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melamori03 [73]
3 years ago
13

Consider the following list. list = {24, 20, 10, 75, 70, 18, 60, 35} Suppose that list is sorted using the insertion sort algori

thm as discussed in the book. What is the resulting list after two passes of the sorting phase; that is, after three iterations of the for loop?a. list = {10, 18, 20, 24, 75, 70, 60, 35}b. list = {10, 20, 24, 75, 70, 18, 60, 35}c. list = {10, 18, 20, 24, 35, 70, 60, 75}d. list = {10, 20, 20, 18, 35, 70, 60, 75}
Engineering
1 answer:
Greeley [361]3 years ago
8 0

Answer:

Option B

10,20,24,75,70,18,60,35

Explanation:

The first, second and third iteration of the loop will be as follows

insertion sort iteration 1: 20,24,10,75,70,18,60,35

insertion sort iteration 2:10,20,24,75,70,18,60,35

insertion sort iteration 3: 10,20,24,75,70,18,60,35

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A) Consider an air standard otto cycle that has a heat addition of 2800 kJ/kg of air, a compression ratio of 8 and a pressure an
Angelina_Jolie [31]

Answer:

a) i) The maximum pressure is approximately 122.37 bar

ii) The thermal efficiency is approximately 56.47%

iii) The mean effective pressure is approximately 20.974 bar

b) (b) Four types of internal combustion engine includes;

1) The diesel engine

2) The Otto engine

3) The Brayton engine

4) The Wankel engine

Explanation:

The parameters of the Otto cycle are;

The heat added, Q_{in} = 2,800 kJ/kg

The compression ratio, r = 8

The beginning compression pressure, P₁ = 1 bar

The beginning compression temperature, T₁ = 300 K

Cp = 1.005 kJ/kg·K

Cv = 0.718 kJ/kg·K

R = 287 kJ/kg·K

K = Cp/Cv = 1.005 kJ/kg·K/(0.718 kJ/kg·K) ≈ 1.4

T₂ = T₁×r^(k - 1)

∴ T₂ = 300 K×8^(1.4 - 1) ≈ 689.219 K

\dfrac{P_1\cdot V_1}{T_1}  = \dfrac{P_2\cdot V_2}{T_2}

P_2 = \dfrac{P_1\cdot V_1 \cdot T_2}{T_1 \cdot V_2}  = \dfrac{V_1}{V_2} \cdot  \dfrac{P_1 \cdot T_2}{T_1 } = r \cdot  \dfrac{P_1 \cdot T_2}{T_1 }

∴ P₂ = 8 × 1 bar × (689.219K)/300 K ≈ 18.379 bar

Q_{in} = m·Cv·(T₃ - T₂)

∴ Q_{in} = 2,800 ≈ 0.718 × (T₃ - 689.219)

T₃ = 2,800/0.718 + 689.219 = 4588.94 K

P₃ = P₂ × (T₃/T₂)

P₃ = 18.379 bar × 4588.94K/(689.219 K) = 122.37 bar

The maximum pressure = P₃ ≈ 122.37 bar

(ii) The thermal efficiency, \eta_{Otto}, is given as follows;

\eta_{Otto} = 1 - \dfrac{1}{r^{k - 1}}

Therefore, we have;

\eta_{Otto} = 1 - \dfrac{1}{8^{1.4 - 1}} \approx 0.5647

The thermal efficiency, \eta_{Otto} ≈ 0.5647

Therefore, the thermal efficiency ≈ 56.47%

(iii) The mean effective pressure, MEP is given as follows;

MEP = \dfrac{\left(P_3 - P_1 \cdot r^k \right) \cdot \left(1 - \dfrac{1}{r^{k-1}} \right)}{(k -1)\cdot (r - 1)}

Therefore, we get;

MEP = \dfrac{\left(122.37 - 1 \times 8^{1.4} \right) \cdot \left(1 - \dfrac{1}{8^{1.4-1}} \right)}{(1.4 -1)\cdot (8 - 1)} \approx 20.974

The mean effective pressure, MEP ≈ 20.974 bar

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1) The diesel engine; Compression heating is the source of the ignition, with constant pressure combustion

2) The Otto engine which is the internal combustion engine found in cars that make use of gasoline as the source of fuel

The Otto engine cycle comprises of five steps; intake, compression, ignition, expansion and exhaust

3) The Brayton engine works on the principle of the steam turbine

4) The Wankel it follows the pattern of the Otto cycle but it does not have piston strokes

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Explanation:

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Answer:

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