(a)
Interpretation:
The term "Quenching" needs to be explained.
Concept Introduction:
Quenching is one of the important concepts in material science and metallurgy. In quenching the fast cooling of workpiece takes place in order to obtain required material characteristics. In the cooling process, the cooling fluids used are water, air, oil. Quenching is the advantageous in case of low-temperature process like phase transformation.
(b)
Interpretation:
The term "Tempering" needs to be explained.
Concept Introduction:
To change the undesirable properties of steel as well as cast iron into desirable and commercially acceptable, various methods are used in which tampering is one of them, it is a heat treating, progress the toughness especially of iron-based alloys.
Tempering method is done by heating the metal to particular temperature, but the temperature must be below the critical point for a certain period of time.
(c)
Interpretation:
The term "retained austenite" needs to be explained.
Concept Introduction:
To change the undesirable properties of steel as well as cast iron into desirable and commercially acceptable, various methods are used in which retained austenite is one of them, In the retained austenite process, austenite that does not completely transform to martensite quenching.
This retained austenite microconstituent occurs when the steel is not quenched to a temperature low enough to form 100% martensite.
(d)
Interpretation:
The term "mar-quenching or "martempering" needs to be explained.
Concept Introduction:
Martempering is also called as interrupted quenching or stepped quenching. The steel above the transformation range or critical point is heated in this process.
In this method, the austenite is a transformed to martensite at stepped quenching, at a rate fast sufficient to avoid the formation of ferrite, bainite or pearlite.
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Essentials of Materials Science and Engineering, SI Edition
- Chapter 12 - Lecture Notes.pptx: (MAE 272-01) (SP25) DY... Scoresarrow_forward%9..+ ۱:۱۹ X خطأ عذرا ، الرقم الذي أدخلته خاطئ. يرجى إدخال رقم بطاقة الشحن الصالحة والمحاولة مرة أخرى. رصيد هاتفك قم بمسح الرمز = رقم بطاقة التعبئة 7794839909080 رمز مكون من 13 او 14 رقماً طريقة إعادة التعبئة قم باعادة تعبئة الرصيد إعادة تعبئة الإنترنت إعادة تعبئة الرصيد O >arrow_forwardProblem 10.14 A hexagonal plane of side 30 mm has a corner in the V.P. The surface of the plane is inclined at 45° to the V.P. and perpendicular to the H.P. Draw its projections. Assume that the diagonal through the corner in the V.P. is parallel to the H.P. d' a 2 b b.f C' c.e b 'C' H.P. (a) V.P E HEX 30 e' O' d' a a' b' C' b' X y a b,f c,e d b,f (b) c,earrow_forwardConsider a large 6-cm-thick stainless steel plate (k = 15.1 W/m-K) in which heat is generated uniformly at a rate of 5 × 105 W/m³. Both sides of the plate are exposed to an environment at 30°C with a heat transfer coefficient of 60 W/m²K. Determine the value of the highest and lowest temperature. The highest temperature is The lowest temperature is °C. °C.arrow_forwardSketch and explain a PV Diagram and a Temperature Entropy Diagram for a 4 stroke diesel engine please, please explain into detail the difference bewteen the two and referance the a diagram. Please include a sketch or an image of each diagramarrow_forwardProblem 10.18 A 60° set-square has the shortest edge of 40 mm lying in the V.P. The surface is in- clined to the V.P. and perpendicular to the H.P. such that the front view appears as an isosceles triangle. Draw the projections of the set-square and determine its inclination with the V.P. Construction Refer to Fig. 10.18. A 60° set-square inclined to the V.P. and per- pendicular to the H.P. can appear as an isosceles triangle in the front view, when the shorter edge is in the V.P. 1. First stage Draw a right angled triangle 40 a' a' b' c' b' X 40 C' a,b C a,b a'b'c' keeping 40 mm long a'b' perpen- dicular to xy. Project the corners to xy and obtain ac as the top view. 2. Second stage Draw another right angled triangle a'b'c' on the horizontal locus line from points a', b' and c' of the first stage such that length of b'c' is equal to that of a'b'. Project a'b' to meet xy at ab. Draw an arc with centre a and radius equal to ac of the first stage to meet the vertical projector of c' at…arrow_forward□ 40 a' 12 o' a O d'arrow_forwardDraw left view of the first orthographic projectionarrow_forwardSketch and Describe a timing diagram for a 2 stroke diesel engine emphasis on the 2 stroke as my last answer explained 4 stroke please include a diagram or sketch.arrow_forwardA 4 ft 200 Ib 1000 Ib.ft C 2 ft 350 Ib - за в 2.5 ft 150 Ib 250 Ib 375 300 Ib Replace the force system acting on the frame. shown in the figure by a resultant force (magnitude and direction), and specify where its line of action intersects member (AB), measured from point (A).arrow_forwardTL = 85 D. Problem 9.12 The top view of a line measures 60 mm. The line is parallel to the V.P. and inclined at 45° to the H.P. One end of the line is 25 mm in front of the V.P. and lies on the H.P. Draw its projections and determine the true length. Interpretation Let the line be PQ parallel to the V.P. The front view has true length and the top view is parallel to xy. Construction Refer to Fig. 9.12. 1. Draw a reference line xy. Mark point p' on xy and point p 25 mm below xy. 2. Draw a 60 mm long line pq parallel to xy. This repre- sents the top view. 3. Draw line from point p', inclined at 45° to xy to meet the projector from point q at point q'. Join p'a' to represent the front view. Measure length of p'a' as true length of line PQ. Here T.L. = 85 mm. Result True length of line PQ is p'q' = 85 mm. 5 Fig. 9.12 p 60 σarrow_forwardProblem 9.13 A 70 mm long line PQ does not have H.T. and V.T. One end of the line is 30 mm in front of the V.P. and 20 mm above the H.P. Draw its projections. Interpretation As the line PQ does not have H.T. and V.T., it is parallel to both H.P. and V.P. Construction Refer to Fig. 9.13. 1. Draw a reference line xy. Mark point p' 20 mm above xy and point p 30 mm below xy. 2. Draw a 70 mm long line p'a' parallel to xy to repre- sent the front view. X 20 p Fig. 9.13 3. Also, draw a 70 mm long line pq parallel to xy to represent the top view. 70 q yarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_iosRecommended textbooks for you
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