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HACTEHA [7]
3 years ago
11

Ammonia enters an adiabatic compressor operating at steady state as saturated vapor at 300 kPa and exits at 1400 kPa, 140◦C. Kin

etic and potential energy effects are negligible. Determine:
a. power input required [kJ/kg]
b. isentropic compressor efficiency
c. rate of entropy production per unit mass [kJ/kg K] in the compressor

Engineering
1 answer:
hammer [34]3 years ago
8 0

Answer:

a. 149.74 KJ/KG

b. 97.9%

c. 0.81 kJ/kg K

Explanation:

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There are 20 forging presses in the forge shop of a small company. The shop produces batches of forgings requiring a setup time
Aleksandr-060686 [28]

Answer:

Considering the guidelines of this exercise.

The pieces produced per month are 504 000

The productivity ratio is 75%

Explanation:

To understand this answer we need to analyze the problem. First of all, we can only produce 2 batches of production by the press because we require 3 hours to set it up. So if we rest those 6 hours from the 8 of the shift we get 6, leaving 2 for an incomplete bath. So multiplying 2 batches per day of production by press we obtain 40 batches per day. So, considering we work in this factory for 21 days per month well that makes 40 x 21  making 840 then we multiply the batches for the pieces 840 x 600 obtaining 504000 pieces produced per month. To obtain the productivity ratio we need to divide the standard labor hours meaning 6 by the amount of time worked meaning 8. Obtaining 75% efficiency.

4 0
3 years ago
g A pump is required to deliver 100 gpm at a head of 100 ft, but the pump rated capacity is 150 gpm at a head of 100 ft. If the
Thepotemich [5.8K]

Answer: valving the pump discharge to reduce the flow will only result in an increase in the water velocity according to the laws of continuity of flow.

Q = AV = constant.

The pump speed of 150 gpm is, and will remain constant. Valving simply reduces flow area A, which is balanced out by increased velocity V of water through the pipe.

This does not affect the pump speed Q (flow rate) and hence it remains the same.

8 0
3 years ago
What is an advantage of a nuclear-fission reactor?.
Kobotan [32]

Answer:

Nuclear fission is almost 8,000 times more efficient than traditional fossil fuels at producing energy. That's a lot of energy packed into a small space. Nuclear energy is more efficient, which means it uses less fuel to power the plant and produces less waste.


advantages:
-produces no polluting gases
-does not contribute to global warming
-very low fuel costs
-Low fuel quantity reduces mining and transportation effects on environment
-High technology research required benefits other industries
-Power station has very long lifetime

Disadvantages:
-Waste is radioactive and safe disposal is very difficult and expensive
-Local thermal pollution from wastewater affects marine life
-Large-scale accidents can be catastrophic
-Public perception of nuclear power is negative
-Costs of building and safely decommissioning are very high
-Cannot react quickly to changes in electricity demand

4 0
2 years ago
Select the best answer to the questo
Norma-Jean [14]

Answer:

C

Explanation:

7 0
3 years ago
Read 2 more answers
Using the celsius_to_kelvin function as a guide, create a new function, changing the name to kelvin_to_celsius, and modifying th
aleksandr82 [10.1K]

Answer:

# kelvin_to_celsius function is defined

# it has value_kelvin as argument

def kelvin_to_celsius(value_kelvin):

   # value_celsius is initialized to 0.0

   value_celsius = 0.0

   

   # value_celsius is calculated by

   # subtracting 273.15 from value_kelvin

   value_celsius = value_kelvin - 273.15

   # value_celsius is returned

   return value_celsius

   

# celsius_to_kelvin function is defined

# it has value_celsius as argument

def celsius_to_kelvin(value_celsius):

   # value_kelvin is initialized to 0.0

   value_kelvin = 0.0

   

   # value_kelvin is calculated by

   # adding 273.15 to value_celsius

   value_kelvin = value_celsius + 273.15

   # value_kelvin is returned

   return value_kelvin

   

value_c = 0.0

value_k = 0.0

value_c = 10.0

# value_c = 10.0 is used to test the function celsius_to_kelvin

# the result is displayed

print(value_c, 'C is', celsius_to_kelvin(value_c), 'K')

value_k = 283.15

# value_k = 283.15 is used to test the function kelvin_to_celsius

# the result is displayed

print(value_k, 'is', kelvin_to_celsius(value_k), 'C')

Explanation:

Image of celsius_to_kelvin function used as guideline is attached

Image of program output is attached.

4 0
3 years ago
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