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3 years ago

Products that are in the process of being manufactured but are not yet complete are called:

Engineering

1 answer:

vazorg [7]3 years ago

6 0

Answer:

Those products are generally called Work in Process WIP

Explanation:

Work in process (WIP), or work in progress (WIP), goods in process, or in-process inventory in a manufacturing industry/company refer to the company's partially finished goods waiting for completion and eventual sale or the value of these items.

These items are either just being produced or require further processing (like purification, separation, packaging or handling) in a queue or a buffer storage.

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The alternator must be operated with the battery disconnected or with the terminals at the back of the alternator

QveST [7]

Answer:

true

Explanation:

4 0

3 years ago

Read 2 more answers

A fuel cell vehicle draws 50 kW of power at 70 mph and is 40% efficient at rated power. You are asked to size the fuel cell syst

svetoff [14.1K]

Answer:

a) fuel cell weight and volume: 50 kg, 33.3 L

b) fuel tank weight and volume: 321 kg, 643 L

Explanation:

Fuel Cell

Delivery of 50 kW from a source with a power density of 1.5 kW/L requires a source that has a volume of ...

(50 kW)/(1.5 kW/L) = 33.3 L

The weight of the power source is 1 kW/kg, so will be ...

(50 kW)/(1 kW/kg) = 50 kg

Fuel Tank

400 miles at 70 mph will take (400/70) h ≈ 5.71429 h. In that time, the energy used by the vehicle power plant is ...

(50 kW)(5.71429 h)(3600 s/h) = 1028.57 MJ

Since the power plant is 40% efficient, it must be supplied with 2.5 times that amount of energy, or 2571.43 MJ.

The tank volume and mass will then be ...

volume = 2571.43 MJ/(4 MJ/L) = 642.9 L

mass = 2571.43 MJ/(8 MJ/L) = 321.4 kg

_____

Comment on fuel tank volume

It appears the fuel tank would need to be equivalent in size to a sphere about 1.1 m in diameter.

4 0

3 years ago

A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th

Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588 = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ

b) $\Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)$

Where:

$x_A$ and $x_B$ are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, $x_A$ = 248 /(248 + 42.8) = 0.83

$x_B$ = 42.8/(248 + 42.8) = 0.1472

∴ $\Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472)$ = 1066.0279 J/K

5 0

3 years ago

A system consists of N very weakly interacting particles at a temperature T sufficiently high so that classical statistical mech

algol [13]

Answer:

the restoring force is = 3/4NKT

Explanation:

check the attached files for answer.

7 0

4 years ago

Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi

qaws [65]

Answer:

$\Delta V = 209.151\,L$ , $\Delta C = 217.517\,USD$

Explanation:

The drag force is equal to:

$F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A$

Where $C_{D}$ is the drag coefficient and $A$ is the frontal area, respectively. The work loss due to drag forces is:

$W = F_{D}\cdot \Delta s$

The reduction on amount of fuel is associated with the reduction in work loss:

$\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s$

Where $F_{D,1}$ and $F_{D,2}$ are the original and the reduced frontal areas, respectively.

$\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s$

The change is work loss in a year is:

$\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)$