Answer:
Base temperature is 46.23 °C
Explanation:
I've attached explanations
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:
- def median(l):
- if(len(l) == 0):
- return 0
- else:
- l.sort()
- if(len(l)%2 == 0):
- index = int(len(l)/2)
- mid = (l[index-1] + l[index]) / 2
- else:
- mid = l[len(l)//2]
- return mid
-
- def mode(l):
- if(len(l)==0):
- return 0
-
- mode = max(set(l), key=l.count)
- return mode
-
- def mean(l):
- if(len(l)==0):
- return 0
- sum = 0
- for x in l:
- sum += x
- mean = sum / len(l)
- return mean
-
- lst = [5, 7, 10, 11, 12, 12, 13, 15, 25, 30, 45, 61]
- print(mean(lst))
- print(median(lst))
- print(mode(lst))
Explanation:
Firstly, we create a median function (Line 1). This function will check if the the length of list is zero and also if it is an even number. If the length is zero (empty list), it return zero (Line 2-3). If it is an even number, it will calculate the median by summing up two middle index values and divide them by two (Line 6-8). Or if the length is an odd, it will simply take the middle index value and return it as output (Line 9-10).
In mode function, after checking the length of list, we use the max function to estimate the maximum count of the item in list (Line 17) and use it as mode.
In mean function, after checking the length of list, we create a sum variable and then use a loop to add the item of list to sum (Line 23-25). After the loop, divide sum by the length of list to get the mean (Line 26).
In the main program, we test the three functions using a sample list and we shall get
20.5
12.5
12
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Answer:
Valvular stenosis , Valvular prolapse , Regurgitation,
Explanation:
