Answer:
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Answer:
This question is comprising many parts (a to r). That is impossible to answer in one sheet. Following are attached images having answers to most of the parts.
I hope it will help you a lot.
Explanation:
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.
Answer:https://global.oup.com/us/companion.websites/9780199385423/student/ch6/mcq/ just go here
Explanation:
Answer:
battery life in year = 9 years and 48 days
Explanation:
given data
Battery Ampere-hours = 1.5
Pulse voltage = 2 V
Pulse width = 1.5 m sec
Pulse time period = 1 sec
Electrode heart resistance = 150 Ω
Current drain on the battery = 1.25 µA
to find out
battery life in years
solution
we get first here duty cycle that is express as
duty cycle =
...............1
duty cycle = 1.5 × 
and applied voltage will be
applied voltage = duty energy × voltage ...........2
applied voltage = 1.5 ×
× 2
applied voltage = 3 mV
so current will be
current =
................3
current = 
current = 20 µA
so net current will be
net current = 20 - 1.25
net current = 18.75 µA
so battery life will be
battery life = 
battery life = 80000 hours
battery life in year = 
battery life in year = 9.13 years
battery life in year = 9 years and 48 days