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Naily [24]
3 years ago
7

CHALLENGE ACTIVITY 2.8.1: Using constants in expressions. The cost to ship a package is a flat fee of 75 cents plus 25 cents per

pound. 1. Declare a const named CENTS_PER_POUND and initialize with 25. 2. Get the shipping weight from user input storing the weight into shipWeightPounds. 3. Using FLAT_FEE_CENTS and CENTS_PER_POUND constants, assign shipCostCents with the cost of shipping a package weighing shipWeightPounds.
Engineering
1 answer:
mihalych1998 [28]3 years ago
7 0

Answer:

Weight(lb): 10

Flat fee(cents): 75

Cents per pound: 25

Shipping cost(cents): 325

Explanation:

we run this as a jave programming language

import java.util.Scanner;

public class Shipping Calculator {

   public static void main (String [] args) {

       int shipWeightPounds = 10;

       int shipCostCents = 0;

       final int FLAT_FEE_CENTS = 75;

       

        final int CENTS_PER_POUND = 25;

       shipCostCents = FLAT_FEE_CENTS + CENTS_PER_POUND * shipWeightPound

       /* look up the solutioin above */

       System.out.println("Weight(lb): " + shipWeightPounds);

       System.out.println("Flat fee(cents): " + FLAT_FEE_CENTS);

       System.out.println("Cents per pound: " + CENTS_PER_POUND);

       System.out.println("Shipping cost(cents): " + shipCostCents);

   }

}

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Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 5208C.
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Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 520°C. Steam exiting the first-stage turbine at 6 MPa is reheated at constant pressure to 500°C. Each turbine stage has an isentropic efficiency of 78% while the pump has an isentropic efficiency of 82%. Saturated liquid exits the condenser that operates at constant pressure of 6 kPa.

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- the thermal efficiency is 36.05%  

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Specific entropy s1 = 5.9566 kJ/kg.K  

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get the enthalpy at state 2 using isentropic turbine efficiency of the turbine. nT1 = (h1 - h2) / (h1 - h2s)

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h2 = 2903.6 kJ/kg

get the enthalpy at state 3 at pressure p2 = p3 = 6 MPa and T3 = 500°C

h3 = 3422.2 kJ/kg

s3 = 6.8803 kJ/kg.K

Process 3 to 4s is isentropic expansion process in the turbine

S3 = S4s

get the enthalpy at state 4s at pressure p4s = p4 = 6 kPa and S4s = 6.8803 kJ/kg.K

h4s = 2118.8 kJ/kg

get the enthalpy at state 4 using isentropic turbine efficiency of the turbine. nT2 = (h3 - h4) / (h3 - h4s)

0.78 = (3422.2 - h4) / ( 3422.2 - 2118.8 )

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get the enthalpy at state 6 using isentropic pump efficiency of the turbine, at

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np = v5( p6 - p5) / (h6 - h5)

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= (Wt1 + Wt2 - Wp) / (Q61 + Q23)

=  [(h1 - h2) + (h3 - h4) - (h6 - h5)] / [(h1 - h6) + (h3 - h2)]

[(3192.3 - 2903.6) + (3422.2 - 2405.5) - (185.89 - 151.53)] / [(3192.3 - 185.89) + (3422.2 - 2903.6)]

= 0.3605

n = 36.05%  

therefore the thermal efficiency is 36.05%  

3 0
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