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:
total time = 304.21 s
Explanation:
given data
y = 50% = 0.5
n = 1.1
t = 114 s
y = 1 - exp(-kt^n)
solution
first we get here k value by given equation
y = 1 - ...........1
put here value and we get
0.5 = 1 - e^{(-k(114)^{1.1})}
solve it we get
k = 0.003786 = 37.86 ×
so here
y = 1 -
1 - y =
take ln both side
ln(1-y) = -k ×
so
t = .............2
now we will put the value of y = 87% in equation with k and find out t
t =
total time = 304.21 s
Answer: the absolute static pressure in the gas cylinder is 82.23596 kPa
Explanation:
Given that;
patm = 79 kPa, h = 13 in of H₂O,
A sketch of the problem is uploaded along this answer.
Now
pA = patm + 13 in of H₂O ( h × density × g )
pA= 79 + (13 × 0.0254 × 9.8 × 1000/1000)
pA = 82.23596 kPa
the absolute static pressure in the gas cylinder is 82.23596 kPa
