Equiaxed MAR-M 247 alloy is to support a stress of 276 MPa (Fig. 7.31). Determine thetime to stress rupture at 850°C.Larsen-Miller parameter, P = [TCC) + 273] [20+ log()] × 10-³23.3 24.4 25.5 26.7 27.8 28.922.23031.1DS MAR-M 247 longitudinal MFB,1221°C/2 t/GFO, 982°C/5 h/AC.871°C/20 h/AC, 1.8 mm diameterproduction dataEquiaxed MAR-M 247 MFB.982°C/5 W/AC+871°C/20 W/ACand 871 C/20 h/AC only.1.8 mm diameter specimensStress (MPa)690552414276207DS CM 247 LC, longitudinal.1232°C/2h+1260°C/20 W/AC982°C/5 h/AC, 871°C/20 V/AC3.2 and 4.1 mm diameter specimens138100806040302010DS CM 247 LC longitudinal,"1232°C/2h+ 1260°C/2 MAC.T1079 C/4 WAC, 871°C/20 W/ACand 1232°C/2h+1260°C/20/AC.1050°C/16h/AC, 871°C/20 WAC3.2 and 4.1 mm diameter specimens40485052545642 44 46Larsen-Miller parameter, P = [T(F) + 460] [20+ log()] x 10-3Figure 7.31Stress (ksi)

Draw the diagram and write the Larsen miller parameter respect to stress of 276 MPa.

Equiaxed MAR-M 247 alloy is to support a stress of 276 MPa (Fig. 7.31). Determine the time to stress rupture at 850°C. Larsen-Miller parameter, P=[T(C) + 273] [20+ log()] x 10³ 23.3 22.2 690 24.4 25.5 26.7 27.8 28.9 30 31.1 552 DS MAR-M 247 longitudinal MFB. 1221°C/2 l/GFO, 982C/5 W/AC. 871°C/20 AC, 1.8 mm diameter production data 414 276 Equiaxed MAR-M 247 MFB. 982°C/5 W/AC+871°C/20 V/AC and 871 C/20 h/AC only. 207 1.8 mm diameter specimens Stress (MPa) DS CM 247 LC, longitudinal 1232C/2h+1260°C/20 W/AC 982C/5 W/AC, 871°C/20 V/AC 3.2 and 4.1 mm diameter specimens 1.38 69 100 80 60 40 DS CM 247 LC longitudinal, 1232C/2h+1260°C/2 MAC. 1079 C/4 MAC, 871C/20 MAC and 1232C/2h+ 1260°C/20/AC. 1050°C/16h/AC, 871°C/20W/AC, 3.2 and 4.1 mm diameter specimens 40 42 46 48 50 52 54 56 44 Larsen-Miller parameter, P- [T(F) + 460] [20+ log(r)] x 10-3 S 20 10 Stress (ksi)

Equiaxed MAR-M 247 alloy is to support a stress of 276 MPa (Fig. 7.31). Determine the time to stress rupture at 850°C. Larsen-Miller parameter, P=[T(C) + 273] [20+ log()] x 10³ 23.3 22.2 690 24.4 25.5 26.7 27.8 28.9 30 31.1 552 DS MAR-M 247 longitudinal MFB. 1221°C/2 l/GFO, 982C/5 W/AC. 871°C/20 AC, 1.8 mm diameter production data 414 276 Equiaxed MAR-M 247 MFB. 982°C/5 W/AC+871°C/20 V/AC and 871 C/20 h/AC only. 207 1.8 mm diameter specimens Stress (MPa) DS CM 247 LC, longitudinal 1232C/2h+1260°C/20 W/AC 982C/5 W/AC, 871°C/20 V/AC 3.2 and 4.1 mm diameter specimens 1.38 69 100 80 60 40 DS CM 247 LC longitudinal, 1232C/2h+1260°C/2 MAC. 1079 C/4 MAC, 871C/20 MAC and 1232C/2h+ 1260°C/20/AC. 1050°C/16h/AC, 871°C/20W/AC, 3.2 and 4.1 mm diameter specimens 40 42 46 48 50 52 54 56 44 Larsen-Miller parameter, P- [T(F) + 460] [20+ log(r)] x 10-3 S 20 10 Stress (ksi)

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Thermal Properties of Materials

In mechanical engineering, the thermal property is represented as the physical quantity that is relevant to the application of heat. The thermal properties are evaluated by variations in any specific material due to low heat or high heat. The quantities that influence the thermal property of the materials are mass, density, temperature, and many others.

Question

Equiaxed MAR-M 247 alloy is to support a stress of 276 MPa (Fig. 7.31). Determine the
time to stress rupture at 850°C.
Larsen-Miller parameter, P = [TCC) + 273] [20+ log()] × 10-³
23.3 24.4 25.5 26.7 27.8 28.9
22.2
30
31.1
DS MAR-M 247 longitudinal MFB,
1221°C/2 t/GFO, 982°C/5 h/AC.
871°C/20 h/AC, 1.8 mm diameter
production data
Equiaxed MAR-M 247 MFB.
982°C/5 W/AC+871°C/20 W/AC
and 871 C/20 h/AC only.
1.8 mm diameter specimens
Stress (MPa)
690
552
414
276
207
DS CM 247 LC, longitudinal.
1232°C/2h+1260°C/20 W/AC
982°C/5 h/AC, 871°C/20 V/AC
3.2 and 4.1 mm diameter specimens
138
100
80
60
40
30
20
10
DS CM 247 LC longitudinal,"
1232°C/2h+ 1260°C/2 MAC.T
1079 C/4 WAC, 871°C/20 W/AC
and 1232°C/2h+1260°C/20/AC.
1050°C/16h/AC, 871°C/20 WAC
3.2 and 4.1 mm diameter specimens
40
48
50
52
54
56
42 44 46
Larsen-Miller parameter, P = [T(F) + 460] [20+ log()] x 10-3
Figure 7.31
Stress (ksi)

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