Explanation:
Diesel cycle:
All diesel engine work on diesel cycle .In diesel cycle there are four process .These processes are as follows
1. Adiabatic reversible compression
2.Heat addition at constant pressure
3.Adiabatic reversible expansion
4.Constant volume heat rejection
In general compression ratio in diesel engine is high as compare to petrol engine.But the efficiency of diesel cycle is less as compare to petrol cycle for same compression ratio.
Applications of diesel cycle:
Generally diesel cycle used for heavy vehicle or equipment because heavy vehicle or equipment is required high initial torque.So this cycle have lots of applications such as in industrial machining,in trucks,power plant,in mining ,in defense or military,large motors ,compressor and pump etc.
Answer:
pipefitters design systems whereas plumbers maintain systems
Complete question:
A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.
Answer:
Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection
so that critical flow is subject to detection
Explanation:
We are given:
Plane strain fracture toughness K 
Yield strength Y = 860 MPa
Flaw detection apparatus = 3.0mm (12in)
y = 1.0
Let's use the expression:

We already know
K= design
a = length of surface creak
Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.
Therefore
![a= \frac{1}{pi} * [\frac{k}{y*a}]^2](https://tex.z-dn.net/?f=%20a%3D%20%5Cfrac%7B1%7D%7Bpi%7D%20%2A%20%5B%5Cfrac%7Bk%7D%7By%2Aa%7D%5D%5E2%20)
Substituting figures in the expression above, we have:
![= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B1%7D%7Bpi%7D%20%2A%20%5B%5Cfrac%7B98.9%20MPa%20%5Csqrt%7Bm%7D%7D%20%7B10%20%2A%20%5Cfrac%7B860MPa%7D%7B2%7D%7D%5D%5E2)
= 0.0168 m
= 17mm
Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection
Answer:
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Explanation:
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