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4 years ago

Zionjasean17 zionjasean17

Engineering

2 answers:

yuradex [85]4 years ago

8 0

Me ksidhejdndhshdjdnhdhdnnd

Julli [10]4 years ago

4 0

Answer:

Zion

Explanation:

Zionjasean17 is your answer because of the capitalization. Good article btw.

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A piston–cylinder device contains 0.78 kg of nitrogen gas at 140 kPa and 37°C. The gas is now compressed slowly in a polytropic

DiKsa [7]

Answer:

The entropy change of nitrogen during this process. is - 0.32628 kJ/K.

Explanation:

Solution

Given that:

A piston cylinder device contains =0.78 kg of nitrogen gas

Temperature = 37°C

The nitrogen gas constant of R = 0.2968 kJ/kg.K

At room temperature cv = 0.743 kJ/kg.K

Now,

We assume that at specific condition the nitrogen can be treated as an ideal gas

Nitrogen has a constant volume specific heat at room temperature.

Thus,

From the polytropic relation, we have the following below:

T₂/T₁ =(V₁/V₂)^ n-1 which is,

T₂ = T₁ ((V₁/V₂)^ n-1

= (310 K) (2)^1.3-1 = 381.7 K

So,

The entropy change of nitrogen is computed as follows:

ΔSN₂ = m (cv₁ avg ln T₂/T₁ + ln V₂/V₁)

= (0.78) ((0.743 kJ/kg .K) ln 381.7 K/310K + (0.2968 kJ/kg. K) ln (0.5))

= 0.57954 * 0.2080 + (-0.2057)

= 0.12058 + (-0.2057) = -0.32628

Therefore the entropy change of nitrogen during this process. is - 0.32628 kJ/K.

6 0

3 years ago

A pentagonal (five-sided) sign tells you there is a __________ nearby.

Studentka2010 [4]

It’s a school sign meaning you are a near a school which means watch for children

7 0

3 years ago

Two parallel surfaces move in opposite directions relative to each other at a velocity of 64 in/sec and are separated by a gap o

Illusion [34]

Answer:

$\mu$ = 2.6906 × $10^{-3}$ lb-s/in²

Explanation:

given data

velocity V = 64 in/sec

separated by a gap x = 0.41 in

relative motion by shear stress $\tau$ = 0.42 lb/in²

solution

we know that shear stress is directly proportional to rate of change of velocity as per newton's law of viscosity.

$\tau = \mu \times \frac{du}{dy}$ ....................1

so here $\mu$ coefficient of dynamic viscosity and $\frac{du}{dy}$ is velocity gradient

and

$\tau = \mu \times \frac{v1 - v2 }{h2-h0}$

put here value and we get

0.42 = $\mu \times \frac{64}{0.41}$

$\mu$ = 2.6906 × $10^{-3}$ lb-s/in²

4 0

3 years ago

Cryogenic liquid storage. Liquid oxygen is stored in a thin-walled spherical container, 96 cm in diameter, which is further encl

Anvisha [2.4K]

Answer:

The answer is "26.55 V"

Explanation:

Given values:

$d_i= 0.96m\\d_o= 1m\\\epsilon = 0.05\\T_0= 280k\\T_i= 95k\\$

For Answer (a) please find the attachment.

Answer (b):

$q_{i-0}= \frac{\sigma (T_{0}^4)-(T_{i}^4)}{\frac{1-\epsilon i }{\epsilon_{i} A_{i}}+ \frac{1 }{\ f_{i o} A_{i}} +\frac{1-\epsilon_{0}}{\epsilon_{0} A_{0}}}$

$f_{i0}= 1 \ it \ is \ fully \ inside \ the \ large \ sphero \\$

$q_{i-0}= \frac{\sigma A_i (T_{0}^4)-(T_{i}^4)}{\frac{1 }{\epsilon_{i}} - 1+ 1 +\frac{1-\epsilon_{0}}{\epsilon_{0}} \times \frac{A_i}{A_0}}\\\\q_{i-0}= \frac{\sigma A_i (T_{0}^4)-(T_{i}^4)}{\frac{1 }{\epsilon_{i}} +\frac{1-\epsilon_{0}}{\epsilon_{0}} \times (\frac{d_i}{d_0})^2}\\\\q_{i-0}= \frac{\sigma (\pi d^2_i) (T_{0}^4)-(T_{i}^4)}{\frac{1 }{\epsilon_{i}} +\frac{1-\epsilon_{0}}{\epsilon_{0}} \times (\frac{r_i}{r_0})^2}\\\\$

$q_{i-0}= \frac{5.67 \times 10^{-8} \times 3.14 \times 9.62 \times 9.62 \times (280^4-94^4)}{\frac{1 }{0.05} +\frac{1-0.05}{0.05} \times (\frac{0.96}{1})^2}\\\\\ After \ solve the \ equation \ the \ answer \ is:\\\\q_{i-0} = 26.55 \ V$

4 0

3 years ago

Implement this C program by defining a structure for each payment. The structure should have at least three members for the inte

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

balance = balance - monthPrinciple;

}

int main()

{

float principle,rate;

int termYear;

printf("Enter the loan amount: $");

scanf("%f",&principle);

printf("Enter the intrest rate :%");

scanf("%f",&rate);

printf("Enter the loan duration in years: ");

scanf("%d",&termYear);

output_amortized(principle,rate,termYear);

}

Explanation:

see output

6 0

3 years ago