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
= 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
Answer:
i)ω=3600 rad/s
ii)V=7059.44 m/s
iii)F=1245.8 N
Explanation:
i)
We know that angular speed given as
We know that for one revolution
θ=2π
Given that time t= 2 hr
So
ω=θ/t
ω=2π/2 = π rad/hr
ω=3600 rad/s
ii)
Average speed V
Where M is the mass of earth.
R is the distance
G is the constant.
Now by putting the values
V=7059.44 m/s
iii)
We know that centripetal fore given as
Here given that m= 200 kg
R= 8000 km
so now by putting the values
F=1245.8 N
Answer:
Explanation:
Assumptions is that
1. The flow is an unsteady one
2. Bubbles diameter is constant
3. The bubble velocity is slow
4. There is no homogenous reaction
5. It has a one dimensional flux model along the radial direction
Answer:
Explanation:
= 113.27 g - 49.31 g = 63.96 g
= 100.06 g - 49.31 g = 50.75
Moisture /water content, w = =
= 0.26 = 26%
Void ratio = water content X specific gravity of solid
= 0.26 X 2.80 =0.728