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kramer
4 years ago
9

A crude fermenter is set up in a shed in the backyard of a suburban house. Under anaerobic conditions with ammonia as the nitrog

en source, about 0.45 g ethanol are formed per g glucose consumed. At steady state, the ethanol production rate averages 0.4 kg h21. The owner of this enterprise decides to reduce her electricity bill by using the heat released during the fermentation to warm water as an adjunct to the household hot water system. Cold water at 10C is fed into a jacket surrounding the fermenter at a rate of 2.5 litres h21. To what temperature is the water heated? Heat losses from the system are negligible. Use a biomass composition of CH1.75 O0.58 N0.18 plus 8% ash.
Engineering
1 answer:
Aleks04 [339]4 years ago
5 0

Answer:

using calculations Heat losses will be 4512 J

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Klio2033 [76]

Answer:

#include<stdio.h>

#include<math.h>

void output_amortized(float loan_amount,float intrest_rate,int term_years)

{

  int i,j;                       //Month

  int payments;                   //Number of payments  

  float loanAmount;               //Loan amount

  float anIntRate;               //Yealy interest Rate

  float monIntRate;               //Monthly interest rate

  float monthPayment;           //Monthly payment

  float balance;                   //Balance due

  float monthPrinciple;           //Monthly principle paid

  float monthPaidInt;           //Month interest paid

 

  balance=loan_amount;

  //Calculations

  //Monthly interest rate

  monIntRate = ((intrest_rate/(100*12)));

  //Monthly payment

  payments=term_years;  

  monthPayment = (loan_amount * monIntRate * (pow(1+monIntRate, payments)/(pow (1+monIntRate, payments)-1)));

  monthPaidInt = balance * monIntRate;

  //Amount paid to principle

  monthPrinciple = monthPayment-monthPaidInt;

  //New balance due

  balance = balance - monthPrinciple;

 

  printf("\n\nMonthly payment should be :%.2f\n\n",monthPayment);

  printf("============================AMORTIZATION SCHEDUAL==========================\n");

  printf("#\tPayment\t\tIntrest\t\tPrinciple\t\tBalance\n");

 

  for(i=0;i<payments;i++)

  {

      printf("%d%9c%.2f%9c%.2f%16c%.2f%14c%.2f\n",(i+1),'$',monthPayment,'$',monthPaidInt,'$',monthPrinciple,'$',balance);

      monthPaidInt = balance * monIntRate;

      //Amount paid to principle

      monthPrinciple = monthPayment-monthPaidInt;

      //New balance due

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  }

}

int main()

{

  float principle,rate;

  int termYear;

  printf("Enter the loan amount: $");

  scanf("%f",&principle);

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  scanf("%d",&termYear);

  output_amortized(principle,rate,termYear);

}

Explanation:

see output

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3 years ago
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