1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
NARA [144]
3 years ago
8

Sketch T-s and p-v diagrams for the Diesel cycle.

Engineering
1 answer:
labwork [276]3 years ago
3 0

Answer:

Diesel cycle:

    All diesel engine works on diesel cycle.It have four processes .These four processes are as follows

1-2.Reversible adiabatic compression

2-3.Heat addition at constant pressure

3-4.Reversible adiabatic expansion

4-1.Heat addition at constant volume

When air inters in the piston cylinder after that it compresses and gets heated due to compression after that heat addition take place at constant pressure after that power is produces when piston moves to bottom dead center.

From the diagram of P-v And T-s we can understand so easily.

You might be interested in
Question 64 (1 point)
Xelga [282]

Answer: c fine sand aggregate, portland cement,fine sand

Explanation:

7 0
3 years ago
In highways the far left lane is usually the _____
Ivan
Fastest


(Known as the fast lane)
8 0
3 years ago
Read 2 more answers
Why is it reasonable to say that no system is 100% efficient?​
Virty [35]

Generally, frictional losses are more predominant for the machines being not 100% efficient. This friction leads to the loss of energy in the form of heat, into the surroundings. Some of the supplied energy may be utilised to change the entropy (measure of randomness of the particles) of the system.

5 0
3 years ago
For a flow rate of 212 cfs find the critical depth in (a) a rectangular channel with ????=6.5 ft, (b) a triangular channel with
Fofino [41]

Answer:

A. 3.21ft

B. 3.51ft

C. 2.95ft

D. 1.5275ft

Explanation:

A) Q =212 cu.f/s

Formula for critical depth of rectangular section is: dc =[(Q^2) /(b^2(g))]^1/3

Where dc =critical depth, ft

Q= quantity of flow or discharge, ft3/s

B= width of channel, ft (m)

g = acceleration due to gravity which is 9.81m/s2 or 32.185ft/s2

Now, from the question,

Q = 212 cu.f/s and b=6.5ft

Therefore, the critical depth is: [(212^2)/(6.5^2 x32. 185)]^(1/3)

To give ; critical depth= (44,944/1359.82)^(1/3) = 3.21ft

B. Formula for critical depth of a triangular section; dc = (2Q^2/gm^2)^(1/5)

From the question, Q =212 cu.f/s and m=1.6ft while g= 32.185ft/s2

Therefore, critical depth = [(212^2) /(1.6^2 x32. 185)] ^(1/5) = (44,944/84.466)^(1/5) = 3.51ft

C. For trapezoidal channel, critical depth(y) is derived from (Q^2 /g) = (A^3/T)

Where A= (B + my)y and T=(B+2my)

Now from the question, B=6.5ft and m=5ft.

Therefore, A= (6.5 + 2y)y and T=(6. 5 + 2(5y))= 6.5 + 10y

Now, let's plug the value of A and T into the initial equation to derive the critical depth ;

(212^2 /32.185) = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Which gives;

1396.43 = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Multiply both sides by 6.5 + 10y to get;

1396.43(6.5 + 10y) = [((6.5 + 2y)^3)y^3]

Factorizing this, we get y = 2. 95ft

D) Formula for critical depth of a circular section; dc =D/2[1 - cos(Ѳ/2)]

Where D is diameter of pipe and Ѳ is angle at critical depth in radians.

Angle not given, so we assume it's perpendicular angle is 90.

Since angle is in radians, therefore Ѳ/2 = 90/2 = 45 radians ; converting to degree, = 2578. 31

Therefore, dc = (6.5/2) (1 - cos (2578.31))

dc = 3.25(1 - 0.53) = 3.25 x 0.47 = 1.5275ft

8 0
3 years ago
A 4KJ of energy is supplied to a machine used for lifting a mass. The force required is 800N. If the machine has an efficiency o
navik [9.2K]

The height at which the mass will be lifted is; 3 meters

<h3>How to utilize efficiency of a machine?</h3>

Formula for efficiency is;

η = useful output energy/input energy

We are given

η = 60% = 0.6

Input energy = 4 KJ = 4000 J

Thus;

0.6 = useful output energy/4000

useful output energy = 0.6 * 4000

useful output energy = 2400 J

Work done in lifting mass(useful output energy) = force * distance moved

Useful output energy = 800 * h

where h is height to lift mass

Thus;

800h = 2400

h = 2400/800

h = 3 meters

Read more about Machine Efficiency at; brainly.com/question/3617034

#SPJ1

8 0
2 years ago
Other questions:
  • System grounding on a power system means electrically connecting the __?__ of every wye-connected transformer or generator to ea
    12·1 answer
  • What Degree Do You Need To Become a Solar Engineer?<br> (2 or more sentences please)
    13·1 answer
  • What considerations are included in the Preliminary Floodproofing/Retrofitting Preference Matrix?
    7·1 answer
  • Evaporation in Double-Effect Reverse-Feed Evaporators. A feed containing 2 wt % dissolved organic solids in water is fed to a do
    14·1 answer
  • Explain why change is inevitable in complex systems and give examples (apart from prototyping and incremental delivery) of softw
    6·1 answer
  • A three-phase transformer bank consists of 3 single-phase transformers to handle 400 kVA witha 34.5kV/13.8kV voltage ratio. Find
    7·1 answer
  • Q#3:(A)Supose we extend the circular flow mode to add imports and export copy the circular flow digram onto a sheet paper and th
    15·1 answer
  • Problem 89:A given load is driven by a 480 V six-pole 150 hp three-phase synchronous motor with the following load and motor dat
    11·1 answer
  • Which type of blade is used with a demolition saw?
    11·1 answer
  • Contrast moral and immoral creativity and innovation<br>​
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!