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Anarel [89]
3 years ago
14

We have a parallel-plate capacitor, with each plate having a width W and a length L. The plates are separated by air with a dist

ance d. Assume that L and W are both much larger than d. The maximum voltage that can be applied is limited to Vmax = Kd, in which K is called the breakdown strength of the dielectric. Derive an expression for the maximum energy that can be stored in the capacitor in terms of K and the volume of the dielectric. If we want to store the maximum energy per unit volume, does it matter what values are chosen for L, W, and d? What parameters are important?
Engineering
1 answer:
Vesnalui [34]3 years ago
8 0

Answer:

k

Explanation:

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Air enters the compressor of a cold air-standard Brayton cycle with regeneration and reheat at 100 kPa, 300 K, with a mass flow
yanalaym [24]

Answer:

a. 47.48%

b. 35.58%

c. 2957.715 KW

Explanation:

T_2 =T_1 + \dfrac{T_{2s} - T_1}{\eta _c}

T₁ = 300 K

\dfrac{T_{2s}}{T_1} = \left( \dfrac{P_{2}}{P_1} \right)^{\dfrac{k-1}{k} }

T_{2s} = 300 \times (10) ^{\dfrac{0.4}{1.4} }

T_{2s} = 579.21 K

T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K

T₃ = T₂ + \epsilon _{regen}(T₅ - T₂)

T₄ = 1400 K

Given that the pressure ratios across each turbine stage are equal, we have;

\dfrac{T_{5s}}{T_4} = \left( \dfrac{P_{5}}{P_4} \right)^{\dfrac{k-1}{k} }

T_{5s} = 1400×\left( 1/\sqrt{10}  \right)^{\dfrac{0.4}{1.4} }  = 1007.6 K

T₅ = T₄ + (T_{5s} - T₄)/\eta _t = 1400 + (1007.6- 1400)/0.8 = 909.5 K

T₃ = T₂ + \epsilon _{regen}(T₅ - T₂)

T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K

T₆ = 1400 K

\dfrac{T_{7s}}{T_6} = \left( \dfrac{P_{7}}{P_6} \right)^{\dfrac{k-1}{k} }

T_{7s} = 1400×\left( 1/\sqrt{10}  \right)^{\dfrac{0.4}{1.4} }   = 1007.6 K

T₇ = T₆ + (T_{7s} - T₆)/\eta _t = 1400 + (1007.6 - 1400)/0.8 = 909.5 K

a. W_{net \ out} = cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg

Heat supplied is given by the relation

cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg

Thermal efficiency of the cycle = (Net work output)/(Heat supplied)

Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%

b. bwr = \dfrac{W_{c,in}}{W_{t,out}}

bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)]  = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%

c. Power = 6 kg *492.9525 KJ/kg  = 2957.715 KW

3 0
3 years ago
Describe two characteristics that bridges and skyscrapers have in common.
Elan Coil [88]

Answer:

skyscrapers and bridges both have structural builds accounting for the influence of wind and they both cost big money

Explanation:

7 0
3 years ago
Chemical milling is used in an aircraft plant to create pockets in wing sections made of an aluminum alloy. The starting thickne
Lelu [443]

Answer:

a) metal removal rate is 1915.37 mm³/min

b) the time required to etch to the specified depth is 500 min or 8.333 hrs

Explanation:

Given the data in the question;

starting thickness of one work part of interest = 20 mm

depth of series of rectangular-shaped pockets = 12 mm

dimension of pocket = 200 mm by 400 mm

radius of corners of each rectangle = 15 mm

penetration rate = 0.024 mm/minute

etch factor = 1.75

a)

To get the metal removal rate MRR;

The initial area will be smaller compare to the given dimensions of 200mm by 400mm and the metal removal rate would increase during the cut as area is increased. so'

A = 200 × 400 - ( 30 × 30 - ( π × 15² ) )

= 80000 - ( 900 - 707 )      

= 80000 - 193

A = 79807 mm²

Hence, metal removal rate MRR = penetration rate × A

MRR = 0.024 mm/minute × 79807 mm²

MRR = 1915.37 mm³/min

Therefore, metal removal rate is 1915.37 mm³/min

b) To get the time required to etch to the specified depth;

Time to machine ( etch ) =  depth of series of rectangular-shaped pockets / penetration rate

we substitute

Time to machine ( etch ) = 12 mm / 0.024 mm/minute

Time to machine ( etch ) = 500 min or 8.333 hrs

Therefore, the time required to etch to the specified depth is 500 min or 8.333 hrs

3 0
3 years ago
Which utility program reads an assembly language source file and produces an object file?
iragen [17]

Answer:

Assembler

Explanation:

An assembler can be define as a computer utility program that read, interpret and convert software programs written in low level assembly language into an object file, machine language, code and instruction that can be understood and executed by a computer.

5 0
3 years ago
A plant has ten machines and currently operates two 8-hr shift per day, 5 days per week, 50 weeks per year. the ten machines pro
Zarrin [17]

Answer:

  83.6%

Explanation:

<h3>(a)</h3>

On its current schedule, the plant can theoretically produce ...

  (30 pc/h/machine)(10 machine)(8 h/shift)(2 shift/day)(5 day/wk)(50 wk/yr)

  = (30)(10)(8)(2)(5)(50) pc/yr = 1,200,000 pc/yr

__

<h3>(b)</h3>

On the proposed schedule, this production becomes ...

  (30 pc/mach)(10 mach)(8 h/sh)(3 sh/da)(6 da/wk)(51 wk/yr)

  = (30)(10)(8)(3)(6)(51) pc/yr = 2,203,200 pc/yr

The increase in capacity is ...

  (2,203,200/1,200,000 -1) × 100% = 83.6% . . . capacity increase

_____

<em>Additional comment</em>

The number of parts per shift did not change. Only the number of shifts per year changed. It went up by a factor of (3/2)(6/5)(51/50) = 1.836. Hence the 83.6% increase in capacity.

We have to assume that maintenance and repair are done as effectvely as before in the reduced down time that each machine has.

3 0
2 years ago
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