Answer:
when 5% excess air is supplied, moles of air supplied/moles of fuel =
Explanation:
Equivalence ratio = 0.6
Equivalence ratio = Actual air to fuel ratio (AAFR)/ stoichiometric air to fuel ratio SAFR
combustion reaction of propane is
From above reaction, 1 mole of propane, from the reaction, 5 moles of oxygen required,
we know that air contains 21% O_2 and 79% N_2,
Therefore, moles of air based on stoichiometry
Theoretical air to fuel ratio
Given
Actual Air Fuel Ratio
when 5% excess air is supplied, moles of air supplied/moles of fuel =
Answer:
5.833
Explanation:
Coefficient of Perfomance (COP) is the ratio of refrigeration effect to power input.
where RE is refrigeration effect and P is power input
Here, the power input is given as 30 kW
We also know that 1 ton cooling is equivalent to 3.5 kW hence for 50 tons, RE=50*3.5=175 kW
Now the
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.
