Suppose an assembly requires five components from five different vendors. To guarantee starting the assembly on time with 90 per
1 answer:
Answer:
The service level for each component must be 97.91%
Explanation:
If we want a 90% confidence of starting on time, that means we need
As the probability of each component being ready is independent from the others, that means that the probability of the 5 components being ready is equal to multiply each probability:
The probability of being ready on time is equal to the service level (in fraction), and all 5 are equal so we can say:
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Answer:
where and
be the inner radius, outer radius of the annalus.
Explanation:
Let ,
and
be the inner radius, outer radius and length of the given annulus.
Temperatures at the inner surface, and at the outer surface,
.
Let q be the rate of heat transfer at the steady-state.
Given that, the heat flux at r=3cm=0.03m is
.
This heat transfer is same for any radial position in the annalus.
Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.
Now, according to Fourier's law:
where,
K=Thermal conductivity of the material.
T= temperature at any radial distance r.
A=Area through which heat transfer is taking place.
Here,
Variation of temperature w.r.t the radius of the annalus is
Putting the values from the equations (ii) and (iii) in the equation (i), we have
[as
, and
]
This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.
Answer:
T = 167 ° C
Explanation:
To solve the question we have the following known variables
Type of surface = plane wall ,
Thermal conductivity k = 25.0 W/m·K,
Thickness L = 0.1 m,
Heat generation rate q' = 0.300 MW/m³,
Heat transfer coefficient hc = 400 W/m² ·K,
Ambient temperature T∞ = 32.0 °C
We are to determine the maximum temperature in the wall
Assumptions for the calculation are as follows
- Negligible heat loss through the insulation
- Steady state system
- One dimensional conduction across the wall
Therefore by the one dimensional conduction equation we have
During steady state
= 0 which gives
From which we have
Considering the boundary condition at x =0 where there is no heat loss
= 0 also at the other end of the plane wall we have
hc (T - T∞) at point x = L
Integrating the equation we have
from which C₁ is evaluated from the first boundary condition thus
0 = from which C₁ = 0
From the second integration we have
From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows
→ C₂ =
T(x) = and T(x) = T∞ +
∴ Tmax → when x = 0 = T∞ +
Substituting the values we get
T = 167 ° C
Answer: maximum length of the nanowire is 510 nm
Explanation:
From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K
Thermal conductivity of silicon carbide k = 30 W/m.K
Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m
lets consider the equation for the value of m
m = ( (hP/kAc)^1/2 ) = ( (4h/kD)^1/2 )
m = ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04
now lets find the value of h/mk
h/mk = 10⁵ / ( 942809.04 × 30) = 0.00353
lets consider the value θ/θb by using the equation
θ/θb = (T - T∞) / (T - T∞)
θ/θb = (3000 - 8000) / (2400 - 8000)
= 0.893
the temperature distribution at steady-state is expressed as;
θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)] / [cosh mL+ (h/mk) sinh mL]
θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)] / [cosh mL+ (h/mk) sinh mL]
θ/θb = [ 1 ] / [cosh mL+ (h/mk) sinh mL]
so we substitute
0.893 = [ 1 ] / [cosh (942809.04 × L) + (0.00353) sinh (942809.04 × L)]
L = 510 × 10⁻⁹m
L = 510 nm
therefore maximum length of the nanowire is 510 nm
