Question

Consider the wall of Gypsum, Insulation, Concrete and Brick. It is located in a city for which the CDD is 800°C.days. The total area of the wall is 600 m2. The interior space is occupied and maintained at constant temperature 24 hours a day for a year. use the information below to answer these questions.

T_inside= 21°C

T_outside= 18°C

Gypsum: x=0.01 m, C=20 W/m2∙°C

Brick: x=0.12 m, k= 1 W/°C

Insulation: x=0.025 m, k= 0.04 W/°C

Block: x=0.20 m, C=5 W/m2∙°C

How much energy is required to cool the building over a year in GJ?

What is the cost to provide this energy if the cooling system is electrically and electricity costs $0.10/kWh with The COP for the cooling system is 2.4?

Answer 1

For heat flow by conduction, heat transferred per unit area is

Q (W) / A (m²)= [k (W/m C) *  T (C)] / L (m)= T(C) / R where R = L (m ) / k (W/m C) ...... eqn (1)

For heat flow by convection, heat transferred per unit area is

Q (W) / A (m²) = h (W/m² C) *  T (C) = T (C) / R where R = 1 / h (W/m² C) ...... eqn (2)

Solution

Gypsum: x=0.01 m, h=20 W/m2∙°C; hence from eqn (2) R_g = 1/ 20 = 0.05 m2∙°C/ W

Brick: x=0.12 m, k= 1 W/m°C ; hence from eqn (1) R_br = 0.12 / 1 = 0.12 m2∙°C/ W

Insulation: x=0.025 m, k= 0.04 W/°C; hence from eqn (1) R_i = 0.025 / 0.04 = 0.625 m2∙°C/ W

Block: x=0.20 m, C=5 W/m2∙°C; hence from eqn (2) R_bl = 1/ 5 = 0.2 m2∙°C/ W

Total resistance = R_t = R_g + R_br + R_i + R_bl = 0.05 + 0.12 +0.625 + 0.2 = 0.995 m2∙°C/ W

Hence from eqn (1) Q_t = A * T_t / R_t in respective units .... eqn (3)

From data T_t (C) = (21-18) = 3 C and  A = 600 m^{2 ;} From calculation R_t = 0.995 m2∙°C/ W

Hence after substituting in eqn (3) we get Q_t = 1809 W = 1809 J/s. This is the heat loss through the system.

This heat loss need to be compensated from outside if the temperature in the building needs to be maintained at 21 C.

Time of maintaining the building under cooling over a period of 365 days= t= 365 day/year* 24 hr/day* 60 min/hr* 60 s/hr = 31536000 s

a) With 365 days in a year, energy required to cool the building =Q₃₆₅₌ Q_t* t = 1809 J/ s * 31536000 s = 57.05 * e9 J = 57.05 GJ

b) COP for heating system = Actual heat supplied per hour / Heat to be removed per hour

2.4 = Q_A / (Q_t * 60 * 60) in consistent units

Hence Q_A (W) = 2.4 * (Q_t * 60 * 60) = 2.4 * 1809 * 60 * 60 = 15629760 W

Q_A (W) = 15629760 W-h = 15629.8 kWh

Unit Cost is $0.10/kWh . Hence energy cost for one hour = $0.10/kWh * 15629.8 kWh = $1563 per hour

For 365 days or one year the total energy cost =   $1563 per hour * 24 h/day * 365 day/ year = $13.7 million