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stepan [7]
3 years ago
5

Three communications channels in parallel have independent failure modes of 0.1 failure per hour. These components must share a

common transceiver. Determine the MTTF of the transceiver in order that the system has a reliability of 0.85 to support a 5-hr mission. Assume constant failure rates.
Engineering
1 answer:
ozzi3 years ago
7 0

Answer:

the MTTF of the transceiver is 50.17

Explanation:

Given the data in the question;

failure modes = 0.1 failure per hour

system reliability = 0.85

mission time =  5 hours

Now, we know that the reliability equation for this situation is;

R(t) = [ 1 - ( 1 - e^{-0.1t )³] e^{-t/MTTF

so we substitute

R(5) = [ 1 - ( 1 - e^{-0.1(5) )³] e^{-5/MTTF = 0.85

[ 1 - ( 1 - e^{-0.5 )³] e^{-5/MTTF = 0.85

[ 1 - ( 0.393469 )³] e^{-5/MTTF = 0.85

[ 1 - 0.06091 ] e^{-5/MTTF = 0.85

0.9391 e^{-5/MTTF = 0.85  

e^{-5/MTTF = 0.85 / 0.9391

e^{-5/MTTF = 0.90512

MTTF = 5 / -ln( 0.90512 )

MTTF = 50.17

Therefore, the MTTF of the transceiver is 50.17

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Explanation:

Step1

In the stress-strain curve of any material, the yield stress is the maximum stress at which material starts yielding.

Step2

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3 years ago
malott, m. e. (2003). paradox of organizational change: engineering organizations with behavioral systems analysis.
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7 0
1 year ago
A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60
Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

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- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

                     T_2 = T_1 * r^(k-1/k)

                     T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

                     T_7 = T_6 * r^(-k-1/k)

                     T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

                     flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

                     flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

                     flow(m) = 7.941 lbm/s

- The heat input is as follows:

                     Q_in = c_p*(T_6 - T_5)

                     Q_in = 0.24*(1400 - 1022.84)

                     Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

                     Q_out = c_p*(T_10 - T_1)

                     Q_out = 0.24*(711.744 - 520)

                    Q_out = 56.01856 Btu/lbm

                                           

                     

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Learn more about the torque at brainly.com/question/28220969

#SPJ4

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1 year ago
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