Solar Thermal Systems Questions and Answers

Q: Passive solar heating technique can be used in building design in mid by latitude having :

Passive solar heating uses large windows facing south.In the mid-latitudes of the northern hemisphere, the winter sun stays low in the southern sky. Large south-facing glazing admits that sunshine, and thermal mass inside the building stores the heat. Overhangs shade the same windows from the high summer sun.North-facing windows get almost no direct sun in winter and lose heat. Small south windows admit too little energy, and east or west windows receive only low-angle morning or evening sun and are hard to shade in summer.

Passive solar heating technique can be used in building design in mid by latitude having :

A.Large windows facing South
B.Large windows facing North
C.Small windows facing South
D.Windows facing East or West

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

Passive solar heating uses large windows facing south.

In the mid-latitudes of the northern hemisphere, the winter sun stays low in the southern sky. Large south-facing glazing admits that sunshine, and thermal mass inside the building stores the heat. Overhangs shade the same windows from the high summer sun.

North-facing windows get almost no direct sun in winter and lose heat. Small south windows admit too little energy, and east or west windows receive only low-angle morning or evening sun and are hard to shade in summer.

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In a salt gradient solar pond, the salinity generally varies from top to bottom of the pond as

A.\(\lt\) 5% to ~ 20%
B.~ 10% to ~ 30%
C.~ 20% to ~ 35%
D.~ 20% to ~ 40%

Exam Year(s)

UGC NET - 2013,

Answer: A

In a salt-gradient solar pond the salinity rises with depth, from under about 5% at the top to about 20% at the bottom.

The dense, very salty bottom layer absorbs sunlight and heats up, but it cannot rise because it is heavier than the layers above. The middle layer acts as a non-convecting insulating zone, which traps the heat in the bottom layer so that it can reach 70 to 80 °C. The other ranges start at a higher surface salinity, which would remove the large density contrast the pond depends on.

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For a solar flat plate collector the following data is given : Useful heat gain = 28 watts/m\(^{2}\) per hour, solar radiation intensity = 350 watts/m\(^{2}\) per hour and the factor to convert beam radiation to that on the plane of the collector = 1.2. The collector efficiency is

A.~ 6.6 %
B.~ 4.8 %
C.~ 12.2 %
D.~ 15.2 %

Exam Year(s)

UGC NET - 2013,

Answer: A

Collector efficiency is the useful heat gain divided by the solar radiation incident on the plane of the collector, which is the intensity multiplied by the conversion factor.

\[ \eta = \frac{28}{350 \times 1.2} \]

\[ \eta = \frac{28}{420} = 0.0667 \approx 6.6\% \]

This is option A. The values 4.8, 12.2 and 15.2 percent do not result from this data.

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A solar pond has electricity generating capacity of 600 MWe. If the efficiency of solar energy to electric generation process was 2% and solar energy supply rate was 300 W/m\(^{2}\), what is the area of solar pond ?

A.100 km\(^{2}\)
B.90 km\(^{2}\)
C.60 km\(^{2}\)
D.180 km\(^{2}\)

Exam Year(s)

UGC NET - 2013,

Answer: A

The thermal energy needed is the electric power divided by the efficiency, and the area is that power divided by the solar supply rate.

\[ P_{solar} = \frac{600 \text{ MW}}{0.02} = 30000 \text{ MW} \]

\[ A = \frac{3 \times 10^{10}}{300} = 10^{8} \text{ m}^2 \]

\[ A = 100 \text{ km}^2 \]

So the answer is 100 km\(^{2}\), option A. The other areas do not follow from the data.

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