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3 years ago

9

Tech A says that a mechanical pressure regulator exhausts excess fluid back to the transmission pan. Tech B says that if the tra

nsmission pan is removed, the magnet must be replaced. Who is correct?

1 answer:

Roman55 [17]3 years ago

4 0

Answer:

A

Explanation:

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A ten-station transfer machine has an ideal cycle time of 30 sec. The frequency of line stops is 0.075 stops per cycle. When a l

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Explanation:

We are given that

A ten-station transfer machine has ideal cycle time=30 sec= $\frac{30}{60}=0.5 min$

Frequency of line=0.075 stops per cycle

Average time=4 min

We have to determine the line efficiency

$T_p=0.5+0.075(4)=0.5+0.3=0.8$

Line efficiency= $\frac{0.5}{0.8}\times 100=$ 62.5%

Hence, the line efficiency=62.5%

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Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is be

zheka24 [161]

Answer:

as the exercise is incomplete, I add the information that is missing:

Consider a packed bed of 75-mm-diameter aluminum spheres ? = 2700 kg/m3, and c = 950 J/kg

Answer: The copper can store 34.1% more thermal energy at 272.42ºC

Explanation:

Given:

Diameter packed bed = 75 mm

Density = 2700 kg/m³

specific heat = 950 J/kg K

initial temperature sphere = 25ºC

thermal conductivity = 240 W/m K

temperature of gas = 300ºC

convection coefficient = 75 W/m²K

The characteristic length is

$L_{c} =\frac{r}{3} =\frac{0.0375}{3} =0.0125m$

The Biot number is equal to

$Bi=\frac{hL_{c} }{k} =\frac{75*0.0125}{240} =3.9x10^{-6}$

Bi < 1, we will use the lumped capacitance method

The energy transfer is equal to:

$\frac{Q}{Q_{max} } =(1-exp(\frac{t}{t_{t} } ))$ (eq. 1)

$t_{t} =\frac{pV}{hA} =\frac{p\frac{\pi D^{3} }{6} c}{h\pi D^{2} } =\frac{pDc}{6h} \\t_{t}=\frac{2700*0.075*950}{6*75} =427.5s$

Replacing in eq. 1

$0.9=1-exp(\frac{-t}{t_{t} } )\\0.1=exp(\frac{-t}{t_{t} } )\\\frac{-t}{t_{t} } =ln(0.1)\\t=2.3t_{t}=2.3*427.5=983.25s$

The temperature at the center of sphere is

$T=T\alpha +(T_{i} -T\alpha )exp(\frac{-6ht}{pDc} )=300+(25-300)exp(\frac{-6*75*983.25}{2700*0.075*950} )=272.42C$

We obtain the percentage increase

Cpcu = 385 J/kg K

(pCp)cu = 8933 * 385 = 3439205 J/m³K

from water:

(pCp)wa = 2700 * 950 = 2565000 J/m³K

the percentage increase is

%P = (3439205 - 2565000)/2565000 = 34.1%

The copper can store 34.1% more thermal energy.

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3 years ago