Energy and Work

Energy has many forms, including: mechanical, electrical, thermal, chemical, and nuclear. These types of energy are used in many ways, such as doing work directly, keeping things warm or cool, powering machinery, or carrying information. The various forms of energy are frequently converted from one form to another; for example, when electricity is used to warm up a room.

Energy can be defined as the capacity for doing work. Whenever a force moves something, work is done. If an object has energy, it is capable of doing work by itself. For example, a truck coasting along a roadway possesses energy due to its motion. The truck is moving its load from one point to another through a distance without its engine. It is doing work.

There are two general forms of energy that can do work:

• Energy due to motion is called kinetic energy. In the above example, the truck can do work because it is moving. Before it comes to rest, it can push against and displace almost any object with which it collides. Also, it will do work as it is being stopped.

• Energy due to position is called gravitational potential energy. An object that can fall has this form of energy. A pile driver which is at the top of its guide possesses potential energy. As it falls, potential energy changes to kinetic energy. This is then converted to work as it forces the pile into the ground.

Energy Conversion - A Thermal Example

In a thermal power plant, electricity is produced using the chemical energy stored within fuels, such as wood waste, coal, oil, and natural gas. Figure 1 illustrates the generation of electricity using wood waste as fuel.

Wood waste is conveyed into the plant and stored in hoppers. The wood waste contains large quantities of stored chemical energy. In the power plant, this energy is released as follows:

• Using pressurized air, the fuel is blown into the boiler furnace where it is burned; this releases large quantities of thermal energy which appears as heat.

• This heat energy is then transferred from the boiler furnace into water which is used to produce steam.

• The high-pressure steam contains thermal energy by virtue of its high temperature. It also contains potential energy due to its high pressure.

• When directed through the blades of a turbine, the temperature and pressure of the steam drop as the steam uses energy to do work on the blades.

• This work appears as rotary mechanical energy which is used to perform work as it drives the generator to produce electricity.

Power

The terms work and energy are commonly interchanged. Power is a reference to the rate at which work is done or energy is used. If 100 joules of work is performed in one second (using 100 joules of energy), the power is 100 watts.

Larger equipment associated with the operation of plants is usually rated in kilowatts. This includes most mechanical drives, such as steam turbines, gas turbines, larger diesel engines, and electric motors. The kilowatt is a larger unit of measure than horsepower.

Energy Measurement

In plant operations, it is important to know how much energy a device or system will consume or produce.

A typical plant purchases large quantities of electrical energy from the local utility company. The amount of electrical energy used or produced is measured by estimating the amount of power used over a period of time. A machine that uses one kilowatt of power for one hour consumes one kilowatt hour (1 kWh) of electrical energy.

Plants also purchase chemical energy in the form of fuels, such as natural gas, fuel oils, coal, wood, agricultural, and municipal wastes.

Heat and Temperature

Heat is one common form of energy. Heat energy added to a body makes it hotter. The removal of heat energy from a body makes it colder. The addition of heat energy can also melt solids into liquids and change liquids into vapours or gases.

The addition of heat to ice raises its temperature so that eventually it melts into water. Continued heating eventually causes the water to boil and change to its vapour form, steam. Heat may thus be described as energy moving from one body to another. This energy flow is due to a temperature difference between the two bodies.

The temperature of an object or substance determines the direction of heat flow. Heat always flows from a body which is at a higher temperature to a body which is at a lower temperature. An important factor in the rate of heat transfer is the temperature difference between the two bodies. The greater the temperature spread, the faster the flow of heat from one body to the other.

Higher temperature objects do not always contain more heat than lower temperature ones. It is possible for an object at very high temperature to contain less heat than an object at a lower temperature.

The quantity of heat contained by a body depends on its mass, its temperature, and its specific heat. An object may have a very small mass and at the same time have a very high temperature. Compared to an object with a very large mass at a lower temperature, the first object may have a relatively small amount of heat energy.

Heat Units

The joule is the basic unit of all energy, including heat. Units of heat are therefore expressed in joules or multiples of the joule.

Temperature Units

The Fahrenheit and Celsius scales are most often used in plant operations. They are related in the following manner:

• One degree Fahrenheit is written as 1°F. On this scale (at atmospheric pressure of 14.7 psi) the freezing point of water is 32°F and the boiling point is 212°F. An equivalent measure of the Fahrenheit scale is the Rankine scale i.e. 1°F = 1°R. The difference between the two is that the Rankine scale starts at absolute zero, which is at approximately -460°F (i.e. 0 °R = -460°F).

• One degree Celsius is written as 1°C. On this scale (at atmospheric pressure of 101.3 kPa) the freezing point of water is 0°C and the boiling point is 100°C. An equivalent measure of the Celsius scale is the Kelvin scale i.e. 1°C = 1 K. The difference between the two is that the Kelvin scale starts at absolute zero, which is at approximately -273.15°C (i.e. 0 K = 273.15°C).

Mechanical Equivalent of Heat

One joule is equivalent to the work done by a force of one newton moving through a distance of one metre in the direction in which the force is applied. Therefore, the work done is one newton metre and this is equal to one joule.

In other words, the unit of work is numerically equal to the unit of heat.

Specific Heat

The specific heat of a substance is the quantity of heat required to raise the temperature of a unit mass of the substance by one degree. Expressed in SI units, the specific heat of a substance is the quantity of heat (in kJ) required to raise 1 kg of the substance 1 K or 1°C. For example, the specific heat of fresh water is 4.2 kJ/kg/°C (4.2 kilojoules per kilogram per degree Celsius), and often written as 4.183 kJ/kg°C.

Other substances require different amounts of heat energy to have one kg of their mass raised one degree Celsius in temperature. Table 1 identifies several examples (temperatures of solids at 0°C) at:

One Btu of heat will raise 1 lb of water through 1°F. However, if 1 lb of a substance other than water is heated through 1°F, then it will be found that the amount of heat required will be either more or less than 1 Btu. In the Imperial system, the specific heat of a substance is the number of Btus necessary to raise the temperature of one pound of the substance by 1°F.

Water is often used as a standard measure of specific heat. This is called the water equivalent. It is the product of the mass of a substance and its specific heat equal numerically to the mass of water that is equivalent in thermal capacity to the substance. In simpler terms, it is the amount of water that would absorb the same amount of heat as the substance per degree rise in temperature.

Sensible Heat Versus Latent Heat

Sensible heat is heat transferred within a system that changes the temperature of the system without changing physical variables such as volume, pressure, or state.

Latent heat is the energy absorbed or released by a thermodynamic system during a constant temperature process.

A common example of these two forms of heat occurs when heating of an ice cube. If heat is applied to an ice cube at temperatures below the freezing point, the temperature rises but other physical properties remain the same. This heat is called sensible heat. If heat is applied at the freezing point, the energy is used to change the phase of the water from solid to liquid. The temperature does not rise until all the water is converted. The heat applied during this second process is called latent heat.

Calculating Heat

The quantity of heat absorbed by a substance when it is increasing in temperature depends on three factors:

1. The temperature rise

2. The mass of the substance

3. The specific heat of that substance

It can be expressed in the form of a formula as follows:

where

Note: Temperature can be expressed in either Celsius or Kelvin but both t1 and t2 must be in the same units.

Example 1

Find the quantity of heat required to raise the temperature of one thousand litres of river water from 10°C to 90°C, if the specific heat of fresh water is 4.2 kJ/kg°C. Assume that 1 litre of fresh water has a mass of 1 kg.

Solution 1

Heat Capacity of Mixtures

The specific heat capacity of a material represents its ability to store energy. Mixtures have both individual heat capacities from each component but also a collective heat capacity for the mixture as a whole.

Within a mixture having N components, the total energy of the system is the sum of individual components:

In a mixture, the total energy is related to the heat capacity as follows:

These two equations can be related in the following example using a three component mixture.

The temperatures may be eliminated since the system conditions remain the same.

Therefore,

Rearranging and isolating the heat capacity for the mixture (cT)

Example 2

d) Find the quantity of heat required to raise the temperature of one thousand kilograms of brass from 10°C to 90°C.

e) Mix the brass intimately with 500 kilograms of aluminum. What is the heat capacity of the mixture?

Solution 2a

Solution 2b

Theory of Thermal Expansion

When heat is transferred to a material, the heat is transformed into internal energy (kinetic energy plus potential energy) by each molecule of the material. As the molecules gain energy and vibrate more rapidly, the average space between molecules increases. The result is a change in the overall dimensions of the material.

If a long, thin material has heat transferred to it, the material will expand mainly in one dimension, that of length. Flat materials, having relatively large surface areas, will expand in both length and width, thus undergoing mainly an expansion in area. In actual fact, solids expand in all directions, and therefore also undergo a change in volume. Liquids and gases expand in volume when heated, because they must assume the shape of their containers.

Factors Affecting the Expansion Rates of Solids

There are three main factors that determine the extent to which the dimensions of an object will change when the object is heated. These factors include:

1. The material from which the object is made; different materials expand at different rates.

2. The original dimensions of the object before heating; a larger object will expand more than a smaller object made of the same material.

3. The temperature change which the object undergoes; the greater the temperature change, the greater the change in dimensions of an object.

Linear Expansion

To calculate the change in length of a solid material, the following formula is used:

Before doing any calculations, the term alpha (α) must be clearly understood. Alpha is simply a constant value, assigned to a material, which indicates by how much a unit length of the material will expand for each degree of temperature change. Given that the average coefficient of linear expansion for steel is given as 12.0 × 10-6 per Celsius degree the following points can be considered:

1. A steel rod that is 1 m long will expand in length by 12.0 × 10-6 m when its temperature is increased by 1°C. Other units may be substituted for metres. For example, a rod of steel that is 1 mm long will expand (lengthen) by 12.0 × 10-6 mm when heated by 1°C, and so on.

2. Conversely, a 1 m long steel rod will decrease in length by 12.0 × 10-6 m when its temperature is decreased by 1°C.

The units for α are usually written as /°C.

The coefficient of linear expansion varies because the values assigned to various materials are not constant at all temperatures. The rates at which materials expand or contract vary with the temperatures of the materials; therefore, an average is used. Table 2 lists values for the coefficient of linear expansion (alpha) for some common materials.

The following are examples of problems that deal with linear expansion.

Example 3

Calculate the change in length of a brass bar 10 m long when its temperature is increased from 25°C to 170°C.

Solution 3

Example 4

A steel pipe is 100 m long at 10°C. It transports steam at 200°C. Calculate the length of the pipe in its fully expanded condition.

Solution 4

Example 5

A steel shaft is 25 m long and its temperature is at 30°C. When raised to operating temperature, the shaft lengthens by 42.3 mm. Calculate the operating temperature.

Solution 5

Let t1 represent the starting temperature and t2 the operating temperature:

Expansion of the Area of a Solid

Up to this point, the expansion of a material in one linear dimension, such as length or thickness, has been discussed. Now consider a square piece of flat material, before thermal expansion, as shown in Figure 2.

The top surface area of the square plate is calculated as: A1 = l2, where A1 is the area before thermal expansion and l is the length of a side.

After thermal expansion, the length of each side has expanded, and can be represented by l + ∆ l. This is shown in Figure 3.

The area of the material after expansion, A2, can be calculated as:

The value of ∆l is small, compared to l (note that the sketch in Figure 2 is drawn out of proportion for clarity). For practical purposes, the value of (∆l)2 is too small to be significant; thus it is omitted from the formula, which leaves A2 = l2 + 2l ∆l

Then, the change in area of the material can be written:

Substituting l × α × ∆t for ∆l:

Substituting l2 = A:

or, rearranging terms:

Comparing this equation to the one used for calculating change in length, it is apparent that the coefficient of area expansion is twice the coefficient of linear expansion.

The Greek letter gamma (γ) is sometimes used as the symbol for the average coefficient of area expansion. However, in the following examples, 2α will be used.

Example 6

A steel plate has an area of 2.00 m2 at 205°C. Calculate its area at 25°C.

Solution 6

Example 7

At 10°C (t1), the radius of a circular plate is 1.85 m. When heated to 450°C (t2), the area of the plate increases by 0.18 m2 (∆A). Calculate the coefficient of linear expansion (α) for the material from which the plate is made.

Solution 7

Rearranging and isolating α gives the following equation:

Expansion in the Area of a Hole in a Material

There are occasions when it is useful to know how a hole in a material changes in size when the material surrounding the hole is heated. For example, when an engine cylinder expands, how is its volume affected?

An example of a hole in a material is the common flat washer, in the shape of an annulus. A simple view is shown in Figure 4. Any flat metal plate with a hole drilled or cut into it will behave in the same way.

When a flat metal plate is heated, its outer width (diameter) increases. The diameter of the inner hole will either increase or decrease. Consider an aluminum flat plate, as shown in Figure 5. The following example illustrates what happens to the dimensions of the metal plate as its temperature is raised by 150°C.

After thermal expansion, the width (W) will have increased by:

Regardless of the hole size, the outside diameter of the material will increase by:

The new diameter of the hole can now be calculated:

The hole has increased in diameter, by

Now consider a solid, circular piece of aluminum, 254 mm in diameter, and calculate its final diameter after a temperature increase of 150°C.

The result indicates that the hole in the aluminum has expanded exactly as though it were made of aluminum. This generally accepted rule is often interpreted as, the area of a hole in a piece of material expands as though the hole itself were made of the same material, regardless of the shape of the hole or the material.

Example 8

A piece of silver has a 25 mm diameter hole in it. The silver is heated from 15°C to 60°C. Calculate (a) the diameter of the hole after heating, and (b) the area of the hole after heating.

Solution 8

The above principle can be used to tightly fit one cylinder inside of another. The inner cylinder is made slightly larger than the inside diameter of the outer cylinder. The outer cylinder is then heated, and the inner cylinder cooled. Eventually the inner cylinder fits into the outer cylinder. As the inner cylinder expands again upon warming, and the outer cylinder contracts as it cools, a tight fit is formed. When this process is carried out, the cylinders are said to have an interference fit or a shrink fit.

Volumetric Expansion of Solids

Consider a solid cube of a material, where each side of the cube has a length of 1 unit. Since the volume of a cube is calculated by multiplying the length times the width times the height, the volume of this one-unit cube can be considered as:

Now consider the cube after it has been heated through 1°C. The change in the length of each side can now be calculated using the formula:

Since l and ∆t are each one unit:

Thus, the length of each side after heating is 1 + α, and the new volume is now (1 + α)3.

Expanding this gives:

The change in volume = volume after heating − volume before heating

The values of 3α2 and α3 are extremely small, relative to α. They are often ignored for practical purposes.

Since the original cube had side lengths of 1 unit, and a volume of (1 unit)3, 3α represents the increase in volume per unit volume per degree increase in temperature. Stated another way, 3α is the value of the coefficient of volumetric (or cubical) expansion, which is 3 times the value of the coefficient of linear expansion.

Therefore, to calculate the change in volume of a material, the following formula is used:

The Greek letter beta (β) is often used to represent the average coefficient of volumetric expansion. However, here we will use 3α.

Example 9

At 15°C, a cast iron cube measures 50 mm by 50 mm by 50 mm.

Calculate its volume at 90°C.

Solution 9

The area of a hole in a material expands as though the hole were made of the same material. The same is true for the volume of a hole within a material. For example, the cavity within a steel tank will expand as though the cavity were made of steel, when the tank is heated.

Example 10

A steel tank has a circumference of 10 m and is 25 m high at 100°C. Calculate its volume at 20°C.

Solution 10

And:

And:

Then:

Convection

In the convection mode of heat transfer, heat is transferred by the actual displacement of molecules within the system that are at a higher temperature to a region of lower temperature. Since convection requires movement of a medium, it can take place only in a fluid. Convection currents in gases and liquids are often caused by variations in density due to the temperature changes. If a fluid is in contact with a hot surface, the portion adjacent to the surface will be heated and will expand, and thus the fluid density will be decreased. If the hot surface is at the bottom of the tank, the heated liquid, which is generally less dense than cooler liquid, rises in the tank, strikes the tank roof, and spreads out. The cooler (and denser) liquid at the bottom of the tank moves in and fills up the space vacated by the hot liquid. The cooler and denser liquid, in turn is heated and displaced. In this manner, circulation of the liquid (or gas) will be established.

Figure 6 illustrates a simplified example of convection currents in a boiler drum. The portion of the water in contact with the heating surface will be heated and then displaced by cooler water, which in turn is heated and displaced.

This is an example what is called “natural” convection. The movement of the fluid is brought about without using mechanical devices. “Forced” convection may also be used in the transfer of heat. In this case, the movement of the fluid is brought about by means of a pump or a fan. Examples of forced convection are:

a) using a pump to circulate hot water through a building heating system

b) using a fan to circulate hot air through the system

A formal definition of heat transfer by convection may be: “The transfer of heat through a fluid by movement of heated particles of the fluid.”

Conduction

When the temperature of a substance is increased by the addition of heat, the additional energy causes molecular vibration to increase. As individual molecules are affected, each one will transfer some energy to adjacent molecules. The vibration of the adjacent molecules will then increase until all the molecules have the same energy and intensity of vibration. The transfer or “flow” of energy (heat) from molecule to molecule in this manner is called “conduction.”

Conduction may occur within a substance, or from one substance to another (as long as they are in physical contact with each other).

A formal definition of conduction may be: “The transfer of heat from one part of a substance to another part of the same substance, or from one substance to another substance in direct physical contact, without significant movement of the molecules.”

The heat flow is from an area of high temperature to an area of low temperature.

The quantity of heat transferred by conduction will vary directly with:

a) Surface area: more heat will be transferred when a large surface area exists.

b) Temperature difference: more heat will be transferred when a large temperature difference exists.

c) Time: more heat will be transferred as the time of heat transfer increases.

The quantity of heat transferred by conduction varies inversely with:

a) Thickness: the longer the path through which the heat must travel, the less heat will flow.

Consider a section of wall as illustrated in Figure 7, of area (A) and thickness (d). An amount of heat (Q) will flow through the wall when the wall is exposed to the temperature gradient T1 − T2.

These concepts can be expressed in formula form:

The constant (λ) is known as the thermal conductivity constant, or the coefficient of thermal conductivity.

A standard form of the thermal conductivity formula is often shown as:

Transposing the formula, and solving the units for “λ”:

After cancelling (m and m2) and substituting W (watts) for J/s,

Conductive Materials

The thermal conductivity constant varies with the ability of the substance to conduct heat, which is directly related to the composition of the substance (for example, number of air pockets). Generally, metals are good conductors of heat. Insulators, however, are poor conductors of heat. Some examples of insulators are air and other gases, asbestos, fiberglass, and cork. Table 3 lists average values of coefficients of thermal conductivity for some substances.

Example 11

Calculate the heat transfer per minute through a solid wood wall which is 10 m long, 4 m high, and 127 mm thick. The outer surface temperature is 5°C and the inner surface temperature is 24°C.

Solution 11

Heat Flow through a Composite Wall

Walls are commonly built from a number of materials. An example might be a wall consisting of an outer wooden panel attached to inner wallboard. The wall system illustrated in Figure 8 is referred to as a composite wall because the wall components are layered on top of each other.

The total temperature difference from the inner to outer surface is represented as TT. The temperature difference between the hotter and cooler surface of each layer (represented by thickness d1, and d2 respectively) will contribute to limiting the heat flow. Note, however, that any amount of heat that flows through one layer must also flow through the other layers. A certain portion of the total temperature difference will occur across each layer, depending upon its insulating properties as represented by its coefficient of thermal conductivity (λ1 and λ2).

For each layer of the wall:

Rearranging and isolating for ∆T creates the following equation;

For the total temperature difference TT (T1 − T3), Q, A and t will be the same:

In general, if “n” represents the number of layers and TT represents the total temperature difference:

Therefore, for a composite wall that has four layers, all of which have the same area:

Example 12

An insulated wall of a storage room is 8 m wide by 3.048 m high. The wall has an outer steel plate that is 63.5 mm thick and an inner wooden wall that is 127 mm thick. Between the steel and wood layers is a 25.4 mm intermediate layer of glass fibre insulation. If the temperature difference between the inner and outer surfaces is 40°C, calculate:

(a) the heat transfer per hour

(b) the temperature difference across the steel wall

(c) the temperature difference across the glass fibre layer

(d) the temperature difference across the wood wall

Solution 12

(a) From Table 3:

Isolating for Q:

Note that, due to the relatively high coefficient of thermal conductivity of steel, the temperature drop across the steel is practically negligible. Almost all of the temperature drop occurs across the glass fibre and wood layers.

Heat Flow through Cylindrical Walls

It is often necessary to determine the heat loss through the cylindrical walls of vessels or pipelines. Figure 9 illustrates a typical wall configuration of an insulated pipe carrying heated liquid.

Because the fluid within the pipe is above ambient temperature, there will be heat transfer from the inner liquid to the outside. When the pipe is insulated, there is often very little temperature drop across the vessel wall, due to its relatively high coefficient of thermal conductivity. This concept was demonstrated in Example 10. It is usual to neglect the temperature drops across the metal elements of the pipe, and to focus on the temperature drop across the insulation instead.

The following formula is often used to calculate the heat flow through cylindrical walls.

Example 13

A steam line has an external pipe diameter of 152.4 mm and is surrounded by glass fibre insulation which has a thickness of 50.8 mm. The pipe is 60 m long and carries steam at a temperature of 280°C. Assuming the outer temperature of the insulation is 20°C, calculate the heat loss from the pipeline, per hour. Neglect any temperature change across the pipe wall (i.e. in this case the external pipe diameter is considered as equal to the inner diameter (d)).

Solution 13

Radiation

Heat transfer by conduction and convection requires a material media; however, heat is also transmitted through areas that do not require a medium. For example, heat reaches the earth directly from the sun through the near vacuum of space. This heat flow process, called radiation, may be defined as: “The transfer of heat energy from one body to another through space by rays of electromagnetic waves.” While radiated heat transfer does not require a material media, heat can travel through the media. For example, radiative heat passes through air and glass, and it must be considered in air-conditioning load calculations. When a material absorbs radiation, it is heated.

The rate at which objects radiate or absorb heat depends on the temperature and surface of each object. As an object radiates heat, its molecules lose energy, and the object begins to cool. When an object absorbs radiation, the heat energy transmits to the molecules of the object and increases its internal energy and temperature.

A general requirement for identifying a good absorber of radiation is one that has a surface which is black and rough. A good absorber of radiation must also be a good emitter. If not, the object would absorb more energy than it emits; its energy and temperature would increase uncontrolled until the object eventually melts or blows up. When the temperature and internal energy of an object are constant, the object must emit and absorb radiation at the same rate since energy is conserved. A blackbody radiator is defined as an object that perfectly absorbs and emits all radiation.

The relationship between the radiative properties of a black body and its surface conditions is shown below in the Stefan-Boltzmann law. This law shows that the rate of heat energy radiated by a black body (Q/t) is directly proportional to its surface area (A) and the absolute temperature (in K) raised to the fourth power. The emissivity (ε) of a body is “the ratio of heat emitted by the body to the heat emitted by a perfect black body for the same surface area, temperature and time.” The emissivity of a perfect black body is thus one, while that of the other extreme would be zero. Emissivity can thus range from zero to one.

Example 14

The temperature of the flame in a natural gas fired furnace is 1950°C. The metal temperature of the water wall is 280°C. The wall dimensions are 3.4 m by 7 m. Calculate the maximum theoretical quantity of heat energy radiated per minute to the wall area.

Solution 14

For maximum possible heat energy radiated, ε = 1

Surface Heat Transfer

In many practical applications, heat is transferred from one fluid to another through a solid (usually metal) plate. In the system illustrated in Figure 10, heat is transferred from fluid A (hotter) through the solid plate to fluid B (cooler).

Note: The temperature profile is superimposed on the sketch. The total temperature drop (TD) consists of the temperature drop from fluid A to the plate (T1 − T2), the temperature drop through the plate (T2 − T3), and the temperature drop from the plate to fluid B (T3 − T4).

At each interface between fluid and plate there is a layer of fluid, which is almost stagnant. This layer is very thin, and is referred to as the “fluid film.” Heat transfer through the fluid film occurs by:

• conduction through the fluid film

• convection currents within the fluid film

• to a lesser degree, radiation of energy from the plate to the fluid

Calculation of heat quantities transferred through the fluid film, via each different mode, could be completed, and the individual results added. However, this type of calculation would be complex, with questionable accuracy. Much empirical data on the quantity of heat transferred through a fluid film has been gathered, and the use of this empirical data is accepted as the standard method of calculation.

The heat transferred through the fluid film varies with area, time, temperature difference, and an empirical factor known as the surface heat transfer coefficient. The quantity of heat transferred through a surface fluid film can be found using:

The values of the surface heat transfer coefficient vary with the velocity of the fluid, the condition of the surface, and the properties of the fluid. Presented in Table 4 are a few typical values for surface heat transfer coefficients.

Note: The order of magnitude differences between the values of “h” for gases and those for liquids is due to the much higher coefficient of thermal conductivity of a liquid compared to a gas.

Example 15

Calculate the heat flow per hour through the fluid film from a 200 m2 flat horizontal floor of a warehouse at 19.0°C, to an ambient air temperature of 15.5°C. Use 2 m/s conditions.

Solution 15

From Table 4:

1. Define the term specific heat.

2. A bar containing a mixture of 40 kg of brass and 50 kg of aluminum at 15°C receives 5000 kJ of heat. Calculate the final temperature of the mixture.

3. Calculate the increase in length of a steel beam that is 20 m long at 20°C, if it is heated to 90°C.

4. A solid glass cylinder, 3.00 m long at 90°C, contracts by 2.45 mm when left outdoors. Calculate the outdoor air temperature to the nearest degree.

5. Calculate the coefficient of linear expansion for the following material. At 15°C, the length of one edge of a cube made of the material is 3.65 m. When heated to 200°C, the area of the face increases by 0.072 m2.

6. At 250°C, the hole in a piece of steel has a diameter of 18 mm. Calculate the area of the hole at 12°C.

7. A steel tank is 15 m in circumference and 10 m high at 100°C. Calculate its volume at 18°C.

8. At 70°C, the diameter of an aluminum sphere is 70 mm. After heating to 200°C, what is the final volume?

9. Identify the three modes of heat transfer. Define each mode.

10. An insulated wall of a storage room is 8 m wide by 3.048 m high. It has an outer wooden layer that is 50.8 mm thick and an inner wallboard wall that is 12.7 mm thick. Between the wooden and wallboard layers is a 25.4 mm intermediate layer of still air acting as insulation. If the temperature difference between the inner and outer surfaces is 30°C, calculate:

a) the heat transfer per hour

b) the temperature difference across the wooden wall

c) the temperature difference across the air layer

d) the temperature difference across the wallboard wall

11. Calculate the heat flow per minute through the fluid film from a 12 m2 flat vertical wall of a warehouse, at 20.1°C, to an ambient air temperature of 16°C. Use still (no wind) conditions.

12. Heat is transferred from one fluid to another through a solid plate by means of what processes? Describe how each process occurs.

Self-Assessment

Question 1

The temperature of the flame in a natural gas fired furnace is 1530°C. The metal temperature of the water wall is 250°C. The wall dimensions are 4.8 m by 12 m. Calculate the maximum theoretical quantity of heat energy radiated per minute to the wall area.

2056 MJ

3406 MJ

3604 MJ

1056 MJ

Question 2

A piece of platinum has a 76 mm diameter hole in it. The platinum is heated from 25°C to 145°C. Calculate the diameter of the hole after heating.

75.177 mm

76.078 mm

76.177 mm

77.178 mm

Question 3

Find the quantity of heat required to raise the temperature of 2700 litres of fresh water from 10°C to 70°C, if the specific heat of fresh water is 4.2 kJ/kg°C. Assume that 1 litre of fresh water has a mass of 1 kg.

334 640 kJ

680 400 kJ

580 400 kJ

380 640 kJ

Question 4

Calculate the change in length of a nickel bar 12 m long when its temperature is increased from 40°C to 190°C.

2.67 x 10 -2 m

2.34 x 10 -2m

2.34 m

2.67 m

Question 5

Find the quantity of heat required to raise the temperature of 500 kg of mild steel from 20°C to 100°C, if the specific heat of steel is 0.494 kJ/kg°C.

19.760 kJ

1976.0 kJ

19 760 kJ

197 600 kJ