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3 years ago

Question 3 (5 points)

Engineering

1 answer:

Aleks [24]3 years ago

6 0

Answer:

Technician A is corect

Explanation:

Because he said the correct reason

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The steel 4140 steel contains 0.4% C, however, it shows higher yield strength and ultimate strength than that of the 1045 (0.45%

Aleonysh [2.5K]

Answer:

4140 steel contains 0.4% C having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

Explanation:

we have given 4140 steel contains 0.4% C

we know here that 4140 steel is low steel alloy , and it have low amount of chromium , manganese etc alloying element

and these elements which are present in 4140 steel they increase yield strength and ultimate strength of steel

while in 1045 steel contains 0.45 % c is plain carbon steel

and it do not contain any alloying element

so that 4140 steel contains 0.4% C having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

4 0

3 years ago

An unknown relative passes away and bequeaths upon you a small tract of land in Amherst. You decide to build a two-story storage

Ratling [72]

Answer: It does make sense, because I've been involved in these careers and have a long family line of them. And other questions?

Explanation:

7 0

3 years ago

Consider a modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each repl

Ulleksa [173]

Answer:

The answers to the question are

(1) Process 1 to 2

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

(2) Process 2 to 3

W = 0

Q = 1135.376 kJ/kg

(3) Process 3 to 4

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

(4) Process 4 to 3

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency = 49.9 %

Explanation:

(a) Volume compression ratio $\frac{v_1}{v_2} = 10$

Initial pressure p₁ = 1 bar

Initial temperature, T₁ = 310 K

cp = 1.005 kJ/kg⋅K

Temperature T₃ = 2200 K from the isentropic chart of the Otto cycle

For a polytropic process we have

$\frac{p_1}{p_2} = (\frac{v_2}{v_1} )^n$ Therefore p₂ = p₁ ÷ $(\frac{v_2}{v_1} )^n$ = (1 bar) ÷ $(\frac{1}{10} )^{1.3}$ = 19.953 bar

Similarly for a polytropic process we have

$\frac{T_1}{T_2} = (\frac{v_2}{v_1} )^{n-1}$ or T₂ = T₁ ÷ $(\frac{v_2}{v_1} )^{n-1}$ = $\frac{310}{0.1^{0.3}}$ = 618.531 K

The molar mass of air is 28.9628 g/mol.

Therefore R = $\frac{8.3145}{28.9628}$ = 0.287 kJ/kg⋅K

cp = 1.005 kJ/kg⋅K Therefore cv = cp - R = 1.005- 0.287 = 0.718 kJ/kg⋅K

1). For process 1 to 2 which is polytropic process we have

W = $\frac{R(T_2-T_1)}{n-1}$ = $\frac{0.287(618.531-310)}{1.3 - 1}$ = 295.16 kJ/kg

Q = $(\frac{n-\gamma}{\gamma - 1} )W$ = $(\frac{1.3-1.4}{1.4-1} ) 295.16 kJ/kg$ = -73.79 kJ/kg

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

2). For process 2 to 3 which is reversible constant volume heating we have

W = 0 and Q = cv×(T₃ - T₂) = 0.718× (2200-618.531) = 1135.376 kJ/kg

W = 0

Q = 1135.376 kJ/kg

3). For process 3 to 4 which is polytropic process we have

W = $\frac{R(T_4-T_3)}{n-1}$ = Where T₄ is given by $\frac{T_4}{T_3} = (\frac{v_3}{v_4} )^{n-1}$ or T₄ = T₃ × $0.1^{0.3}$

= 2200 × $0.1^{0.3}$ T₄ = 1102.611 K

W = $\frac{0.287(1102.611-2200)}{1.3 - 1}$ = -1049.835 kJ/kg

and Q = 262.459 kJ/kg

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

4). For process 4 to 1 which is reversible constant volume cooling we have

W = 0 and Q = cv×(T₁ - T₄) = 0.718×(310 - 1102.611) = -569.09 kJ/kg

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency is given by

$\eta = 1-\frac{T_4-T_1}{T_3-T_2}$ = $1-\frac{1102.611-310}{2200-618.531}$ = 0.499 or 49.9 % Efficient

$p_{m} = \frac{p_1r[(r^{n-1}-1)(r_p-1)]}{ (n-1)(r-1)}$ where r = compression ratio and $r_p$ = $\frac{p_3}{p_2}$

However p₃ = $\frac{p_2T_3}{T_2}$ = $\frac{(19.953)(2200)}{618.531}$ =70.97 atm

$r_p$ = $\frac{p_3}{p_2}$ = $\frac{70.97}{19.953} = 3.56$

Therefore $p_m =\frac{1*10*[(10^{0.3}-1)(3.56-1)]}{0.3*9}$ = 9.44 bar

Please find attached generalized diagrams of the Otto cycle

8 0

3 years ago

How might an operations manager alter operations to meet customer demand? Name at least two ways.

Citrus2011 [14]

One way is manager changes itself and the other one is the same thing i think.

4 0

3 years ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$