Starting from the results of Examples 5.4 and 5.5, estimate the number of cubic meters of water that would flow each second thro
1 answer:
Answer: Hello Results of examples 5.4 and 5.5 are missing from your question but I was able to find it online and use it to resolve your question
answer :
a) 1780 m^3/sec
b) 23.81 m
Explanation:
<u>a) calculate the number of m^3 of water flowing through an OTEC plant charge </u>
<em>assumption : assuming an ideal heat engine</em>
Q / t = m*c*∆T * (eff) / t
1e^9 J/s = 1g/cc *V(1cal/g°C) * 2°C (6.7%)*(4.184J/cal) / t
Hence: V / t ( number of m^3 of water ) = 1.78e^9 cc / sec = 1.78 * 10
= 1780 m^3/sec
<em><u>b) Determine the diameter of pipes </u></em>
Given that the water flows at 4m/s ( velocity )
1780 m^3 /sec = A * V ------ ( 1 )
A = π*d^2/4 = 1780 / 4 = 445 m^2 -------- ( 2 )
v = 4 m/s
Back to equation 2
d^2 = ( 445 * 4 ) / π
≈ 567
∴ necessary diameter ( d ) = = 23.81 m
You might be interested in
Answer:
Speed of the boat, v = 4.31 m/s
Explanation:
Given that,
Height of the bridge, h = 32 m
The model boat is 11 m from the point of impact when the key was released, d = 11 m
Firstly, we will find the time needed for the boat to get in this position using second equation of motion as :
Here, u = 0 and a = g
t = 2.55 seconds
Let v is the speed of the boat. It can be calculated as :
v = 4.31 m/s
So, the speed of the boat is 4.31 m/s. Hence, this is the required solution.