Answer: 1.137*10^7 Btu/h.
Explanation:
Given data:
Efficiency of the plant = 4.5percent
Net power output of the plant = 150kw
Solution:
The required collection rate
QH = W/n
= 150/0.045 * 0.94782/ 1 /60 */60 Btu/h.
= 3333.333 *3412.152Btu/h.
= 11373840 Btu/h
= 1.137*10^7 Btu/h.
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Explanation:
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Answer:
Coiled tubing is often used to carry out operations similar to wire lining.
