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Mariulka [41]
3 years ago
13

Two Carnot engines operate in series such that the heat rejected from one is the heat input to the other. The heat transfer from

the high-temperature reservoir is 500 kJ. The overall temperature limits are 1000K and 400K. Both engines produce equal work. What is most nearly the intermediate temperature between the two engines?
Engineering
1 answer:
kykrilka [37]3 years ago
8 0

Answer:

Given:

high temperature reservoir T_{H} =1000k

low temperature reservoir T_{L} =400k

thermal efficiency n_{1}= n_{2}

The engines are said to  operate on Carnot cycle which is totally reversible.

To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as

n_{1} =1-\frac{T}{T_{H} }

The thermal efficiency of second heat engine can be written as

n_{2} =1-\frac{T_{L} }{T}

The temperature of intermediate reservoir can be defined as  

1-\frac{T}{T_{H} } =1-\frac{T_{L} }{T} \\T^2=T_{L} T_{H} \\T=\sqrt{T_{L} T_{H} }\\T=\sqrt{400*1000} =632k

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390.242 MPa

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Attached is the full solution.

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The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Si
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Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

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Explanation:

<u>A) Classifying the soil according to USCS system</u>

 ( using 2nd image attached below )

<em>description of sand</em> :

The soil is a coarse sand since  ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

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Group symbol as per the 2nd image attached below = SW

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5 0
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3. A 4-m × 5-m × 7-m room is heated by the radiator of a steam-heating system. The steam radiator transfers heat at a rate of 10
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14.52 minutes

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14 minutes and 31 seconds

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Specific heat at constant volume at 27°C = 0.718 kJ/kg*K

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Final temperature (required) of room = 293.15 K

Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg

Heat required at constant volume: 0.718 * (change in temp) * (mass of air)

Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ

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