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Kobotan [32]
2 years ago
15

Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr

essure is 100 kPa and produces 630 kJ of work output. Assume air has constant specific heats evaluated at 300 K.
Determine the entropy change of the air.
Engineering
1 answer:
son4ous [18]2 years ago
4 0

Answer:

The entropy change of the air is 0.240kJ/kgK

Explanation:

T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa

T_{2}  is unknown

we can apply the following expression to find T_{2}

-w_{out} =mc_{v} (T_{2} -T_{1} )

T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }

now substitute

T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K

To find entropy change of the air we can apply the ideal gas relationship

Δs_{air}=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }

Δs_{air} =1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )

Δs_{air} =0.240kJ/kgK

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2 years ago
A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an
Anarel [89]

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here  thermodynamic equation that

energy equation that is

h1 + \frac{v1}{2}  + q = h2 + \frac{v2}{2}  + w

put here value with

turbine is insulated so q = 0

so here

3015.4 *1000 + \frac{10^2}{2}  =  2431.7 * 1000 + \frac{90^2}{2}  + w

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

7 0
3 years ago
A single-phase transformer circuit feeds a motor and lighting load of 50 kilowatts. At a power factor of .8, the KVA rating of t
AveGali [126]

The KVA rating of the step down transformer at the given power factor would be 62.5 kVA.

<h3>What is power factor of a transformer?</h3>

Power factor (PF) is the ratio of working power, measured in kilowatts (kW), to apparent power, measured in kilovolt amperes (kVA).

PF = working power / apparent power

PF =  kW/kVA

kVA = kW/PF

kVA = 50 kW/0.8

kVA = 62.5 kVA

Thus, the KVA rating of the step down transformer at the given power factor would be 62.5 kVA.

Learn more about power factor here: brainly.com/question/7956945

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Identify the Levels of Biological Organization below. Copy and answer each statement. Write
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Answer:

no idea

Explanation:

gvghvvvghhbbgggggggig gjgjg gjgbg gig gg bg GB g GB GB GB GB g GB GB GB of me and I am a diary and crewmate hide if you see a diary or not the other person is a non-electrolyte or no one vent the imposter win a diary or you can get any more confidently than you are in a spectacular place is the most important points to be mo to get skins and crewmate for a few days to be sure it will be the same for me it will take a while for a couple to get the right angle bu the other one is come to the end and then you have birds to get a lot more than a diary of a candle or no other one in a few weeks or no time to do it for me and my friend who has no idea how much time I have been to a new episodes and the best of all the time I am going sleep

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3 years ago
A heat pump cycle whose coefficient of performance is 2.5 delivers energy by heat transfer to a dwelling at a rate of 20kW.
12345 [234]

Answer:

a) 8kW

b) $128

Explanation:

Given the coefficient of performance of the heat pump cycle to be 2.5

Energy delivered by the heat pump = 20kW

a) net power required to operate the heat pump = Energy delivered / coefficient of performance

Net power required = 20/2.5

= 8kW

b) Given the cost of electricity is $0.08 for 1kWhour

Since net power required to operate heat pump = 8kW

If the heat pump operate for 200hours, total power required for a month = 8kW×200hours = 1600kWhour

since 1kWh of electricity costs $0.08, cost of electricity used in a month when the pump operates for 200hour will be 1600kWh×$0.08 which is equivalent to $128

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