Question

Consider a house with a 1 ton air conditioner, run- ning 500 h/year at full power.

(a) What is the annual thermal energy delivered to the house?
(b) What is the corresponding electricity consump- tion, if the air conditioner delivers 2.0 J, of cool- ing for 1.0 J, of electricity (COP = 2.0)?
(c) Suppose one stores winter ice in an ice tank for summer cooling. How large a volume is needed if there are no losses from storage? Consider only the latent heat of melting, and take the density of ice as 0.9 ton/m³. (Such sea- sonal storage has been proposed and tested in various places.)
(d) How much is the cooling energy of 1 ton (2205 lbm) of ice worth? %3D

Answer 1

a) annual thermal energy delivered to the house is 6.3 × 10⁹ J

b) corresponding electricity consumption is 3.15 × 10⁹ J

c) volume is needed if there are no losses from storage is 20.895 m³

d) E = 335 MJ

Step-by-step explanation:

a)

Given that 1 ton refrigeration is 3.5 kw

so Annual thermal energy delivered to the house is

Energy = power × time

E = 3.5 × ( 500×3600)

we converted 500 hrs to seconds

E = 6.3 × 10⁶ kJ

E = 6.3 × 10⁹ J

therefore annual thermal energy delivered to the house is 6.3 × 10⁹ J

b)

corresponding electricity consumption {Cop = 2}

Electricity consumption = Refrigeration effect / work input

∴ EC = 6.3 × 10⁹J / 2

EC = 3.15 × 10⁹ J

∴ corresponding electricity consumption is 3.15 × 10⁹ J

c)

we know that latent heat of ice is 335 kj/kg = 335 × 10³ J/kg

now let m represent the mass of ice needed for required refrigeration

E = mL

6.3 × 10⁹ j = m × (335 × 10³J/kg)

m = 6.3 × 10⁹ J / 335 × 10³J/kg

m = 18805.97 kg

Given that density of ice = 0.9ton/m³ = 900 kg/m³

NOW

Volume of ice needed V = mass / density

v = 18805.97 kg / 900 kg/m³

v = 20.895 m³

volume is needed if there are no losses from storage is 20.895 m³

d)

cooling energy of 1 ton ( m = 1000 kg) ice

we know L = 335 kJ

E = mL

E = 1000 × 335KJ

E = 335 MJ