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Degger [83]
3 years ago
8

3. A power cycle operates between hot and cold reservoirs at 500 K and 310 K respectively. At steady state the power output deve

loped is: 0.1 MW. What is the minimum rate of energy rejection by heat transfer to the cold reservoir, in MW
Engineering
1 answer:
grin007 [14]3 years ago
4 0

Answer:

0.163 MW

Explanation:

To get the minimum rate of energy rehection by heat transfer to cold reservoir, it implies that the power cycle operates in reversible cycle. The efficiency of reversible cycle whose value will be same as efficiency of power cycle will be given by

Efficiency of reversible cycle

n= (Th-Tc)/Th where T represent temperature, n efficiency, subscripts h and c hot and cold respectively

Substituting 500 and 310 for hot and cold temperatures respectively then efficiency

n=(500-310)/500=0.38

Efficiency of power cycle, n= Power output/Qh

Qh= 0.1/0.38= 0.263

Net power output, W= Qh- Qc

Qc=Qh-W= 0.263-0.1=0.163 MW

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

\Delta W = 2.043\times 10^{9}\,J

\Delta W = 2.043\times 10^{6}\,kJ

The change in chemical energy from gasoline is:

\Delta E = \frac{\Delta W}{\eta}

\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}

\Delta E = 6.81\times 10^{6}\,kJ

The changes in gasoline consumption is:

\Delta m = \frac{\Delta E}{L_{c}}

\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }

\Delta m = 154.772\,kg

\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }

\Delta V = 209.151\,L

Lastly, the money saved is:

\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )

\Delta C = 217.517\,USD

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3 years ago
A group of n Ghostbusters is battling n ghosts. Each Ghostbuster carries a proton pack, which shoots a stream at a ghost, eradic
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Answer:

Using the above algorithm matches one pair of Ghostbuster and Ghost. On  each side of the line formed by the pairing, the number of Ghostbusters and Ghosts are  the same, so use the algorithm recursively on each side of the line to find pairings. The  worst case is when, after each iteration, one side of the line contains no Ghostbusters  or Ghosts. Then, we need n/2 total iterations to find pairings, giving us an P(n^{2} lg n)-  time algorithm.

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3 years ago
A cone penetration test was carried out in normally consolidated sand, for which the results are summarized below: Depth (m) Con
Cerrena [4.2K]

Answer:

hello your question is incomplete attached below is the missing equation related to the question  

answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°

Explanation:

<u>Determine the friction angle at each depth</u>

attached below is the detailed solution

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also inverse of Tan = Tan^-1

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2 meters = 40.389°

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5 meters = 38.022°

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slamgirl [31]

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<u>Explanation:</u>

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3 years ago
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Answer:

(a)  The force sustained by the matrix phase is 1802.35 N

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Explanation:

Find attachment for explanation

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3 years ago
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