Fill in the blank to correctly complete the statement below.

Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh

Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

$R_a=e^{-\left ( \frac{T-r}{n}\right )B}$

$R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}$

$R_a=e^{-4.4}=0.01227$

For B

$R_b=e^{-\left ( \frac{300-336}{100}\right )3}$

$R_b=e^{1.08}=2.944$

B is better for 100 hours

(b)For 750 hours

$R_a=e^{-0.5866}=0.55621$

$R_b=e^{0.144}=1.154$

So here B is more Reliable.

3 0

3 years ago

A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas (both gases are monatomic) at 27

lidiya [134]

Answer:

Q = 62 ( since we are instructed not to include the units in the answer)

Explanation:

Given that:

$n_{HCl} = 3 \ kmol\\n_{Ar} = 7 \ k mol$

$T_1 = 27^0 \ C = ( 27+273)K = 300 K$

$P_1 = 200 \ kPa$

Q = ???

Now the gas expands at constant pressure until its volume doubles

i.e if $V_1 = x\\V_2 = 2V_1$

Using Charles Law; since pressure is constant

$V \alpha T$

$\frac{V_2}{V_1} =\frac{T_2}{T_1}$

$\frac{2V_1}{V_1} =\frac{T_2}{300}$

$T_2 = 300*2\\T_2 = 600$

mass of He =number of moles of He × molecular weight of He

mass of He = 3 kg × 4

mass of He = 12 kg

mass of Ar =number of moles of Ar × molecular weight of Ar

mass of He = 7 kg × 40

mass of He = 280 kg

Now; the amount of Heat Q transferred = $m_{He}Cp_{He} \delta T + m_{Ar}Cp_{Ar} \delta T$

From gas table

$Cp_{He} = 5.9 \ kJ/Kg/K\\Cp_{Ar} = 0.5203 \ kJ/Kg/K$

∴ Q = $12*5.19*10^3(600-300)+280*0.5203*10^3(600-300)$

Q = $62.389 *10^6$

Q = 62 MJ

Q = 62 ( since we are instructed not to include the units in the answer)

5 0

3 years ago

Read 2 more answers

Determine (with justification) whether the following systems are (i) memoryless, (ii) causal, (iii) invertible, (iv) stable, and

lina2011 [118]

Answer:

a.

y[n] = x[n] x[n-1] x[n+1]

(i) Memory-less - It is not memory-less because the given system is depend on past or future values.

(ii) Causal - It is non-casual because the present value of output depend on the future value of input.

(iii) Invertible - It is invertible and the inverse of the given system is $\frac{1}{x[n] . x[n-1] x[n+1]}$

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.

b.

y[n] = cos(x[n])

(i) Memory-less - It is memory-less because the given system is not depend on past or future values.

(ii) Causal - It is casual because the present value of output does not depend on the future value of input.

(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .

For example - for x[n] = 0 , y[n] = cos(0) = 1

for x[n] = 2 $\pi$ , y[n] = cos(2 $\pi$ ) = 1

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.

3 0

2 years ago

A cylindrical specimen of a metal alloy 45.8 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 378 MPa ca

Sliva [168]

Answer:

390.242 MPa

Explanation:

Attached is the full solution.

8 0

2 years ago

Where are the ar manufacturers not fitting the engine in the high end sport cars

fomenos

Answer:

it depends on the but i would recommend check in the front next to the turbo intake.

8 0

2 years ago