Hi Stefan, yes what you say is correct. In the last issue we looked at temperature, heat, latent and sensible heat.

Let us move on to the quantity of sensible heat

When an amount of heat is transferred to or from a mass of substance, which results in a change in temperature, the heat added is called sensible heat. The resultant change in temperature can be calculated from:

Q = m cDT

where Q = The amount of heat transferred in kJ

m = The mass of the substance in kg

c = The specific heat capacity in kJ/kg.K .

DT = The temperature difference in degrees Kelvin.

D is the Greek symbol “delta” which is used to denote “difference”.

When the rate of heat transfer is required, the mass flow rate must be used instead.

Thus the formula then becomes:

. .

Q = m c D T (the dot above the units means “per second”).

or, heat transfer rate = mass flow rate x specific heat capacity x temperature change.

.

Q = The amount of heat transferred in kJ/s which = kW

.

m = The mass flow rate in kg/s

c = The specific heat capacity in kJ/kg. K .

DT = The temperature difference or change in degrees Kelvin.

Quantity of latent heat.

When an amount of heat is transferred to or from a mass or substance, which results in a change of phase, the heat added or removed is called latent heat.

The amount of heat required to change the phase of a mass or substance can be calculated from the formula.

** **Q = m. h_{f g}

Where Q = The amount of heat required in kJ

m = The mass of the substance in kg

h_{f g} = The latent heat required to change the phase of 1 kg of the substance in kJ/kg

Where the heat transfer rate is required, the formula changes to

** ****.**** .**

** **Q = m.h_{f g}

** .**

Where Q = The heat transfer rate in kJ/s or kW

** .**

m = The mass flow rate in kg/s

**Calculations involving sensible and latent heat.**

**Example 1.**

Question:

Calculate the amount of sensible heat required to heat 5kg of water from 20°C to 100°C .

The specific heat of water is 4,2 kJ/kg .

Answer:

The sensible heat formula is: Q = mc∆T

The amount of sensible heat required is 5 x 4,2 x (100 – 20) = 1 680 kJ .

**Example 2.**

Question:

How much heat is required to evaporate 5 kg of water at 100°C .

The “Heat of evaporation” for water at 100°C is 2 257 kJ/kg .

Answer:

The latent heat formula is: **Q = m· h _{f g}**

The amount of heat required = 5 x 2 257 = 11 285 kJ .

**Example 3.**

Question:

The geyser at your home contains 200 litres of water and the element installed has a capacity of 3 kW.

How long will it take to heat the water from 20°C to 70°C ?

Answer:

The amount of heat required to heat the water can be calculated from the sensible heat formula. Thus:

Q = m·c·∆T = 200 x 4,2 x (70 – 20) = 42 000 kJ.

The heat is added at a rate of 3 kJ/s \ the time taken to heat the water is:

42 000 ÷ 3 = 14 000 seconds.

This is 14 000 ÷ 60 = 233 minutes which is 3 hours and 53 minutes

(Rounded off).

**Example 4.**

Question:

A hot water storage tank in a large building holds 15 000 litres of water.

This water must be heated up from 40°C to 85°C during the night.

The electrical energy will be available from 10:00 PM to 05:00 AM.

Calculate the capacity of the electric heaters required.

The specific heat of water is 4,2 kJ/kg and 1 litre is equivalent to 1 kg .

Answer.

The amount of heat required to heat the water is calculated from the sensible heat formula.

Thus the amount of heat required is: mc∆T = 15 000 x 4,2 x (85 – 40) = 2 835 000 kJ .

This amount of heat is required and is independent of the amount of time available.

The capacity of the heater elements is given in Joules per second (J/s) . Thus the amount of heat to be added per second must be calculated.

The time available is 7 hours which is 7 x 3 600 seconds = 25 200 seconds.

The amount of heat to be added per second = 2 835 000 ÷ 25 200 = 112,5 kJ/s .

As 1 kW = 1 kJ/s the capacity of the heater must be 112,5 kW .

Often calculations involve both sensible and latent heat.

Examples of this are the humidification of air and the production of ice.

When doing calculations involving both sensible and latent heat you must be very careful to determine which is sensible heat and which is latent heat.

Remember: Sensible heat is what changes the temperature of a substance without changing its phase.

Latent heat changes the phase without changing the temperature.

If you put a pot of water with ice blocks in it on the stove and switch the stove on heat will be added.

Initially this heat will be used to melt the ice. Theoretically the water should stay at 0°C, however because the transfer of heat to the ice is slow the water will increase in temperature as well. The heat added at this stage is mostly “latent heat” because a change in phase is taking place.

Once all the ice has melted the water will increase in temperature; the heat added now is “sensible heat”. It can be sensed, or felt.

When the water reaches boiling point (which depends on the altitude) the temperature will remain constant and the water will start to evaporate. Because now there is no increase in temperature but a change in phase the heat added is “latent heat”.

**Example 5:**

Question:

A pot of water containing 1 litre of water at 0°C and 1 kg of ice, also at 0°C, is put on a stove. We are at sea level altitude.

Calculate the amount of heat required to produce steam at 110°C .

The boiling point of the water at sea level is 100°C .

The heat of fusion of ice is 333,7 kJ/kg .

The specific heat of water is 4,2 kJ/kg and the latent heat of vaporisation of water at 100°C is 2 257 kJ/kg .

The specific heat of water vapour is 1,89 kJ/kg .

Answer:

It is obvious that there are changes of phase and changes in temperature involved.

The sequence of events is as follows:

- The ice melts. This is latent heat
- The water is heated from 0°C to 100°C. This is sensible heat.
- The water evaporates at 100°C. This is latent heat.
- The water vapour is heated from 100 to 110°C. This is again sensible heat.

Calculations:

- To melt 1 kg of ice requires 1 kg x 333,7 kJ/kg = 333,7 kJ
- To heat the water requires 2 kg x 4,2 kJ/kg x (100 – 0 K) = 840,0 kJ
- To evaporate the water requires 2 kg x 2 257 kJ = 4 514,0 kJ
- To heat the water vapour from 100 to 110°C requires

2 x 1,89 x (110 – 100 K) = 37,8 kJ.

—————

Total amount of heat required 5 725,5 kJ

**Example 6:**

Question:

A humidifier must add 5 kg of steam per hour to an air stream.

If the water being supplied to the humidifier is 20°C and the boiling point of water at the site is 95°C .

Calculate the size of element required to produce this steam if the latent heat of vaporisation at this temperature is 2 270 kJ/kg

Answer:

To produce 5 kg of steam 5 kg of water is required.

This water first has to be heated from 20°C to 95°C and then evaporated.

The amounts of heat required are as follows.

To heat the water: 5 x 4,2 x (95 – 20) = 1 575 kJ

To evaporate the water requires 5 x 2 270 = 11 350 kJ

Total amount of heat required = 12 925 kJ

This amount of heat must be added in 1 hour; thus the amount to be added per second is:

12 925 ÷ 3 600 = 3,59 kJ/s . This is equivalent to 3,59 kW.

Note: the size of the electrical element is the amount of heat it can add per second.

**Example 7:**

Question:

Calculate the refrigerating capacity of a plant which produces 2 000 kg of ice in an 8-hour shift if the water supply is at 20°C and the ice is required at -10°C .

Answer:

For 2 000 kg of ice 2 000 kg of water is required.

This water must be cooled from 20°C to 0°C .

The heat to be removed for this = 2 000 x 4,2 x (20 – 0) = 168 000 kJ

Next this water must be changed to ice.

The heat to be removed for this = 2 000 x 333,7 = 667 400 kJ

Next the ice must be cooled from 0°C to -10°C .

The heat to be removed for this = 2 000 x 2,1 x (0 – -10) = 42 000 kJ

——————-

Total amount of heat to be removed in 8 hours = 877 400 kJ

The capacity of the plant is the amount of heat it must be able to remove per second.

This is calculated by dividing the total amount of heat to be removed by the number of seconds in the time allowed. Thus the capacity of the plant is:

877 400 ÷ (8 x 3 600) = 877 400 ÷ 28 800 = 30,47 kJ/s = 30,47 kW .

Thank you for the question Stefan, I hope that this sheds some light on the fundamentals of refrigeration.