternators. It has been discussed in previous modules that both DC and AC generators produce alternating current.

Figure 1
Types of Generators

The main difference between the two types of generators is that the A.C. generator uses slip rings while the DC generator uses a commutator.

Fig. 1 shows the magnetic field to be stationary (stator) while the supply conductors rotate (rotor). This is not a practical design because it is difficult to deliver large amounts of power through slip rings or commutators.

2. Explain the relationship between alternator speed, frequency, and number of pole pairs

The formula that relates number of poles, speed and frequency is:

$$ f = \frac{pN}{60} $$

where \( f \) = frequency

\( p \) = number of pairs of poles

\( n \) = rotor speed in r/min

For example, a generator connected to a hydraulic turbine turns at a speed of 300 RPM and generates 60 Hz power. The frequency formula for generators is:

$$ \begin{aligned} P &= \frac{f \times 60}{N} \\ &= \frac{60 \times 60}{300} \\ &= 12 \text{ pairs of poles} \end{aligned} $$

This would be an application for a salient pole machine.

It is not possible to construct a generator with less than 1 pole-pair ( i.e. 2 poles). If the above formula is considered, this fixes the maximum speed of rotor of a 60 Hz AC generator at 3600 r/min. The speed of a 4-pole machine would be fixed at 1800 r/min. 1800r/min and 3600 r/min are typical speeds for cylindrical rotor machines. Using the above formula, find the speed necessary for a 12 pole machine.

60 Hz systems dictate the constant speed necessary for any given generator. This speed is known as synchronous speed. AC generators are often called synchronous generators. For example, the synchronous speed of a 2- pole generator is 3600 r/min and for a 4-pole generator is 1800r/min.

3. With the aid of a simple sketch, explain the sequence of operation for a brushless excitation system for a modern alternator.

The following is a sequence of operation for a brushless excitation system for a modern alternator. A simple schematic is shown in Fig. 11.

1. The residual magnetism that exists in the stationary exciter field allows an AC voltage to be generated in the 3-phase rotating exciter winding.
2. The rotating diode rectified bridge converts the exciter AC voltage to D.C.

• c) The DC voltage causes a DC current to flow through the rotating main field, which creates the north and south poles necessary to induce voltage in the main stator windings.
• d) Three-phase power is supplied from the main stator field.
• e) The automatic voltage regulator (AVR) senses the voltage in the supply conductors. This will be lower than desired since the residual magnetism in the exciter stator field will only produce a small amount current and therefore a small main field. The AVR will convert the AC alternator output voltage to DC and supply current to the stationary exciter field. The amount of current supplied will be proportional to the difference between the actual, or measured, output voltage and the desired output voltage.

Figure 11
Brushless Excitation System

4. With the aid of a simple sketch, explain the operating principle of a direct-acting type of regulator with rolling contacts.

Fig. 16 illustrates the operating principle of a direct-acting type of regulator with rolling contacts. The two sectors, S1 and S2, are arranged so that clockwise movement of their pivots P1 and P2 cause each of them to roll over a bank of contacts to increase the resistance in the exciter field circuit; anticlockwise movement results in a decrease of resistance. The movement of the pivots is controlled by the drum D which has a restraining torque exerted by springs and an operating torque derived from the machine terminal voltage.

Figure 16
Principle of Direct-acting Rheostatic Regulator

5. Explain the advantages hydrogen cooling has over air cooling for alternators.

Hydrogen gas has several advantages over air:

• • High heat transfer coefficient, which means that it will absorb heat rapidly as it flows through the generator windings
• • Rejects this same heat more rapidly in the cooling stages of the ventilation circuit
• • Less dense than air and requires less energy to drive it through the different generator components
• • Offers less braking effect (windage) to the rotating parts of the alternator.

6. With the aid of a simple sketch, describe the seal oil system that is used for a hydrogen cooled alternator.

The seals prevent hydrogen from escaping outwards by forcing oil inwards against the seal. Seal oil is circulated from the main machine lubricating oil system, through the seals, to the hydrogen detraining tanks and then back to the main oil tank. The hydrogen detraining tanks allows any gas that may become entrained with the circulated oil to be removed before it is returned to the main storage tank. Fig. 40 shows the equipment and piping layout for a typical seal-oil system for a hydrogen cooled alternator.

Figure 40
Seal Oil System for H ₂ Cooled Alternator

1. 7. With the aid of a simple sketch, explain the steps that are taken to synchronize an incoming AC generator to the supply system.

The series of operations required to bring about the above conditions and to close the switch are known as Synchronizing . The process of synchronizing may be illustrated by the following diagrams of the incoming machine and the system voltage waves, as shown on Fig. 41.

Fig. 41(a) shows the existing system voltage wave (one phase only shown).

Referring to Fig. 41(b), the machine voltage wave is shown dotted and is out of phase and frequency.

The generators output voltage is slowly increased to equal the systems maximum voltage. This is accomplished by adjusting the field rheostat.

Fig. 41(c) shows that the machine and system voltages are now equal. The voltages are out of phase but the frequency is being increased by increasing the speed of the prime mover.

In Fig. 41(d), the machine and systems:

• • voltages are equal
• • frequencies are equal
• • are both in phase

The synchroscope shows 12 o'clock and the switch can now be closed.

1. (a) Existing system voltage wave. (One phase only shown.)

1. (b) Machine voltage wave shown dotted. Out of phase and frequency. Being built up to equal the system max. volts by adjustment of field rheostat.

1. (c) Machine voltage now equal to system. Voltage waves out of phase but frequency being increased by increasing speed of prime mover.

1. (d) Machine voltage now equal to system, in phase and with equal frequency. Synchroscope shows 12 o'clock Switch can now be closed.

Figure 41

Steps Taken to Synchronize an Incoming A-C Generator to the Supply System

Chapter 4 Solutions

Alternating Current Motors

1. 1. Describe the principle of induction and the development of torque for a squirrel-cage rotor.

A coil of wire is used as a rotor. As the applied single-phase AC voltage builds up, so does the strength of the stator field. The magnetic field cuts through the conductor. This action induces an emf in the rotor coil, which causes a current to flow in the closed loop. The amount of current depends on the resistance of the coil of wire. The current flowing in the rotor sets up a magnetic field that opposes the stator field. This principle of induction produces the torque that causes the rotor to turn.

1. 2. What is another name for the end rings of a squirrel-cage rotor?

Shorting rings.

1. 3. What construction technique is used to reduce eddy currents in the rotor of AC induction motors?

The rotor core is laminated to reduce eddy currents.

1. 4. What is the speed of a 60 Hz synchronous motor?

$$ \text{Speed} = \frac{\text{frequency} \times 60}{\text{number of pole pairs}} $$

$$ \text{Speed} = \frac{60 \times 60}{2} $$

$$ \text{Speed} = \frac{3600}{2} $$

$$ \text{Speed} = 1800 \text{ rpm (Ans.)} $$

1. 5. What is the FLA of a motor and name one common place to find the FLA of a motor?

The amount of current the motor will draw under full load (torque) conditions is called full-load amps (FLA). It is also known as nameplate amps since this is the figure that appears on the nameplate of the motor.

1. 6. What is the effect upon torque when rotor resistance is increased in a wound rotor motor?

Torque is increased as resistance is increased in a wound rotor motor.

1. 7. What type of single-phase motor works with both AC and DC?
   Universal motor.

1. 8. Match the following motors with the correct characteristics.

Motor:
Universal	( e )
Single-phase	( f )
Shaded-pole	( a )
Three-phase	( d )
Capacitor-start	( b )
Split-phase	( c )

1. 9. What type of AC motor can be used for power factor correction?
   Synchronous motor.

1. 10. What does it mean when a motor is “started across the line”?
   Full voltage is applied to the stator windings.

1. 11. What is the approximate inrush, or starting, current for an AC induction motor?
   600% percent of the full load amps.

Chapter 5 Solutions

Transformers

1. 1. How are the laminations of a transformer core electrically insulated from each other?

The laminations are coated with varnish that electrically isolates it from the adjacent winding.

1. 2. How are the laminations of a transformer core electrically insulated from each other?

Transformer construction falls into two general classifications: core-type and shell-type. In the core type the windings surround the core and in the shell-type, the core surrounds the windings.

1. 3. What is the secondary current of a 10 kVA single-phase transformer with a secondary voltage of 130 volts?

$$ \frac{10 \text{ kVA}}{130 \text{ V}} = \frac{10 \, 000 \text{ VA}}{130} = 76.92 \text{ A} $$

1. 4. How much power can be supplied by three identical 37.5 kVA transformers connected in parallel?

$$ 3 \times 37.5 \text{ kVA} = 112.5 \text{ kVA}. $$

1. 5. What is the primary voltage of a three-phase 10 MVA transformer that has a primary current of 418.38 amps?

$$ \frac{10 \text{ MVA}}{\sqrt{3} \times 418.38 \text{ A}} = \frac{10 \, 000 \, 000 \text{ VA}}{1.732 \times 418.38} = 13 \, 800 \text{ V} $$

1. 6. What is the definition of temperature rise in a transformer?

Temperature rise indicates the maximum temperature rise (in degrees Celsius) the transformer will exhibit at full load conditions.

1. 7. What is the definition of K-factor?

K-factor is an indicator of how well a transformer will tolerate harmonics.

1. 8. Briefly describe copper loss and core loss for a transformer.

Copper loss results from the resistance of the primary winding and secondary winding. Core loss is the energy required to sustain the magnetic field in the steel core.

9. What are the two most widely used cooling mediums for power and distribution transformers?

Air and oil.

10. Describe how most dry-type transformers are cooled.

There are no fans to force air into and out of the enclosure with typically no external fins or radiators. Cooler air enters the lower ports, is heated as it rises past windings, and exits the upper ventilation ports.

11. What other purposes than cooling does transformer oil serve?

Cooling oil also provides insulation and helps extinguish arcs.

12. Why is oil better than air for transformer cooling?

Oil has a greater ability to dissipate heat than air.

13. Describe the difference between a liquid-immersed, forced liquid-cooled transformer and a liquid-immersed, forced liquid-cooled, water cooled transformer.

Liquid-immersed, forced liquid-cooled. This transformer normally has only one rating. The transformer is cooled by pumping oil (forced oil) through a radiator normally attached to the outside of the tank. Also, air is forced by fans over the cooling surface.

Liquid-immersed, forced liquid-cooled, water cooled. This transformer is cooled by an oil/water heat exchanger normally mounted separately from the tank. Both the transformer oil and the cooling water are pumped (forced) through the heat exchanger to accomplish cooling.

14. A current transformer is labelled with a ratio 200:5. What current will flow in the secondary if 150 amps flows in the primary?

$$ \frac{200}{5} = \frac{150}{\text{secondary current}} \rightarrow \text{secondary current} = \frac{150}{40} = 3.75 \text{ amps} $$

15. What is the most common three-phase transformer winding arrangement to supply a 208/120 volt unbalance load?

three-phase four-wire star connected grounded neutral system.

16. What letters are used to designate the high voltage side and low voltage side of a transformer?

H is used to designate the high voltage side and X is used to designate the low voltage side.

17. What does the presence of carbon monoxide in transformer oil indicate?

The presence of carbon dioxide would indicate the degradation of winding insulation.

18. Name two types of overcurrent devices.

Breakers and fuses.

Chapter 6 Solutions

Electrical System Protection

1. 1. Describe the difference between continuous current and interrupting capacity.
   Interrupting capacity is the capability of a device such as a breaker or a fuse to safely open an electrical circuit that is carrying a dangerous amount of current. The continuous rating of a breaker or fuse refers to its ability to continuously carry a certain amount of current.

1. 2. What is the main difference between breakers and fuses?

The main difference between breakers and fuses is that breakers are designed to be reused if a fault occurs.

1. 3. Briefly describe the inverse time principle for breakers and fuses.

The larger the current, the faster the protective device will operate to clear the overcurrent fault or short circuit current.

1. 4. True or False. The inrush current for devices such as motors and transformers is significantly lower than the normal rated current for the devices.

False.

1. 5. Fill in the blanks. _____ fuses are not current limiting and as a result limit the duration of a fault on the electrical system, not the magnitude. _____ fuses are fuses that, when their current responsive elements are melted by a current within the fuse's specified current limiting range, abruptly introduces a high resistance to reduce current magnitude and duration, resulting in subsequent current interruption.

Expulsion fuses are not current limiting and as a result limit the duration of a fault on the electrical system, not the magnitude.

Current limiting fuses are fuses that, when their current responsive elements are melted by a current within the fuse's specified current limiting range, abruptly introduces a high resistance to reduce current magnitude and duration, resulting in subsequent current interruption.

1. 6. What is the maximum interrupting capacity you would expect to find in a 600 volt Class J fuse?

200,000 A.

1. 7. Briefly describe the thermal/magnetic principle commonly used in breakers.

Thermal mechanism is a bimetal strip that heats during an overload condition.

Magnetic mechanism is a solenoid or electromagnet.

1. 8. Current limiting reduce the magnitude of current and clear the fault in less than \( \frac{1}{2} \) cycle. How much time is required for \( \frac{1}{2} \) cycle on a 60 Hz system?
   60 cycles per second = 120 ( \( \frac{1}{2} \) ) cycles per second = \( 1/120 \) seconds = 0.0083 seconds.
2. 9. Describe the principle behind a compressed air circuit breaker.
   The breaker uses an assembly of parts and devices, which provide compressed air for the operation of a circuit breaker. The arc is not stretched in this type of breaker. The use of a high-speed flow of cool air through the unit removes ionized gases that are produced by the arc. Air is the insulating medium for extinguishing the arc. The gases that are removed are replaced by cooler air, which also helps in cooling the arc.
3. 10. Give two functions of the oil used in an oil circuit breaker.
   The oil serves three purposes:
4. • • To insulate the arc from grounded (earthed) metal and to confine it.
   • • To provide adequate cooling for arc the breaker contacts.
   • • To provide some arc control by the natural pressure ( i.e. head) of the oil.
5. 11. SF ₆ is the symbol for sulphur hexafluoride. Is SF ₆ a gas or a liquid?
   SF ₆ is a gas.
6. 12. Is the dielectric of a vacuum higher or lower than ambient air?
   A vacuum has a higher dielectric strength.
7. 13. What is the purpose of an arc chute?
   An arc chute directs the arc and ionized gases away from the current carrying contacts of the breaker.
8. 14. Name the two main categories of electromechanical relays.
   Attraction type and induction type.
9. 15. Name two different types of instrument transformers that are used to provide current and voltage signals to protective relays.
   Current transformers (CTs) and potential transformers (PTs).
10. 16. What type of protective relaying scheme is shown in the figure below?
    Differential current.
11. 17. What type of protective relaying would be used to prevent an alternator from motoring?
    Reverse power.

18. A motor has a service factor of 1.10. What does this mean?

This is a rating above the full load A that the motor can safely operate. For example, many motors have a service fact of 1.10, which means the motor can handle 110% of its full load current.

19. What is meant be the term “selectivity” as it pertains to protective relaying?

The relay must be able to differentiate between a condition that requires immediate attention and a condition that requires a time delay. The relay must also be selective only in the zone it is protecting. In other words, a relay must not take action if a fault occurs in another protected zone.

20. A current transformer has a ratio of 100:5. If 80 A are flowing in the primary of the transformer, what amount of current is flowing in the secondary?

4 A.

Chapter 7 Solutions

Air and Gas Compression

1. 1. What are the two main types of compressors? Describe how they work and provide a further breakdown of each type of compressor.

Compressors can be broken down into two types:

1. 1. Positive displacement
2. 2. Dynamic

Positive displacement compressors function by drawing the fluid into an enclosed space whose volume is then physically decreased. The best known type is the reciprocating compressor. Other types are known as rotary compressors which include screw, sliding vane, lobe and liquid sealed compressors.

Dynamic compressors make use of blading to increase fluid velocity which is converted into pressure by diffusion (that is, a reduction in area). This type of compressor breaks down into two major types: centrifugal and axial.

1. 2. Describe three possible configurations for multi-stage compressor cylinders in reciprocating compressors.

Multi-stage compressor cylinders may be arranged in a number of different ways. They may be:

1. 1. Side by side in either a horizontal or vertical plane
2. 2. In line with the pistons actuated by a common piston rod
3. 3. Radially arranged with the connecting rods coupled to a common crankpin.

1. 3. Into what four types can rotary compressors be divided? Provide a brief description for each one.

From the point of view of classification, rotary air compressors can be conveniently divided into four types, namely:

1. 1. Sliding vane, in which longitudinal vanes slide radially in a rotor mounted eccentrically in a casing
2. 2. Roots or lobe, in which two or more lobed impellers revolve within a casing
3. 3. Screw type, in which two intermeshing screws rotate in a casing
4. 4. Liquid sealed, in which a liquid displaces air within a rotating element

1. 4. Using a simple sketch, explain the operating principle of a centrifugal compressor wheel. Show the gas flow entering and exiting the wheel.

The disc is connected to the shaft, as shown in the sketch below. Air enters at the center or eye of the impeller and is forced out at high speed and increased pressure at the impeller rim because of centrifugal force. After leaving the impeller, the air passes through the casing where further conversion of speed to pressure occurs.

Centrifugal Compressor Wheel

1. 5. Describe the principle and operation of a desiccant dryer. What are the two ways in which moisture is removed?

The desiccant dryer works on the principle that certain materials, called desiccants, can absorb large amounts of moisture in the air. The moisture is then removed by a heatless or heat-activated process.

Reactivation of the desiccant or removal of the moisture is done by a heatless process or by the application of heat. Normally two towers are used with one tower used to dry the compressed air. When it becomes saturated, the compressed air is switched to the second tower while the first one is dried out.

A heatless dryer uses some of the compressed air, known as purge air, to dry out the saturated dryer. As the compressed air expands, its relative humidity is drastically lowered and it is able to reabsorb the moisture from the desiccant. The disadvantage of this method is that usually about 15-18% of the compressed air is required which reduces the capacity by that amount, increasing the cost. One advantage is that they can be pneumatically controlled which makes them suitable for hazardous areas.

Heat reactivated dryers operate by heating the saturated desiccant and then passing a small amount of compressed or purge air over the desiccant bed to carry the moisture away.

6. Explain the functions of receivers, air filters and separators.

A receiver provides a reserve capacity for the compressed air and also serves to dampen out any pressure pulsations, two features which together will result in a steady air flow, gradual pressure changes and a smooth regulation of compressor output.

Air filters ensure that clean air is provided to the compressor. Atmospheric air is never clean, although to the naked eye it might appear to be so. Suspended foreign matter is always present. Very fine particles of dust, sand and grit are always entrained in the atmosphere.

For lubricated compressors, an air-oil separator is installed after the compressor to remove the lubricant injected into the compressor.

7. List four things that should be checked after a refrigeration system has been started.

Regulation systems fall conveniently into four basic types:

1. 1. Start-and-stop control: a compressor so controlled would start up at some predetermined minimum receiver pressure, run at a constant speed and then stop at some predetermined maximum receiver pressure.
2. 2. Variable speed control: a throttling governor regulates the speed of the compressor, and consequently its output, by regulating the speed of the driving unit
3. 3. Constant speed control: constant-speed units require some form of unloading device which will permit the driving unit to run at full speed without delivering more air than is required
4. 4. Dual control: this type of control uses a combination of both start and stop as well as variable speed control.

8. State the three basic functions of the control system and give some examples.

Control systems have three basic functions:

1. 1. Control: control of the compressor by start/stop, variable speed, etc.
2. 2. Protection: protective devices such as relief valves, high/low pressure alarms and shutdowns, low lube oil pressure, high discharge temperature
3. 3. Monitoring: pressure, temperature and oil level monitoring

9. Explain the effect of altitude on air compressors.

The density of air decreases with altitude. Consequently the mass of air handled by a compressor will be reduced with altitude and, in addition the air will need to be compressed through an increased pressure range to maintain the same discharge pressure.

These effects of altitude result in an increased power input to a particular compressor per unit mass of air delivered. Moreover, the increased ratio of compression is accompanied by an increase in air temperature and this may result in excessively high discharge temperatures, an operating factor which is extremely deleterious both to maintenance and lubrication.

10. Describe three possible causes for high discharge temperature on a reciprocating compressor.

High discharge temperature can be caused by (choose any three):

1. 1. Failure of one or more discharge valves
2. 2. Leaky discharge valves resulting from the presence of carbon deposits
3. 3. Inadequate cylinder cooling due to scale in the water jackets, intercooler pipes and other water-cooled parts
4. 4. Inefficient cooling from either the intercooler or the aftercooler.

Chapter 8 Solutions

Refrigeration Systems and Equipment

1. State six properties of an ideal refrigerant.

The ideal refrigerant should possess the following properties:

• • Low boiling temperature at atmospheric pressure
• • High latent heat capacity
• • Moderate condensing temperature at a relatively low pressure
• • Low specific volume, at compressor suction pressure, to minimize the required compressor size
• • Stable chemical composition
• • Inoffensive odour that will not taint stored goods
• • Non-poisonous nature
• • Non-corrosive properties
• • Non-flammable and non-explosive nature when mixed with air
• • Low cost and readily available

2. a) List the four basic components of a closed vapour compression refrigeration system.

b) Use a simple diagram to show the relative pressures and temperatures between each component.

a) A closed-compression system consists of four essential parts:

• • A compressor
• • A condenser and receiver
• • An evaporator or cooling tank
• • Connecting pipes and a regulating valve

b) The following figure shows an actual example using the refrigerant (Freon 12) with the relative pressures and temperatures at various points in the system.

3. Explain the difference between direct and indirect cooling.

With an indirect system of cooling, the evaporating coils are led through a tank containing a liquid such as brine (a strong solution of a salt in water with a low freezing point). The ammonia, in evaporating, extracts heat from and cools the brine. The brine is then pumped through the cold rooms and carries heat from these rooms back to the evaporator.

With direct cooling, the evaporating coils are placed directly in the rooms to be cooled. Liquid refrigerant is evaporated in them without the intervention of brine or any other carrying medium.

4. What two factors does the ammonia absorption system depend on?

Operation of the ammonia absorption system of refrigeration depends upon:

• • The affinity of water and ammonia for each other
• • The process of fractional distillation whereby the application of heat to a mixture of water and ammonia (aqua ammonia) drives off the ammonia as a gas leaving the water behind

5. With the aid of a simple sketch, describe a cascade system.

Cascade Refrigeration Systems

Another means of producing very low temperatures is the cascade system where two refrigerants are used with two separate cooling loops. As shown in Fig. 8, the low-temperature stage is the one that provides the intended cooling. The condenser of the low-temperature stage is connected to the evaporator of the high-temperature stage.

The refrigerant used for the low-temperature stage can be chosen to operate at high pressure and low temperature, while the refrigerant for the high temperature stage can be a normal refrigerant.

For the cascade system to operate properly there must be an overlap between the condenser temperature of the low stage and the evaporator temperature of the high stage. This needs to be somewhat lower, around 5-10°C, to permit adequate heat transfer.

Cascade Refrigeration System

6. Explain what a hermetic system is and why it is used.

Hermetic Systems

Since refrigerant leakage is an undesirable consequence of refrigeration systems, efforts have been made to encapsulate components into sealed containers, and thus produce what are referred to as hermetic systems.

Hermetic design is most often applied to small compressors used for domestic refrigerators, small air conditioners, and other small to medium size refrigeration units. The motor and compressor are contained in a sealed unit that can be welded to other components and piping. Both the compressor and motor are in contact with the refrigerant, but no shaft seals are required. Since ammonia attacks the copper windings of the motor, refrigerants are generally limited to the halocarbon types.

Some compressors are semi-hermetic with covers that allow access to pistons and cylinders for servicing and maintenance. Large compressors are normally of the open type.

1. 7. With the aid of a simple sketch, describe a rotary screw compressor.

Screw Compressor

Screw compressors, also known as helical rotary compressors, have become more common in the past few decades and have at least partially replaced reciprocating compressors in many refrigeration applications. With improvements in their design, they are able to meet high compression requirements with flexible capacity control.

A twin-screw design is the most common arrangement. It consists of two rotational elements, or rotors, that mesh together and provide a decreasing area from one end of the rotor to the other. This action produces a corresponding pressure increase. One rotor has four convex lobes that mesh with six concave lobes on the other rotor. Other combinations, such as 3/5 and 5/7, are also sometimes used. The rotors are located in a pressurized housing.

The compression ratio is determined by the ratio of the volume of the cavity at the discharge end to the volume of the cavity at the suction end. Capacity can be controlled by a sliding valve mechanism that alters the volume of the suction port or by controlling the speed of the rotor using a variable speed drive.

Rotary Screw Compressor

1. 8. Explain what evaporators are used for and describe the two types of evaporators.

Evaporators

The evaporator is the heat exchanger where cooling occurs. The cooling may be direct in which the evaporator is placed inside the room or container that needs to be cooled. It may also be indirect, in which case it cools a fluid that is circulated to the location that needs to be cooled. There are two principal types of evaporators:

• • Dry expansion
• • Flooded

Dry Expansion Evaporators

In this type, the ammonia is evaporated from a liquid into a gas in banks of pipe coils. The refrigerant is entirely evaporated at the exit of the coils.

Flooded Evaporators

The disadvantage of the dry evaporator is that its surfaces are not constantly wetted by the boiling refrigerant. This has led to the development of the flooded type evaporator in which some form of tank or header, called an accumulator, keeps the inside of the evaporator surface flooded with liquid. While this requires a greater initial charge of liquid refrigerant, the higher heat transfer rate per area of evaporator surface is a decided advantage.

9. List the three types of condensers and briefly describe them.

Types of condensers in use include:

• • Double-Pipe Condensers
• • Evaporative (Atmospheric) Condensers
• • Shell-and-Tube Condensers

Double-Pipe Condensers

The double-pipe condenser consists of two pipes, one inside the other, placed in vertical rows. The cooling water circulates inside the smaller pipe, and the ammonia circulates in the annular space between the inner and outer pipes. The ammonia gas enters at the top, and the liquid is drawn off at the bottom. The gas and the cooling water travel in opposite directions so that the coldest water is in contact with the cool liquid ammonia, and the hottest water is in contact with the hot gas.

Evaporative (Atmospheric) Condensers

In the simplest evaporative condenser, water trickles over the coils and drains away to waste or to be cooled and used again where water is scarce. Since the counter-current principle cannot be used (the water must flow from top to bottom), the coils are rearranged so that the hot gas enters at the bottom.

Shell-and-Tube Condensers

The shell-and-tube condenser consists of an outer shell containing a number of tubes expanded into headers at each end. It uses a two-pass design; the water flows through the tubes, and the refrigerant condenses outside the tubes.

This is the most common type, being quite efficient and compact. It is made in a wide variety of sizes, and units may be stacked one above the other in multiples to increase capacity. Vertical units are also made, usually having gravity water-flow down the tubes.

Chapter 9 Solutions

Refrigeration Safety, Control and Operation

1. 1. 1.
      1. a) What is the main code applicable to refrigeration in Canada?
      2. b) Describe two criteria that are used to classify refrigeration requirements as it relates to design.
   
   
   2. a) The primary standard in Canada is CSA B52-99 Mechanical Refrigeration Code which is produced by the Canadian Standards Association.
   
   
   3. b) The following criteria are used to classify refrigeration requirements (any two can be mentioned):
      1. 1. Occupancy – this is a classification of the location of the refrigeration system based upon occupancy is divided into residential, commercial, industrial and mixed occupancy
      2. 2. Type of refrigeration system – types of systems are direct (evaporator or condenser is in direct contact with the air or other substance to be cooled or heated), double direct (two refrigerant systems that are connected with the primary circuit directly cooling the substance such as in a cascade system) or indirect ( a secondary coolant loop is used to cool the substance). Along with this classification is one that deals with the leakage probability or the likelihood that a leakage of refrigerant could enter an occupied area
      3. 3. Classifications of refrigerant – refrigerants are classified into safety groups according to their flammability and toxicity using a matrix developed by ANSI/ASHRAE Standard 34.
2. 
   
3. 2. Describe the purpose of a temperature actuated control (thermostat) and list three methods that can be used to achieve this.

The purpose of the thermostat is to start and stop the compressor or regulate its output when the temperature changes and approaches the upper or lower limit.

Thermostats may be operated by (choose any three):

• • Movement of a bimetallic element
• • Expansion of a fluid (liquid or gas)
• • Vapour pressure of a volatile fluid
• • Electrical resistance
• • Electronic signals

1. 1. 3. a) State the two pressure limits present in a refrigeration system and explain how they function.
   2. b) At what pressures are they set?
2. 1. a) The two pressure limits are the high pressure safety cutout and the low pressure safety cutout.
   2. b) The high pressure limit is set so that the compressor is prevented from producing a pressure in excess of the high side design pressure. It is set at a pressure not more than 90% of the system high side design pressure.

The low pressure operated safety switch is used to protect the system against a lower than normal suction pressure. The low pressure limit is set according to operational requirements.

4. Describe the function and operation of a thermostatic expansion valve.

The thermostatic expansion valve is a liquid refrigerant control valve which maintains a constant pressure in the evaporator during the time the compressor is operating, regardless of load. In addition, the valve automatically shuts off the liquid flow when the compressor stops. A thermostatic element ensures that the vapor leaves at a slightly superheated state to prevent liquids leaving the evaporator.

The valve contains a valve stem attached to a bellows or diaphragm in the underpart of the valve housing. A spring exerting a downward force on the diaphragm tends to open the valve. This force is counteracted by the evaporator pressure acting upward against the diaphragm, which tends to close the valve. The spring tension is adjusted so that during operation the two forces balance each other and the valve is opened sufficiently to allow enough liquid to flow into the evaporator to maintain the desired pressure and, therefore, the desired temperature.

5. List and describe three methods for varying the capacity of compressors.

The following methods are used for varying the capacity of compressors (any three can be mentioned):

1. 1. Unloaders – Cylinder unloaders work by keeping the intake valves of one or more cylinders in the open position, preventing the compression of the vapour drawn in during the suction stroke.
2. 2. Cylinder bypass – Another method of controlling the capacity of reciprocating compressors is by bypassing the discharge from one or more cylinders back to the suction side of the compressor.
3. 3. Hot gas bypass – This method bypasses some of the discharge gas through a solenoid valve located in the bypass line which is energized by the pressure or temperature at the compressor inlet, allowing some hot gas to go directly into the suction line.

1. 4. Speed control – the speed of a compressor can be controlled by varying the speed of the driver.
2. 5. Suction throttling - Suction throttling is accomplished by a butterfly damper installed at the inlet to a centrifugal compressor.
3. 6. Variable guide vanes - Pneumatic vane operators are used on centrifugal compressors to change the angle of the flow at the inlet of the compressor to better match that of the impeller vanes.

6. List four things that should be checked after a refrigeration system has been started.

The following items should be checked (any four can be chosen):

1. 1. 1. Check the whole system over; observe temperature and pressure gages.
   2. 2. Check controls for proper operation and reset if necessary.
   3. 3. Check superheat setting of thermostatic expansion valve and adjust if necessary.
   4. 4. Check the operation of the water regulating valve in the water supply line to the condenser.
   5. 5. If the compressor discharge head pressure is too high, adjust for increased condenser water flow.
   6. 6. Check the liquid refrigerant sight glass for bubbles; if any appear, refrigerant should be added.
   7. 7. Check the oil level in the crankcase after the compressor has run for about 15 to 20 minutes.
   8. 8. If the compressor is pressure lubricated, check the oil pressure.
   9. 9. Check the entire system with a leak detector.
2. 7. a) What are the two types of leak detectors used for halogen refrigerants?
   b) Explain briefly how they work.
3. a) The two types of leak detectors suitable for halogen refrigerants are the halide torch detector and the electronic leak detector.
4. b) The halide torch consists of a special burner connected to an acetylene or propane cylinder. Combustion air is drawn to the burner through a search tube connected to the side of the detector. To search for leaks, the burner is lit and the end of the tube is slowly moved over and around all the joints or any other place where a leak can be suspected. Escaping Freon will be drawn into the tube, travel to the burner and cause the flame to turn a brilliant green.

The electronic leak detector draws vapour through a tube fitted with a sniffer at the end. The sniffer is moved over any area where a leak may be suspected. It measures the electrical resistance of the vapour sample. As long as air is drawn into the detector, it does not react. But as soon as the sample contains refrigerant,

even a minute amount, the change in resistance of the sample and the resulting change in current flow causes the detector to react. It indicates the presence of the refrigerant either on a meter or by activating a light or a buzzer.

8. What are two reasons for drying a refrigeration system?

Two reasons for drying a refrigeration system are:

1. 1. When exposed to low temperatures produced by the system, moisture entrained in the refrigerant causes icing or “freeze-up” at the expansion valve.
2. 2. In systems using Freon as the refrigerant, acid is formed that reacts with oil to produce sludge and corrode metal parts.

9. Describe a problem that can be encountered by:

a) Insufficient refrigerant

b) Overcharging

1. a) The proportions of the charge contained in the condenser and evaporator respectively, when the plant is working must also be correct. The condenser must contain enough refrigerant to ensure that the temperature of the liquid formed is as near as possible to that of the cooling medium. The evaporator must be supplied with sufficient refrigerant to keep the working surface covered with liquid refrigerant, and so maintain the correct degree of superheat. If a plant is undercharged with refrigerant, either the condenser or evaporator or both will not contain enough refrigerant. If the condenser is undercharged, the liquid will leave the unit at a higher than designed temperature, and, if the evaporator is undercharged, the vapor leaving it will be too highly superheated, causing the compressor to run hot and the temperature of the compressor discharge to be abnormally high. Adjustment of the regulator, under these conditions, can do no more than transfer the state of undercharge from the condenser to the evaporator or vice versa. Undercharge of refrigerant may well be due to refrigerant leakage from pipes, pipe joints or shaft seals.
2. b) If the plant is overcharged and the regulator adjusted to make the compressor operate at its normal correct temperature by preventing wet vapor entering it, the condenser pressure may rise to a dangerous extent. Overcharging is likely due to the addition of too much refrigerant when topping up so this needs to be watched carefully when this is done.

Chapter 10 Solutions

Refrigeration Calculations

1. 1. Describe and illustrate the heat engine and the basic refrigeration machine.

The heat engine takes heat from a hot reservoir (usually provided by combustion of a fuel) and extracts work after which the remaining heat is transferred to a cold reservoir. In a refrigeration and heat pump cycle, heat is removed from a reservoir (that is the substance to be cooled), work is done on the refrigerant and heat is expelled to a hot reservoir.

1. 2. Draw a Rankine form of refrigeration cycle on a pressure-enthalpy diagram. Explain each stage as applied to a practical refrigeration machine.

The four stages are:

Stage 1 - 2 Expansion through the expansion valve without change in heat content (kJ/kg of refrigerant).

Stage 2 - 3 Evaporation at constant pressure.

Stage 3 - 4 Adiabatic compression.

Stage 4 - 1 Condensing at constant pressure.

3. Using the diagram drawn for Question 2, construct a heat balance for the refrigeration cycle and from this define the following quantities for this cycle (use enthalpy units):

1. Work done in compression
2. Refrigerating effect
3. Coefficient of Performance

1. The work done in compression will be \( (H_4 - H_3) \) .
2. The heat absorbed by the refrigerant in passing through the evaporator or the refrigerating effect will be given by the difference \( (H_3 - H_2) \) kJ/kg.

$$ \text{Coefficient of Performance} = \frac{\text{Heat absorbed in evaporator}}{\text{Heat equivalent of compression work}} $$

c)

$$ COP = \frac{(H_3 - H_2)}{(H_4 - H_3)} $$

4. The compressor of a refrigerating machine is driven by a motor of 7.5 kW output.

1. If the COP is 3.5, calculate the capacity of the machine expressed in tonnes of ice per day from and at \( 0^\circ\text{C} \) .
2. Find the mass of ice at \( -10^\circ\text{C} \) that this machine can make per day from water at \( 20^\circ\text{C} \) .

Take specific heats of water as 4.183 kJ/kg and ice as 2.135 kJ/kg and latent heat of fusion as 335 kJ/kg.

(a) Capacity of machine

The coefficient of performance is related to the refrigerating effect and the work done by the equation

$$ \begin{aligned}\text{COP} &= \frac{\text{Refrigerating effect/min}}{\text{Compressor work/min}} \\ &= \frac{(H_3 - H_2)}{(H_4 - H_3)}\end{aligned} $$

The motor output of 7.5 kW = \( 7.5 \times 60 \) kJ/min

and COP = 3.5 (given)

Refrigerating effect/min = \( \text{COP} \times \text{Motor output} \)

Refrigerating effect/min = \( 3.5 \times 7.5 \times 60 \)

Refrigerating effect/min = 1575 kJ/min

Machine capacity:

$$ \begin{aligned}\text{Capacity} &= \frac{\text{Refrigerating effect/min}}{\text{Heat absorption rate/tonne}} \\ \text{Capacity} &= \frac{1575 \text{ kJ/min}}{233 \text{ kg/min/t}} \\ \text{Capacity} &= 6.76 \text{ t of refrigeration (Ans.)}\end{aligned} $$

(b) Mass of ice

1 kg of ice at \( -10^{\circ}\text{C} \) from water at \( 20^{\circ}\text{C} \) requires the following heat removal:

$$ \text{Water at } 20^{\circ}\text{C to water at } 0^{\circ}\text{C} = 20^{\circ}\text{C} \times 4.183 \text{ kJ/kg} $$

$$ \text{Water at } 20^{\circ}\text{C to water at } 0^{\circ}\text{C} = 83.66 \text{ kJ/kg} $$

$$ 1 \text{ kg water at } 0^{\circ}\text{C to } 1 \text{ kg ice at } 0^{\circ}\text{C} = 335 \text{ kJ} $$

$$ 1 \text{ kg ice at } 0^{\circ}\text{C to } 1 \text{ kg ice at } -10^{\circ}\text{C} = [0^{\circ}\text{C} - (-10^{\circ}\text{C})] \times 2.135 \text{ kJ/kg} $$

$$ 1 \text{ kg ice at } 0^{\circ}\text{C to } 1 \text{ kg ice at } -10^{\circ}\text{C} = [0^{\circ}\text{C} + 10^{\circ}\text{C}] \times 2.135 \text{ kJ/kg} $$

$$ 1 \text{ kg ice at } 0^{\circ}\text{C to } 1 \text{ kg ice at } -10^{\circ}\text{C} = 10^{\circ}\text{C} \times 2.135 \text{ kJ/kg} $$

$$ 1 \text{ kg ice at } 0^{\circ}\text{C to } 1 \text{ kg ice at } -10^{\circ}\text{C} = 21.35 \text{ kJ/kg} $$

$$ \text{Total heat removed} = 83.66 \text{ kJ/kg} + 335 \text{ kJ/kg} + 21.35 \text{ kJ/kg} $$

$$ \text{Total heat removed} = 440.01 \text{ kJ/kg} $$

$$ \text{Refrigerating effect of this machine} = 1575 \text{ kJ/min} $$

$$ \text{kg ice made/min} = \frac{\text{Refrigerating effect of this machine}}{\text{Total heat removed}} $$

$$ \text{Ice made/min} = \frac{1575 \text{ kJ/min}}{440.01 \text{ kJ/kg}} $$

$$ \text{Ice made/min} = 3.58 \text{ kg} $$

$$ \text{Ice made in 24 h} = \frac{\text{Ice made/min} \times \text{min/hr} \times \text{hours}}{1000 \text{ kg/tonne}} $$

$$ \text{Ice made in 24 h} = \frac{3.58 \text{ kg/min} \times 60 \times 24}{1000 \text{ kg/tonne}} $$

$$ \text{Ice made in 24 h} = \frac{5155.2 \text{ kg}}{1000 \text{ kg/tonne}} $$

$$ \text{Ice made in 24 h} = \mathbf{5.16 \text{ t}} \text{ (Ans.)} $$

1. 5. In a vapour-compression refrigerating system working under ideal conditions with ammonia as the refrigerant, the evaporator pressure is 108 kPa gauge. The ammonia leaves the evaporator with 100% quality to enter compressor suction. The compressor raises the pressure of the vapour at constant entropy to 864 kPa gauge. The ammonia leaves the condenser as a sub-cooled liquid at 20°C.

Find the following:

1. Heat rejected in the condenser
2. Work input of the compressor
3. Heat absorbed in the evaporator
4. COP of the cycle
5. Kg of ammonia to be circulated per tonne of refrigeration
6. kW per tonne of refrigeration needed to run the cycle

Note: Quantities (a), (b) and (c) should be stated in kJ/kg of refrigerant circulated. Atmospheric pressure can be taken as 101.3 kPa. Assume heat at end of compression as 1640.8 kJ/kg.

1. (a) Heat rejected in the condenser

The enthalpy at the entry to the condenser is the same as the exit of the compressor which has been given as 1640.8 kJ/kg. At the exit of the condenser, the enthalpy is the same as that of saturated ammonia at a pressure of 864 kPa gauge or 965.3 kPa abs and is found by interpolating:

$$ \text{Enthalpy at 965.3 kPa} = \text{Enthalpy at 913.4} + \frac{51.9}{58.8} (\text{Enthalpy at 972.2} - \text{Enthalpy at 913.4}) $$

$$ \text{Enthalpy at 965.3 kPa} = 284.6 \text{ kJ/kg} + 0.8827 (294.1 \text{ kJ/kg} - 284.6 \text{ kJ/kg}) $$

$$ \text{Enthalpy at 965.3 kPa} = 284.6 \text{ kJ/kg} + 0.8827 (9.5 \text{ kJ/kg}) $$

$$ \text{Enthalpy at 965.3 kPa} = 284.6 \text{ kJ/kg} + 8.39 \text{ kJ/kg} $$

$$ \text{Enthalpy at 965.3 kPa} = 292.99 \text{ kJ/kg} $$

The temperature at this point is interpolated by the same amount to be 23.8°C. Since the water is sub-cooled to 20°C, the sensible heat to do this is:

$$ \text{Sensible heat required} = (23.8 - 20) \times 4.183 \text{ kJ/kg} $$

$$ \text{Sensible heat required} = 3.8 \times 4.183 \text{ kJ/kg} $$

$$ \text{Sensible heat required} = 15.9 \text{ kJ/kg} $$

The total amount of heat rejected in the condenser:

$$ \text{Heat rejected} = (1640.8 - 292.99) + 15.9 $$

$$ \text{Heat rejected} = 1347.81 + 15.9 $$

$$ \text{Heat rejected} = \mathbf{1363.71 \text{ kJ/kg}} \text{ (Ans.)} $$

(b) Work input of the compressor

The enthalpy at the entry to the compressor is equal to the exit of the evaporator where the refrigerant leaves at a pressure of 108 kPa gauge or 209.3 kPa abs and 100% quality. Again by interpolation, the enthalpy of vapor at this pressure is

$$ \text{Enthalpy at } 209.3 \text{ kPa} = \text{Enthalpy at } 207.7 + \frac{1.6}{18.8} (\text{Enthalpy at } 226.5 - \text{Enthalpy at } 207.7) $$

$$ \text{Enthalpy at } 209.3 \text{ kPa} = 1422.7 \text{ kJ/kg} + 0.0851 (1425.3 \text{ kJ/kg} - 1422.7 \text{ kJ/kg}) $$

$$ \text{Enthalpy at } 209.3 \text{ kPa} = 1422.7 \text{ kJ/kg} + 0.0851 (2.6 \text{ kJ/kg}) $$

$$ \text{Enthalpy at } 209.3 \text{ kPa} = 1422.7 \text{ kJ/kg} + 0.2213 \text{ kJ/kg} $$

$$ \text{Enthalpy at } 209.3 \text{ kPa} = 1422.92 \text{ kJ/kg} $$

Heat equivalent of compression:

$$ \text{Work input} = 1640.8 \text{ kJ/kg} - 1422.92 \text{ kJ/kg} $$

$$ \text{Work input} = \mathbf{217.88 \text{ kJ/kg}} \text{ (Ans.)} $$

(c) Heat absorbed in the evaporator

The enthalpy at the entry to the evaporator is the same as at the exit of the condenser.

The enthalpy of the sub-cooled water is:

$$ \text{Enthalpy} = \text{Enthalpy at exit of condenser} - \text{Sensible heat to subcool the water} $$

$$ \text{Enthalpy} = 293 \text{ kJ/kg} - 15.7 \text{ kJ/kg} $$

$$ \text{Enthalpy} = 277.3 \text{ kJ/kg} $$

This is also the enthalpy at the entry of the evaporator since expansion is at constant enthalpy. Since the enthalpy at the entry of the compressor is 1422.92 kJ/kg, the heat absorbed by the evaporator is:

$$ \text{Heat absorbed by evaporator} = \text{Enthalpy of vapour} - \text{Enthalpy of subcooled water} $$

$$ \text{Heat absorbed by evaporator} = 1422.92 \text{ kJ/kg} - 277.3 \text{ kJ/kg} $$

$$ \text{Heat absorbed by evaporator} = \mathbf{1145.62 \text{ kJ/kg}} \text{ (Ans.)} $$

(d) COP of the cycle

$$ COP = \frac{\text{Heat extracted}}{\text{Work done}} $$

$$ COP = \frac{1145.62 \text{ kJ/kg}}{217.88 \text{ kJ/kg}} $$

$$ COP = 5.26 \text{ (Ans.)} $$

(e) Kg of ammonia to be circulated/hour/tonne of refrigeration:

$$ \text{Refrigerant circulated kg/t} = \frac{\text{Heat absorption rate/tonne}}{(H_3 - H_2)} $$

$$ \text{Refrigerant circulated kg/t} = \frac{233 \text{ kg/min/t}}{1422.92 \text{ kJ/kg} - 277.3 \text{ kJ/kg}} $$

$$ \text{Refrigerant circulated kg/t} = \frac{233 \text{ kg/min/t} \times 60 \text{ min}}{1145.62 \text{ kJ/kg}} $$

$$ \text{Refrigerant circulated kg/t} = \frac{13980 \text{ kg/hr/t}}{1145.62 \text{ kJ/kg}} $$

$$ \text{Refrigerant circulated kg/t} = 12.20 \text{ kg/hr/t (Ans.)} $$

(f) kW per tonne of refrigeration needed to run the cycle

Since the work input and the amount of refrigerant being circulated is known, the power required is:

$$ \text{Power required (kJ/h)} = \text{Work output} \times \text{Refrigerant circulated kg/t} $$

$$ \text{Power required (kJ/h)} = 217.88 \text{ kJ/kg} \times 12.2 \text{ kg/hr/t} $$

$$ \text{Power required (kJ/h)} = 2658.14 \text{ kJ/hr/t} $$

$$ \text{But } 1 \text{ kW} = 3600 \text{ kJ/h} $$

$$ \text{Power required (kJ/h)} = \frac{2658.14 \text{ kJ/hr/t}}{3600} $$

$$ \text{Power required (kJ/h)} = 0.7384 \text{ kW/t (Ans.)} $$