Heat Load Calculation

I no longer have the textbooks from my school days (won't say how long ago that's been!) and I haven't worked in that field in as much time, but if I recall:

Heat load is calculated by f^2 of the wall area with adjustments for insulation level in walls, ceiling, and more adjustments for window area. Then the "normal occupancy" is factored in for commercial/industrial apps. Also needed to be factored are heat producing items like appliances, computers etc. I seem to recall having found a few online calculators but I can't remember where they are. You might try an online search for this.

Sorry I couldn't be more helpful.

charlie

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HEAT LOAD COMPUTATION:

The total Heat Load is a function of many factors viz; the temperature gradient or differential between the ambiance and your set point, the quantity and thermodynamic properties of material - animate and inanimate - in the room, the air exchange rate or air cycles per unit of time, heating appliances in the room such as water heater, heat emission from electronic and electrical devices, the incandescent lamps and other illuminations, frequency at which the door and windows are open, the thermal resistance of the walls (insulation properties).

Before computation you must have some data, for example:

your set point -the temperature U want to achieve = t_s [°C]

the ambient temperature = t_a [°C]

the volume of your room = V_r [m³]

air exchange rate = R [m³/s] or V_r/3600

the thermodynamic properties of the air, average specific heat and density, c [kJkg^-1°C^-1] and kgm^-3],

the thermodynamic and geometric properties of the walls, heat transfer coefficient, k [Wm^-1°C^-1], wall thickness, x [m].

So the heat load is the sum of: (i) product heat - animate and inanimate materials

(ii) Heat Loss thru the walls (iii) Heat generated by various heat and electrical/electronic appliances (iv) Heat required to bring the total voluve of Air from ambient to set point and (v) Miscellaneous heat losses.

HEAT LOSS thru the walls Q_w=qA, where q=specific heat loss per m² of the wall, kWm-² and A is the surface area of the wall.

q = k/x[t_a-t_s], kWm^-1

Q = k/x[t_a-t_s]A, kW

PRODUCT HEAT is a the enthalpy change and a function of temperature.

dq = c_pdT or q = c_p∫dT, for every component and for n component the total heat will be n x q or Q_p. this is in kJ, and to convert this value to KW, U have to divide you result by 3600 i.e. kJ/3600 = kW {note that the parameters have consistent dimensions and are all in SI Units}

A person at rest dissipates/emits about 30 W of heat,

HEAT LOADS DUE TO APPLIANCES CAN BE OBTAINED FORM THE PRODUCTS INFORMATION. they are alway in Watts.

HEAT REQUIRED TO BRING THE AIR VOLUME FROM AMBIENT TO SET POINT

Q_A = G_Ac_p[t_a-t_s],

where, G_A is the air mass flow rate, kgs^-1

G_A = V_o x p, where V_o is volumetric flow rate, ms^-1 and p = density, kgm^-1 of the air respectively.

Addition of all these will give U the Heat Load, but because it is not possible to have an aggregate that works on 100% efficiency, U need to multiply your total heat by a factor of say, 20%. Q_gt = 1.2Q_t. where Q_gt is the grand total heat requirement.

to convert this to tons of refrigeration divide the Q_{gt by} 3.5, 1 ton of Refrigeration is approximately 3.5 kW.

Note that U may get a negative signs in front of your answers (the juxtaposition of your temperatures may cause this), Except there is a cold coming from that source or unit, U must ignore it.

QED,

Soji Oyewole,

Ibadan,

South West, Nigeria.

Refer Refrigiration and Air Conditioning By Arora & Arora

Refrigiration and Air Conditioning By Arora & Domkundawar(Khanna Pub)

Refrigiration and Air Conditioning By P. N. Ananthnarayan(TMH)

In the above books you can find many numericals solved on this topic.

It is totally dependent on the insulation employed, people load, ambient temperature and humidity patterns and durations, tightness of construction (air changes per hour) wall, ceiling and floor areas and coverings, other heat loads, etc. LOTS of factors.

For a 'typical' 3 bedroom, 2 bath, home in Houston Texas, 2 adults, 2 children, 1700 sq. ft. current insulation and vapor barrier standards, a 36000 BTU/hr (3 ton) is actually oversized, but still recomended. The higher insulation standards now (R-40 ceiling, R-18 walls, double pane LOW 'e' , low leakage windows, total wrap with vapor barrier), a 24,000 BTU/hr unit is minimum, with 30,000 BTU/hr (2.5T) adequate. Reasoning behind 3T recomendation is to have reserve capacity for those 100F+ days and another 4-6 people in the house. Works out OK.

In reality, A/C units are 'rated' at 85F outside air temp, which is exceeded for 12-15 hours/day for at least 6 months here. A '3 ton' rated unit is really 2.5 tons here during the time when it is most severely taxed. New installations also use TXV (thermostatic expansion valve) instead of less efficient, less adaptable capillary tube system. The TXV reduces compressor load when cooling load drops and ambient temps are lower, and keep the evaporator temp constant so you can run closer to freezing point, removing hummidity better. Over 50% of typical cooling load here is REMOVING MOISTURE.