Answer:
Explanation:
To solve this problem we use the expression for the temperature film
Then, we have to compute the Reynolds number
Re<5*10^{5}, hence, this case if about a laminar flow.
Then, we compute the Nusselt number
but we also now that
but the average heat transfer coefficient is h=2hx
h=2(8.48)=16.97W/m^{2}K
Finally we have that the heat transfer is
In this solution we took values for water properties of
v=16.96*10^{-6}m^{2}s
Pr=0.699
k=26.56*10^{-3}W/mK
A=1*0.5m^{2}
I hope this is useful for you
regards
Answer:
ΔT= 11.94 °C
Explanation:
Given that
mass of water = 10 kh
Time t= 15 min
Heat lot from water = 400 KJ
Heat input to the water = 1 KW
Heat input the water= 1 x 15 x 60
=900 KJ
By heat balancing
Heat supply - heat rejected = Heat gain by water
As we know that heat capacity of water
Now by putting the values
900 - 400 = 10 x 4.187 x ΔT
So rise in temperature of water ΔT= 11.94 °C
Answer:
non-functional requirement,
Yes they can.
The application loading time is determined by testing system under various scenarios
Explanation:
non-functional requirement are requirements needed to justify application behavior.
functional requirements are requirements needed to justify what the application will do.
The loading time can be stated with some accuracy level after testing the system.
It is study of the relationships between heat, temprature, work and energy
The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%
W = 
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
<h3>What is the workdone by this engine?</h3>
B. W = 420 kj/s
= 420x1000 w
= 4.2x10⁵W
The work done is 4.2x10⁵W
c. 780/308 - 1200/1000
= 2.532 - 1.2
= 1.332kj
The total enthropy gain is 1.332kj
D. Q1 = 1200
T1 = 1000
<h3>Cournot efficiency = W/Q1</h3>
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: brainly.com/question/6364271
