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Aloiza [94]
3 years ago
8

The following data were obtained when a cold-worked metal was annealed. (a) Estimate the recovery, recrystallization, and grain

growth temperatures; (b) recommend a suitable temperature for a stress-relief heat treatment; (c) recommend a suitable temperature for a hot-working process; and (d) estimate the melting temperature of the alloy.
Engineering
1 answer:
Oduvanchick [21]3 years ago
7 0
Sorrry needdddd pointssssss
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Two common types of glue
dolphi86 [110]

Answer:

white craft glue

yellow wood glue

Explanation:

the answer is this!

was it correct?sir!?

8 0
2 years ago
Read 2 more answers
A structural element for a new bridge is designed for a constant load of 1000 psi. Its mean resistance is 1200 psi and the proba
andrew11 [14]

Answer:

i) SF = 0.83

ii) 10 psi

iii) 3.16 psi

iv) 0.0083

Explanation:

Constant load = 1000 Psi

mean resistance = 1200 psi

Probability of failure = 1.9 * 10^-3

<u>Determine </u>

<u>i) Safety factor </u>

S.F =  1 / (mean load / load design ) = 1 / ( 1200 / 1000 ) = 0.83

<u>ii) Standard deviation </u>

1.9 X 10^-3 = e  ( -1/2 ( μ / б) ^3  /  (2.5 * 6  )

0.004756 = e^- 20000 / б^3

hence std = 10 psi

<u>iii) Variance </u>

= \sqrt{10} = 3.16 psi

<u>iv) coefficient of variation </u>

Cv = std / mean resistance

    = 10 / 1200 = 0.0083

<u />

4 0
3 years ago
An insulated piston–cylinder device initially contains 1 m3 of air at 120 kPa and 17°C. Air is now heated for 15 min by a 200-W
irakobra [83]

Answer:

∆S1 = 0.5166kJ/K

∆S2 = 0.51826kJ/K

Explanation:

Check attachment for solution

4 0
3 years ago
A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa
taurus [48]

Answer:

Explanation:

Given

T_h=250^{\circ}C\approx 523\ K

T_L=30^{\circ}C\approx 303\ K

Q_1=6 kW

From Clausius inequality

\oint \frac{dQ}{T}=0  =Reversible cycle

\oint \frac{dQ}{T}  =Irreversible cycle

\oint \frac{dQ}{T}>0  =Impossible

(a)For P_{out}=3 kW

Rejected heat Q_2=6-3=3\ kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K

thus it is Impossible cycle

(b)P_{out}=2 kW

Q_2=6-2=4 kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K

Possible

(c)Carnot cycle

\frac{Q_2}{Q_1}=\frac{T_1}{T_2}

Q_2=3.47\ kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{3.47}{303}=0

and maximum Work is obtained for reversible cycle when operate between same temperature limits

P_{out}=Q_1-Q_2=6-3.47=2.53\ kW

Thus it is possible

6 0
4 years ago
Estimate the energy (head) loss a short length of a pipe conveying 300 litres of water per second and suddenly enlarging from a
Ludmilka [50]

Known :

Q = 300 L/s = 0.3 m³/s

D1 = 350 mm = 0.35 m

D2 = 700 mm = 0.7 m

g = 9.81 m/s²

Solution :

A1 = πD1² / 4 = π(0.35²) / 4 = 0.096 m²

A2 = πD2² / 4 = π(0.7²) / 4 = 0.385 m²

hL = (kL / 2g) • (U1² - U2²)

hL = (kL / 2g) • Q² (1/A1² - 1/A2²)

hL = (1 / 2(9.81)) • (0.3²) • (1/(0.096²) - 1/(0.385²))

hL = 0.467 m

5 0
3 years ago
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