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
Answer:
Theory of Mind : A robotic head has a face that recognizes and simulates emotions.
Self aware : A robot tries to protect itself from harm
Purely Reactive : A door automatically opens when a person steps in front of it.
Limited Memory : A personal assistant software tracks a persons travel routes and suggests shorter routes.
Explanation:
Artificial intelligence is simply a technology which enables automation. It enables system perform task without being explicitly controlled. Purely reactive systems do not store Data in memory, it simply observes what going on at the moment which is what it was programmed to do and takes a step. One the machine detects someone approaching up to a certain distance, it opens.
Limited Memory systems store information about the past and this enhances its Decison making, prediction engines, self driving cars use this kind of artificial intelligence.
Theory of Mind : Here, systems are trained to detect, understand and replicate what is understood. Once the robot identifies an emotion, it replicates it.
Self - Aware : An advanced level of AI, where systems will not only be able to replicate what they see, but also make conscious decisions as to which action to take in different circumstances.
Answer:
Explanation:
First we compute the characteristic length and the Biot number to see if the lumped parameter
analysis is applicable.
Since the Biot number is less than 0.1, we can use the lumped parameter analysis. In such an
analysis, the time to reach a certain temperature is given by the following
From the data in the problem we can compute the parameter, b, and then compute the time for
the ratio (T – T)/(Ti
– T)
Answer:
work is 50 kj
Explanation:
Given data
heat (Q) = 50 kj
To find out
work input for the compression stroke per kilogram of air
Solution
we will apply here "first law of thermodynamics" i.e.
The First Law of Thermodynamics states that heat is a form of energy, subject to the principle of conservation of energy, that heat energy cannot be created or destroyed. It can be transferred from one location to another location. i.e.
ΔU = Q – W ................1
here ΔU is change in internal energy, Q is heat and W is work done
here U = 0 because air compressor the compression takes place at a constant internal energy in question
so that by equation 1
Q = W
and Q = 50
so work will be 50 kj
