Question

In: Mechanical Engineering

1) The Bay of Fundy-Passamaquoddy area near the border between Maine and New Brunswick/Nova Scotia in...

1) The Bay of Fundy-Passamaquoddy area near the border between Maine and New Brunswick/Nova Scotia in Canada, is another prime site for tidal power. The basin area is 700km2 and the average tidal range is 10.8 meters. Calculate the theoretical maximum power generating capability and the estimated realistic power available from this site. (SHOW your WORK)

2) A client wants you to design a roof-mounted solar water heater to supplement the energy input to their radiant floor heating system. They would like to have an indirect solar collection system similar to the one shown in Figure 11.16 of your textbook. The south-facing solar collector has a stated efficiency of 41%, and will be used to INDIRECTLY heat water in a supplemental storage tank that has a capacity of 150 Liters (~ 40 gallons). The circulating pump will operate at the rate of 20 x 10-6 m3/sec. On a typical summer day, there are 5.6 hours of direct sunshine, which creates a 15.5°C temperature rise in the solar collector fluid for that whole time period. Assume the collector fluid is water (heat capacity of 4.184 J/g-°C). What will be the increase in temperature of the water in the storage tank?

Solutions

Expert Solution

I can only answer 1 question at a time so Iam answering 1st.

tidal range of tide at a particular place = 10.8 m
The surface of the tidal energy harnessing plant is 700 km²= 7*108 m2
Specific density of sea water = 1025.18 kg/m3

Mass of the water = volume of water × specific gravity
= (area × tidal range) of water × mass density
= (7 × 108 m2 × 10.8 m) × 1025.18 kg/m3
= 7.75 × 1012 kg (approx)

Potential energy content of the water in the basin at high tide = ½ × area × density × gravitational acceleration × tidal range squared
= ½ × 7*108 m2 × 1025 kg/m3 × 9.81 m/s2 × (10.8 m)2
=4.105*1014 J (approx)

Now we have 2 high tides and 2 low tides every day. At low tide the potential energy is zero.
Therefore the total energy potential per day = Energy for a single high tide × 2
=4.105*1014 J × 2
= 8.2*1014 J

Therefore, the mean power generation potential = Energy generation potential / time in 1 day
= 8.2*1014 J / 86400 s
= 9502 MW

Therefore, theoritical maximum power generating capability = 9502 MW

the estimated realistic power available from this site = Assuming the power conversion efficiency to be 30%: The daily-average power generated = 9502MW * 30% / 100%
= 2850.6 MW (approx)

Please do rate me and mention doubts, if any, in the comments section.


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