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Alecsey [184]
3 years ago
14

g Part 2: The features arrived the grammar Splitting categories and non-terminals gets out of hand fast. An alternative to the p

roliferation of non-terminals is to attach features to the terminals and non-terminals, and use these features to constrain the grammar. Consider the case example in part 1. We split pro into pro_nom and pro_acc, and we split np into np_nom and np_acc. The sequence det n could be either, so we added two rules (one for np_nom, and one for np_acc) with identical right-hand sides. Instead of splitting the symbols of the grammar, we can modify the lexical entries as follows:
Engineering
1 answer:
koban [17]3 years ago
3 0

Answer:

The split is given by including spaces in both tabs

Explanation:

The bracket notation can be used to indicate the split. Here is an example:

String [ ] parts = s. split ( "[/]")

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It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
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Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

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