Answer:
attached below
Explanation:
A mechanical system comprises three subsystems in series with reliabilities of 98, 96, and 94 percent. What is the overall relia
1 answer:
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Answer:
F = 8849 N
Explanation:
Given:
Load at a given point = F = 4250 N
Support span = L = 44 mm
Radius = R = 5.6 mm
length thickness of tested material = 12 mm
First compute the flexural strength for circular cross section using the formula below:
σ
σ = FL / π R³
Putting the given values in the above formula:
σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³
= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³
= 187 / 3.141593 (0.0056)³
= 338943767.745358
= 338.943768 x 10⁶
σ = 338 x 10⁶ N/m²
Now we compute the load i.e. F from the following formula:
= 2 σ
d³/3 L
F = 2σd³/3L
= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)
= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)
= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)
= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)
= 146016 / 125 / 3 ( 44 x 1 / 1000 )
= ( 146016 / 125 ) / (3 ( 11 / 250 ))
= 97344 / 11
F = 8849 N
Answer:
Time period = 41654.08 s
Explanation:
Given data:
Internal volume is 210 m^3
Rate of air infiltration
length of cracks 62 m
air density = 1.186 kg/m^3
Total rate of air infiltration
total volume of air infiltration
Time period
Answer:
a)120C
b)29kg
Explanation:
Hello!
To solve this exercise follow the steps below
1. we will call 1 the initial state, 2 the steam that enters and 3 the final state
2. We find the quality of the initial state, dividing the mass of steam by the total mass.
3 Find the internal energy in the three states using thermodynamic tables
note:Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
u1=IntEnergy(Water;x=0,6(quality);P=200kPa) =1719KJ/kg
u2=IntEnergy(Water;t=350;P=5000kPa) =2808KJ/kg
u3=IntEnergy(Water;x=1;P=200kPa) =2529KJ/kg
4. use the internal energy and pressure to find the temperature in state 3, using thermodynamic tables
T3=Temperature(Water;P=200kPa;u=u3=2529KJ/kg)=120C
5. Use the first law of thermodynamics in the system, it states that the initial energy in a system must be equal to the final
m1u1+m2u2=(m1+m2)u3
where
m1=inital mass=10kg
m2=the mass of the steam that has entered.
solve for m2
(m1)(u1-u3)=(m2)(u3)-(m2)(u2)