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Westkost [7]
3 years ago
14

Suppose an assembly requires five components from five different vendors. To guarantee starting the assembly on time with 90 per

cent confidence, what must the service level be for each of the five components? (Assume the same service level for each component).
Engineering
1 answer:
spayn [35]3 years ago
6 0

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

P_{\mbox{starting on time}}=P_{\mbox{every component being ready on time}}=0.9\\

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:

0.9=P_{\mbox{component 1 ready on time} } * P_{\mbox{component 2 ready on time} } *\\ P_{\mbox{component 3 ready on time} } * P_{\mbox{component 4 ready on time} } *\\P_{\mbox{component 5 ready on time} }

The probability of being ready on time is equal to the service level (in fraction), and all 5 are equal so we can say:

0.9=(\mbox{service level(in fraction)})^5\\\\\sqrt[5]{0.9} =\mbox{service level(in fraction)}=0.9791\\\mbox{In percentage}: \mbox{service level (in fraction)}*100 = 97.91\%

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faltersainse [42]

Answer:

a) 0.489

b) 54.42 kg/s

c) 247.36 kW/s

Explanation:

Note that all the initial enthalpy and entropy values were gotten from the tables.

See the attachment for calculations

4 0
4 years ago
An experiment to determine the convection coefficient associated with airflow over the surface of a thick stainless steel castin
Maksim231197 [3]

Answer:

h = 375 KW/m^2K

Explanation:

Given:

Thermo-couple distances: L_1 = 10 mm , L_2 = 20 mm

steel thermal conductivity k = 15 W / mK

Thermo-couple temperature measurements: T_1 = 50 C , T_2 = 40 C

Air Temp T_∞ = 100 C

Assuming there are no other energy sources, energy balance equation is:

                                               E_in = E_out

                                        q"_cond = q"_conv

Since, its a case 1-D steady state conduction, the total heat transfer rate can be found from Fourier's Law for surfaces 1 and 2

q"_cond = k * (T_1 - T_2) / (L_2 - L_1) = 15 * (50 - 40) / (0.02 - 0.01)

=15KW/m^2

Assuming SS is solid, temperature at the surface exposed to air will be 60 C since its gradient is linear in the case of conduction, and there are two temperatures given in the problem. Convection coefficient can be found from Newton's Law of cooling:

q"_conv = h * ( T_∞  - T_s ) ----> h = q"_conv / ( T_∞  - T_s )

                                                   h = 15000 W / (100 - 60 ) C = 375 KW/m^2K

4 0
3 years ago
A light bulb is switched on and within a few minutes its temperature becomes constant. Is it at equilibrium or steady state.
EleoNora [17]

Answer:

The temperature attains equilibrium with the surroundings.  

Explanation:

When the light bulb is lighted we know that it's temperature will go on increasing as the filament of the bulb has to  constantly dissipates energy during the time in which it is on. Now this energy is dissipated as heat as we know it, this heat energy is absorbed by the material of the bulb which is usually made up of glass, increasing it's temperature. Now we know that any object with temperature above absolute zero has to dissipate energy in form of radiations.

Thus we conclude that the bulb absorbs as well as dissipates it's absorbed thermal energy. we know that this rate is dependent on the temperature of the bulb thus it the temperature of the bulb does not change we can infer that an equilibrium has been reached in the above 2 processes i.e the rate of energy absorption equals the rate of energy dissipation.

Steady state is the condition when the condition does not change with time no matter whatever the surrounding conditions are.

6 0
3 years ago
Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0
3 years ago
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Answer:

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