Photosynthesis converts less than 0.5% of incoming solar radiation into new phytomass. The best annual fuelwood productivity of traditional fast-growing species (poplars, eucalyptus, pines) are no more than 10 t/ha. With the energy density of the dry wood averaging 18 GJ/t. (ha = 104 m²). Now, think about a large eighteenth-century city, which is located near a very large forest, would have required at least 20-30 W/m² of its built-up area for heating, cooking, and artisanal manufacturers. This city was totally dependent on the phytomass fuels. With the annual harvest of 20 t/ha: What is the approximate power density of the harvested phytomass fuel? (t = ton; ha = hectare) 600 (W)/(m²) 1.2 (W)/(m²) 1200 (W)/(m²) 1800 (W)/(m²) 0.6 (W)/(m²)
Photosynthesis converts less than 0.5% of incoming solar radiation into new phytomass. The best annual fuelwood productivity of traditional fast-growing species (poplars, eucalyptus, pines) are no more than 10 t/ha. With the energy density of the dry wood averaging 18 GJ/t. (ha = 104 m²). Now, think about a large eighteenth-century city, which is located near a very large forest, would have required at least 20-30 W/m² of its built-up area for heating, cooking, and artisanal manufacturers. This city was totally dependent on the phytomass fuels. With the annual harvest of 20 t/ha: What is the approximate power density of the harvested phytomass fuel? (t = ton; ha = hectare) 600 (W)/(m²) 1.2 (W)/(m²) 1200 (W)/(m²) 1800 (W)/(m²) 0.6 (W)/(m²)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![Photosynthesis converts less than 0.5% of incoming solar radiation into new phytomass. The
best annual fuelwood productivity of traditional fast-growing species (poplars, eucalyptus, pines)
are no more than 10 t/ha. With the energy density of the dry wood averaging 18 GJ/t. (ha =
104 m²).
Now, think about a large eighteenth-century city, which is located near a very large forest, would
have required at least 20-30 W/m² of its built-up area for heating, cooking, and artisanal
manufacturers. This city was totally dependent on the phytomass fuels. With the annual harvest
of 20 t/ha:
What is the approximate power density of the harvested phytomass fuel? (t = ton; ha =
hectare)
600 (W)/(m²)
1.2 (W)/(m²)
1200 (W)/(m²)
1800 (W)/(m²)
0.6 (W)/(m²)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F744f8ed7-0184-4cca-83d6-12fdf28e3f0b%2F1d89b2eb-0407-487c-96c9-aeeaa19473d1%2Fgoxg36y_processed.png&w=3840&q=75)
Transcribed Image Text:Photosynthesis converts less than 0.5% of incoming solar radiation into new phytomass. The
best annual fuelwood productivity of traditional fast-growing species (poplars, eucalyptus, pines)
are no more than 10 t/ha. With the energy density of the dry wood averaging 18 GJ/t. (ha =
104 m²).
Now, think about a large eighteenth-century city, which is located near a very large forest, would
have required at least 20-30 W/m² of its built-up area for heating, cooking, and artisanal
manufacturers. This city was totally dependent on the phytomass fuels. With the annual harvest
of 20 t/ha:
What is the approximate power density of the harvested phytomass fuel? (t = ton; ha =
hectare)
600 (W)/(m²)
1.2 (W)/(m²)
1200 (W)/(m²)
1800 (W)/(m²)
0.6 (W)/(m²)
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