|

COOLING LOADS QUESTION ANSWERS CLASS NOTES

COOLING LOADS QUESTION

 ANSWERS CLASS NOTES

Heat loads determine the capacity of a

refrigeration plant for a certain

application. These heat loads consist of

sensible and latent heat loads. Sensible

heat is due to difference of temperature.

Latent heat is due to change of state.

Fig. Cooling Load Calculations

 Q1. What are Sensible heat loads? Give examples.

Sensible heat load is due to temperature difference only. It has no connection with the water vapor in the air. Therefore it is also called DRY LOAD.

Examples of sensible heat loads are through/from

(i)  walls, roof and floor

(ii)  glass

(iii)   occupants

(iv) lights

(v)   power equipment’s, bulbs, fluorescent tubes, mixes, food processors, pumps, computers, fans

(vi)  Products (Potatoes, Bananas, apples etc.)

(vii)  infiltrated air

(viii)   additional infiltrated air

(ix)  ventilation air

(x)  supply duct

(xi) return duct

(xiii) miscellaneous sources

Q2. What are latent heat loads? Give examples.

Latent heat load only due to phase change (moisture difference from outside condition to inside room condition)

Examples: Latent heat load from

(i) occupants

(ii)  infiltrated air

(iii)  additional infiltrated air

(iv)  ventilation air

(v)  miscellaneous sources

Q3. What do mean by heat gain from occupants?

 Inside the air-conditioned space, persons are working. These persons are called occupants. Human body is such that it always looses heat, partly as sensible heat and partly as latent heat.

Sensible heat is due to temperature difference.

QSHL =N x SHL/person

N number of persons

SHL/person is activity dependent

Latent heat

QLHL = N x LHL/person

Total heat lost is sum of SH and LH.

This heat lost depends on the sex, activity and surrounding conditions.

An adult female loses 75 % of an adult male.

A child loses 60 % of that of an adult person.

CAUTION

In absence of any specific conditions, one can assume

SHL/adult person = 160 kJ/h 

LHL/adult person= 240 kJ/h 

In summer, the outside temperature is above the body temperature. The body will not be able to loose heat by sensible means. Rather it will be gaining sensible heat from the surroundings. It will become highly uncomfortable unless the total heat is lost as latent heat i.e. by perspiration. In the absence adequate air supply, person will not be able to lose heat and it may lead to fatal health results.

Q4. Differentiate between Ventilation load and Infiltration heat loads.

Ventilation load: heat load added to the room (in turn added to AC) due to fresh air brought in the room to attain the required oxygen content in the room air.

Infiltration Load: Heat load added to the room (in turn to AC) due to the opening of door of the  air conditioned room. Both contain sensible and latent heat loads.

Q5. Define outside design conditions (ODC). Give examples of outside design conditions.

OUTSIDE SUMMER CONDITIONS

These are average conditions occurring for most of the summer. These are different for different places.

City                   DBT                         WBT                       RH

Chandigarh     40.1C                     23.9C                   27 %

Delhi                 40.4C                    23.9C                  26 %

Calcutta            35.3C                      27.9C                      60 %

OUTSIDE WINTER DESIGN CONDITIONS

CITY              DBT                 WBT           % RH                                                                                                               

Delhi               7.2                   5.0                   70 %

Calcutta          13.3                 8.9                   55% 

Q6. Give the cooling load equations.

(i) Transmission load through wall, roof, floor, glass =U A dt

(ii) Solar radiation load through glass = Ag SC SCL

SC Shading coefficient

SCL solar cooling load factor

(iii) Occupancy Load

QSHL = N x SHL/person

QLHL = N x LHL/person

N number of persons

SHL/person is sex, activity and room conditions dependent

(iv) Light Loads

Q Light = Total wattage x Use factor

Infiltration / Ventilation = Density x Volumetric flow rate x Specific heat x dt

dt temperature difference

 

https://mesubjects.net/wp-admin/post.php?post=14356&action=edit                 MCQ Cooling loads

https://mesubjects.net/wp-admin/post.php?post=6047&action=edit             Theory cooling load calculations

Similar Posts