1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kirza4 [7]
3 years ago
6

A fixed mass of saturated water vapor at 400 kpa is isothermally cooled until it is a saturated liquid. Calculate the amount of

heat rejected during this process in kJ/kg.

Engineering
1 answer:
Kazeer [188]3 years ago
6 0
This is the explanation

You might be interested in
6) Describe the differences between the troposphere and stratosphere.
vodka [1.7K]

Answer:

frejfru8758u37

Explanation:

i need points

8 0
2 years ago
un contenedor de 0.01m∧3 se llena con 2kg de nitrogeno a una presion de 15mpa ¿cual es la temperatura del nitrogeno?resolver uti
777dan777 [17]
Ecfñnokg pinogdf gabn Etta r
7 0
3 years ago
A distillation column is initially designed to separate a mixture of toluene and xylene at around ambient temperature (say, 100°
weeeeeb [17]

Answer:

Purchase cost= 87056

Bar module cost= 292725

Explanation:

solution is attached below

6 0
3 years ago
2. BCD uses 6 bits to represent a symbol. a) True b) False​
Goshia [24]

Answer:

true because BCD used 6 bits to represent a symbol .

Explanation:

mark me brainlist

4 0
2 years ago
Consider a regenerative gas-turbine power plant with two stages of compression and two stages of expansion. The overall pressure
iris [78.8K]

Answer: the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

Explanation:

from the T-S diagram, we get the overall pressure ratio of the cycle is 9

Calculate the pressure ratio in each stage of compression and expansion. P1/P2 = P4/P3  = √9 = 3

P5/P6 = P7/P8  = √9 =3  

get the properties of air from, "TABLE A-17 Ideal-gas properties of air", in the text book.

At temperature T1 =300K

Specific enthalpy of air h1 = 300.19 kJ/kg

Relative pressure pr1 = 1.3860  

At temperature T5 = 1200 K

Specific enthalpy h5 = 1277.79 kJ/kg

Relative pressure pr5 = 238  

Calculate the relative pressure at state 2

Pr2 = (P2/P1) Pr5

Pr2 =3 x 1.3860 = 4.158  

get the two values of relative pressure between which the relative pressure at state 2 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.  

Relative pressure pr = 4.153

The corresponding specific enthalpy h = 411.12 kJ/kg  

Relative pressure pr = 4.522

The corresponding specific enthalpy h = 421.26 kJ/kg  

Find the specific enthalpy of state 2 by the method of interpolation

(h2 - 411.12) / ( 421.26 - 411.12) =  

(4.158 - 4.153) / (4.522 - 4.153 )

h2 - 411.12 = (421.26 - 411.12) ((4.158 - 4.153) / (4.522 - 4.153))  

h2 - 411.12 = 0.137

h2 = 411.257kJ/kg  

Calculate the relative pressure at state 6.

Pr6 = (P6/P5) Pr5

Pr6 = 1/3 x 238 = 79.33  

Obtain the two values of relative pressure between which the relative pressure at state 6 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.  

Relative pressure Pr = 75.29

The corresponding specific enthalpy h = 932.93 kJ/kg  

Relative pressure pr = 82.05

The corresponding specific enthalpy h = 955.38 kJ/kg  

Find the specific enthalpy of state 6 by the method of interpolation.

(h6 - 932.93) / ( 955.38 - 932.93) =  

(79.33 - 75.29) / ( 82.05 - 75.29 )

(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )

h6 - 932.93 = 13.427

h6 = 946.357 kJ/kg

Calculate the total work input of the first and second stage compressors

(Wcomp)in = 2(h2 - h1 ) = 2( 411.257 - 300.19 )

= 222.134 kJ/kg  

Calculate the total work output of the first and second stage turbines.

(Wturb)out = 2(h5 - h6) = 2( 1277.79 - 946.357 )

= 662.866 kJ/kg  

Calculate the net work done

Wnet = (Wturb)out  - (Wcomp)in

= 662.866 - 222.134

= 440.732 kJ/kg  

Calculate the minimum mass flow rate of air required to generate a power output of 105 MW

W = m × Wnet

(105 x 10³) kW = m(440.732 kJ/kg)

m = (105 x 10³) / 440.732

m = 238.2 kg/s

therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

4 0
3 years ago
Other questions:
  • Solve the compound inequality. 3x − 4 > 5 or 1 − 2x ≥ 7
    8·1 answer
  • A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs ar
    8·1 answer
  • Reusable refrigerant containers under high-pressure must be hydrostatically tested how often?
    10·1 answer
  • Im passed due someone help meeeeeee
    7·2 answers
  • Find the general solution of the equation<br>a) Tan A = 1/√3​
    11·1 answer
  • How to solve circuit theory using mesh analysis<br>​
    11·1 answer
  • 50 points
    7·1 answer
  • Bob would like to run his house off the grid, therefore he needs to find out how many solar panels and batteries he needs to buy
    12·1 answer
  • Most equipment is cooled by bringing cold air in the front and ducting the heat out of the back. What is the term for where the
    9·1 answer
  • If a material is found to be in the tertiary phase of creep, the following procedure should be implemented:
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!