Answer:
Condition to break: ![H[j] \geq max {H[2j] , H[2j+1]}](https://tex.z-dn.net/?f=H%5Bj%5D%20%5Cgeq%20max%20%7BH%5B2j%5D%20%2C%20H%5B2j%2B1%5D%7D)
Efficiency: O(n).
Explanation:
Previous concepts
Heap algorithm is used to create all the possible permutations with K possible objects. Was created by B. R Heap in 1963.
Parental dominance condition represent a condition that is satisfied when the parent element is greater than his children.
Solution to the problem
We assume that we have an array H of size n for the algorithm.
It's important on this case analyze the parental dominance condition in order to the algorithm can work and construc a heap.
For this case we can set a counter j =1,2,... [n/2] (We just check until n/2 since in order to create a heap we need to satisfy minimum n/2 possible comparisions![and we need to check this:Break condition: [tex]H[j] \geq max {H[2j] , H[2j+1]}](https://tex.z-dn.net/?f=%20and%20we%20need%20to%20check%20this%3A%3C%2Fp%3E%3Cp%3E%3Cstrong%3EBreak%20condition%3A%20%3C%2Fstrong%3E%5Btex%5DH%5Bj%5D%20%5Cgeq%20max%20%7BH%5B2j%5D%20%2C%20H%5B2j%2B1%5D%7D)
And we just need to check on the array the last condition and if is not satisfied for any value of the counter j we need to stop the algorithm and the array would not a heap. Otherwise if we satisfy the condition for each ![j =1,2,.....,[n/2]p](https://tex.z-dn.net/?f=j%20%3D1%2C2%2C.....%2C%5Bn%2F2%5Dp)
then we will have a heap.
On this case this algorithm needs to compare 2*(n/2) times the values and the efficiency is given by O(n).
Answer:
(A) Power output will be 5.55 KW (b) lower temperature will be 315 K
Explanation:
We have given efficiency of heat engine 
= 0.37
Input power = 15 KW
Temperature of heat reservoir 
(A) We know that 
So [text]0.37=\frac{output}{15}[text]
Output = 5.55 KW
(B) We also know that [text]\eta =1-\frac{T_L}{T_H}0.37=\frac{output}{15}[text], here 
is lower temperature and 
is higher temperature
So 
Answer:
He wore his black suit, another color of shirt (not purple) and shoes
Explanation:
Holmes owns two suits: one black and one tweed.
Whenever he wears his tweed suit and a purple shirt, he chooses not to wear a tie and whenever he wears sandals, he always wears a purple shirt.
So, if he wore a bow tie yesterday, it means he wore his black suit, another color of shirt (not purple) and shoes because the shirt color is not purple
Answer:
Explanation:
A plane wall of thickness 2L=40 mm and thermal conductivity k=5W/m⋅Kk=5W/m⋅K experiences uniform volumetric heat generation at a rateq
˙
q
q
˙
, while convection heat transfer occurs at both of its surfaces (x=-L, +L), each of which is exposed to a fluid of temperature T∞=20∘CT
∞
=20
∘
C. Under steady-state conditions, the temperature distribution in the wall is of the form T(x)=a+bx+cx2T(x)=a+bx+cx
2
where a=82.0∘C,b=−210∘C/m,c=−2×104C/m2a=82.0
∘
C,b=−210
∘
C/m,c=−2×10
4
C/m
2
, and x is in meters. The origin of the x-coordinate is at the midplane of the wall. (a) Sketch the temperature distribution and identify significant physical features. (b) What is the volumetric rate of heat generation q in the wall? (c) Determine the surface heat fluxes, q
′′
x
(−L)q
x
′′
(−L) and q
′′
x
(+L)q
x
′′
(+L). How are these fluxes related to the heat generation rate? (d) What are the convection coefficients for the surfaces at x=-L and x=+L? (e) Obtain an expression for the heat flux distribution q
′′
x
(x)q
x
′′
(x). Is the heat flux zero at any location? Explain any significant features of the distribution. (f) If the source of the heat generation is suddenly deactivated (q=0), what is the rate of change of energy stored in the wall at this instant? (g) What temperature will the wall eventually reach with q=0? How much energy must be removed by the fluid per unit area of the wall (J/m2)(J/m
2
) to reach this state? The density and specific heat of the wall material are 2600kg/m32600kg/m
3
and 800J/kg⋅K800J/kg⋅K, respectively.
Answer:
2
Explanation:
So for solving this problem we need the local heat transfer coefficient at distance x,
We integrate between 0 to x for obtain the value of the coefficient, so
Substituing
The ratio of the average convection heat transfer coefficient over the entire length is 2
