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Slav-nsk
2 months ago
11

In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle tempe

rature is 2031K. Determine the thermal efficiency.
Engineering
1 answer:
Daniel [329]2 months ago
7 0
The value obtained is 0.60. From the given conditions, we take k = 1.4 for air and r = 16. We know that for calculations involving diesel engine efficiency, we arrive at 0.60.
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A 10-ft-long simply supported laminated wood beam consists of eight 1.5-in. by 6-in. planks glued together to form a section 6 i
pantera1 [306]
Point B is where Q_B = 101.25 \ in^3 shows the highest Q value at section a–a. The missing diagram that was supposed to accompany this question can be found in the attached file. Based on the given information, we must identify the point with the greatest Q value at section a–a. To achieve this, we're working with the attached image. The image reveals that there are 8 blocks stacked vertically on the right-hand side, totaling 12 inches; thus, each block measures 1.5 inches in height. Additionally, the blocks are divided into sections marked as A, B, C, and D. For point A, we use the formula Q representing the moment of Area A. For point B, we set Q as the moment of Area B, and similarly for points C and D. However, point D has no area, resulting in a moment of 0.
3 0
1 month ago
Suppose we include the lead resistance in the calculation of temperature for a class A RTD. If R3 = 1000 ohms, Ra = 18 ohms, V0
alex41 [359]

Respuesta:

La temperatura máxima que se puede medir (en °C) es 14170.27°C

Explicación:

Los RTDs son termómetros compuestos de metales cuya resistencia aumenta con la temperatura.

Para un RTD de Clase A, l, Alpha = 0.00385.

La fórmula para el RTD es

Rt = Ro ( 1 + alpha x t)

Donde

Rt es la resistencia a la temperatura t°C,

Ro es la resistencia a 0°C

Alpha es un coeficiente de temperatura constante para un RTD de clase A.

Aquí, Rt = 1000ohms,

Ro se considera como Ra = 18Ohms

Por lo tanto,

1000 = 18 ( 1 + 0.00385t)

Dividiendo ambos lados por 18

1 + 0.00385t = 1000/18

0.00385t = 55.55 - 1

0.00385t = 54.55

t = 54.55/0.00385

t = 14170.27°C

5 0
1 month ago
Given a 5x5 matrix for Playfair cipher a. How many possible keys does the Playfair cipher have? Ignore the fact that some keys m
alex41 [359]

Answer:

a. 25! = 2^{84}(Approximately)

b. 24!

Explanation:

a. In a Playfair cipher, there are 25 keys available because it is structured in a 5 * 4 grid. By using permutations to enumerate all potential configurations, we derive: 25! = 1.551121004×10²⁵ = 2^{84}

Although there are 26 letters available, in the Playfair cipher, the letters 'i' and 'j' are treated as a single letter.

b. Considering any configuration of 5x5, each of the four row shifts yields equivalent configurations, amounting to five total equivalencies. Similarly, for each of these five setups, any of the four column shifts also results in equivalent arrangements. Therefore, each configuration corresponds to 25 equivalent arrangements. Consequently, the total count of distinct keys can be expressed as:

25!/25 = 24! = 6.204484017×10²³

6 0
1 month ago
Three return steam lines in a chemical processing plant enter a collection tank operating at steady state at 1 bar. Steam enters
alex41 [359]

Response:

a) 4 kg/s

b) 99.61 °C

Rationale:

Refer to the pictures provided.

5 0
1 month ago
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