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Simora [160]
3 years ago
15

To 3 significant digits, what is the change of entropy of air in kJ/kgk if the pressure is decreased from 400 to 300 kPa and the

temperature is increased from 300 to 900 K? DO NOT ASSUME constant specific heats.

Engineering
1 answer:
Levart [38]3 years ago
5 0

Answer:

The change of entropy is 1.229 kJ/(kg K)

Explanation:

Data

T_1 = 300 K

T_2 = 900 K

p_1= 400 kPa

p_2= 300 kPa

R= 0.287 kJ/(kg K) (Individual Gas Constant for air)  

For variable specific heats  

s(T_2, p_2) - s(T_1, p_1) = s^0(T_2) - s^0(T_1) - R \, ln \frac{p_2}{p_1}

where s^0(T) is evaluated from table  attached

s^0(900 K) = 2.84856 kJ/(kg K)

s^0(300 K) = 1.70203 kJ/(kg K)

Replacing in equation

s(900 K, 300 kPa) - s(300 K, 400 kPa) = 2.84856 kJ/(kg K) - 1.70203 kJ/(kg K) - 0.287 kJ/(kg K) \, ln \frac{300 kPa}{400 kPa}

s(900 K, 300 kPa) - s(300 K, 400 kPa) = 1.229 kJ/(kg K)

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Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0
3 years ago
Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of subst
Kaylis [27]

Answer:

The specific heat capacity of substance A is 1.16 J/g

Explanation:

The substances A and B come to a thermal equilibrium, therefore, the heat given by the hotter substance B is absorbed by the colder substance A.

The equation becomes:

Heat release by Substance B = Heat Gained by Substance A

The heat can be calculated by the formula:

Heat = mCΔT

where,

m = mass of substance

C = specific heat capacity of substance

ΔT = difference in temperature of substance

Therefore, the equation becomes:

(mCΔT) of A = (mCΔT) of B

<u>FOR SUBSTANCE A:</u>

m = 6.01 g

ΔT = Final Temperature - Initial Temperature

ΔT = 46.1°C - 20°C = 26.1°C

C = ?

<u>FOR SUBSTANCE B:</u>

m = 25.6 g

ΔT = Initial Temperature - Final Temperature

ΔT = 52.2°C - 46.1°C = 6.1°C

C = 1.17 J/g

Therefore, eqn becomes:

(6.01 g)(C)(26.1°C) = (25.6 g)(1.17 J/g)(6.1°C)

C = (182.7072 J °C)/(156.861 g °C)

<u>C = 1.16 J/g</u>

5 0
3 years ago
Write a Nios II assembly program that reads binary data from the Slider Switches, SW11-0, on the DE2-115 Simulator, and display
Citrus2011 [14]

Algorithm of the Nios II assembly program.

  • Attain data for simulation from the  SW11-0, on the DE2-115 Simulator
  • The data will be  read from the switches in loop.
  • The decimal output is displayed using the seven-segment displays and done using the loop.
  • The program is ended by the user operating the SW1 switch

and

The decimal equivalent on the seven-segment displays HEX3-0 is

  • DE2-115
  • DE2-115_SW11
  • DE2-115_HEX3
  • DE2-115_HEX4
  • DE2-115_HEX5
  • DE2-115_HEX6
  • DE2-115_HEX7

<h3>The  Algorithm and decimal equivalent on the seven-segment displays HEX3-0</h3>

Generally,  the program will be written using a  cpulator simulator in order to attain best result.

We are to

  • Attain data for simulation from the  SW11-0, on the DE2-115 Simulator
  • The data will be  read from the switches in loop.
  • The decimal output is displayed using the seven-segment displays and done using the loop.
  • The program is ended by the user operating the SW1 switch

This will be the Algorithm of the Nios II assembly program .

Hence, the decimal equivalent on the seven-segment displays HEX3-0 is

  • DE2-115
  • DE2-115_SW11
  • DE2-115_HEX3
  • DE2-115_HEX4
  • DE2-115_HEX5
  • DE2-115_HEX6
  • DE2-115_HEX7

For more information on Algorithm

brainly.com/question/11623795

6 0
2 years ago
La extensión de un resumen debe estar en un rango de _________ del texto inicial. a)25 a 35 % b)10 a 20 % c)15 a 20 % d)20 a 25
mestny [16]

Answer:

b)

Explanation:

because it is correct

3 0
3 years ago
Create a set of questions a company designing a new motorcycle for a younger audience might want to ask external customers to ma
Brut [27]

Answer:

Who would pay for my product or service?

Who has already bought from me?

Am I overestimating my reach?.

What does my network think?.

Am I making assumptions based on my personal knowledge and experience?

What's my revenue model

How will I sell my product or service?

How did my competitors get started?

How will I find my customers?

Is there room to expand my target market?

3 0
2 years ago
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