Question

Q4. A restaurant in Suwaiq requires 85000 Whr of heat during winter season for the month of January to March to retain a constant inside temperature of 295 K. If the solar heating system has 40 % efficiency, design a storage tank and collector with appropriate size based on the given data. Assume the temperature of the hot water at the outlet of the storage tank is 69 0C. Represent the space heating setup with a neat sketch. Under which category this type of space heating falls? justify your answer. Also, discuss and analyze the effects on the collector area and the mass of water, when the solar radiation changes from every month by considering the suitable global solar radiation data?

Answer 1

Answer:

Given data:

Heat required, Q = 85000 Whr = 85000*3600 joules = 306000 kJ

Inside temperature, Ti = 295 K = 295-273 °C = 22 °C

Hot water temperature, Th = 69 °C

Efficiency of solar collector, = 40% = 0.40

Total time period, t = 3 months = 3*30 days = (90 days)*(8 shours/day)

[ Assuming that sufficient sun light is for 8 hours per day].

Hence, t = (720 hours) = 720* 3600 s = 2592000 s

Assumptions:

1.Specific heat of the water, c = 4.18 kJ/kg°C

2.Density of water, = 1000 kg/m^3

3.Assuming diameter of sorage tank, d = 1 m

4.Assuming ideal heating of resturatnt by storage water

5.Suitably average value global solar radiation, E = 1360 W/m^2

6.Assume solar collector is of square shape

Analysis:

(a) the mass of water, m:

From Newton's law of cooling,

Q = m*c*(To-Ti)..................................................(i)

substituting the value of known data in equation (i)

306000*10^3 = m*4.18*(10^3)*(69 - 22)

m = 1557.57 kg Ans(a)

Comment: This much amount is needed for the desired heating purpose

for the given period of time. But water can be recycled very easily, hence

net requirement of water can be reduced significantly.

(b) size of the storage tank, dia (d) and height (h):

volume of water required = volume of storage tank

......................................................(ii)

substituting all the known parameters into the above relation, we get

h = 1.98 m = 2m

Therefore; for storage tank

Diameter, d = 1 m (assumed value)  

and, Height, h = 2 m (approximately) Ans(b)

(c) size of the solar collector (it is assumed of square shape), with a side length a:

Efficiency = Heat energy output/ Radiation energy input

here, Heat energy output = Heat Required = Q,

Radiation energy input = solar constant* total time period*area of solar collector = E*t*a^2

thus,

  ....................................................(iii)

by substituting known data in the above equaition and solving for siz eof collector, we finally obtained,

a = 0.466 m = 46.6 cm Ans(c)

Comment : Total area of the solar collector is 2171.56 square cm.

(d) the space heating setup with a neat sketch:

(e) Category of the system: Broadly it is simply a solar heating system but precisley it can be termed as the

SWH: Solar Water Heating.

Fundamently solar energy is being converted  into heat by solar thermal collectors and further this heat

energy is used to heat water which finally transfer this energy to the main system being used for human

comfort and other services.