Answer:
Explanation:
a) for shifting reactions,
Kps = ph2 pco2/pcoph20
=[h2] [co2]/[co] [h2o]
h2 + co2 + h2O + co + c3H8 = 1
it implies that
H2 + 0.09 + H2O + 0.08 + 0.05 = 1
solving the system of equation yields
H2 = 0.5308,
H2O = 0.2942
B) according to Le chatelain's principle for a slightly exothermic reaction, an increase in temperature favors the reverse reaction producing less hydrogen. As a result, concentration of hydrogen in the reformation decreases with an increasing temperature.
c) to calculate the maximum hydrogen yield , both reaction must be complete
C3H8 + 3H2O ⇒ 3CO + 7H2( REFORMING)
CO + H2O ⇒ CO2 + H2 ( SHIFTING)
C3H8 + 6H2O ⇒ 3CO2 + 10 H2 ( OVER ALL)
SO,
Maximum hydrogen yield
= 10mol h2/3 molco2 + 10molh2
= 0.77
⇒ 77%
Answer:
The dependent variable is MEDV - Median value of owner-occupied homes in $1000's
Explanation:
The median value of the house has to be predicted, based on its properties and neighborhood properties, this can be done by using a linear regression model.
The dependent variable in Machine Learning is the output variable that we want to predict.
Therefore, according to the question given "MEDV" is the dependent variable.
Answer:
Cc= 12.7 lb.sec/ft
Explanation:
Given that
m = 22 lb
g= 32 ft/s²

x= 4.5 in
1 in = 0.083 ft
x= 0.375 ft
Spring constant ,K

K= 58.66 lb/ft
The damper coefficient for critically damped system


Cc= 12.7 lb.sec/ft
Answer:
a) Please see attached copy below
b) 0.39KJ
c) 20.9‰
Explanation:
The three process of an air-standard cycle are described.
Assumptions
1. The air-standard assumptions are applicable.
2. Kinetic and potential energy negligible.
3. Air in an ideal gas with a constant specific heats.
Properties:
The properties of air are gotten from the steam table.
b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.
P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K
T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K
Qin=m(u₂₋u₁)=mCv(T₂-T₁)
=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ
Qout=m(h₃₋h₁)=mCp(T₃₋T₁)
=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ
Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ
c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰
Answer:
Complete question is:
write the following decorators and apply them to a single function (applying multiple decorators to a single function):
1. The first decorator is called strong and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <strong> and </strong> to the argument of the decorator. The return value of the wrapper should look like: return “<strong>” + func() + “</strong>”
2. The decorator will return the wrapper per usual.
3. The second decorator is called emphasis and has an inner function called wrapper. The purpose of this decorator is to add the html tags of <em> and </em> to the argument of the decorator similar to step 1. The return value of the wrapper should look like: return “<em>” + func() + “</em>.
4. Use the greetings() function in problem 1 as the decorated function that simply prints “Hello”.
5. Apply both decorators (by @ operator to greetings()).
6. Invoke the greetings() function and capture the result.
Code :
def strong_decorator(func):
def func_wrapper(name):
return "<strong>{0}</strong>".format(func(name))
return func_wrapper
def em_decorator(func):
def func_wrapper(name):
return "<em>{0}</em>".format(func(name))
return func_wrapper
@strong_decorator
@em_decorator
def Greetings(name):
return "{0}".format(name)
print(Greetings("Hello"))
Explanation: