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Rashid [163]
3 years ago
7

A rigid copper tank, initially containing 1 m3 of air at 295 K, 5 bar, is connected by avalve to a large supply line carrying ai

r at 295 K, 15 bar. The valve is opened only as long as required tofill the tank with air to a pressure of 15 ,bar at which point the air in the tank is at 310 .K The copper tank,which has a mass of 20 ,kg is at the same temperature as the air in the tank, both initially and finally. Thespecific heat of the copper is c=0.385 kJ/kg K. Assuming ideal gas behavior for the air, determine:
(a) the initial and final mass of the air within the tank, each in kg, and
(b) the heat transfer to the surroundings from the tank and its contents, in kJ, ignoring kinetic and potential energy effects.
Engineering
1 answer:
asambeis [7]3 years ago
6 0

Answer:

Initial mass = 5.91 kg

Final mass = 16.8 kg

Heat transfer Q = - 625.9 KJ

Explanation:

Given data

Tank volume V = 1 m^{3}

Entering outside air temperature T_{i} = 295 K

Entering outside air pressure P_{i} = 15 bar

Initial tank pressure P_{1} = 5 bar

Initial tank temperature T_{1} = 295 K

Final pressure P_{2} = 15 bar

Final temperature T_{2} = 310 K

We know that

P V = m R T

(a). Initial mass is given by

m_{1} = \frac{P_{1} V_{1} }{R T_{1} }

Put all the values in given equation

m_{1} = \frac{(500000)(1)}{(287)(295)}  = 5.91 \ kg

(b). Final mass is given by

m_{2} = \frac{(1500000)(1)}{(287)(310)}

m_{2} = 16.8 \ kg

This is the final volume of the tank.

Δ U = Δ q + h_{i} Δ m_{cv}

m_{2} u_{2} - m_{1} u_{1} = Q + h_{i} (m_{2} - m_{1} )

Specific internal energy at initial temperature & pressure u_{1} = 210.5 \frac{KJ}{Kg}

Specific internal energy at final temperature & pressure u_{2} = 221.25 \frac{KJ}{Kg}

Specific enthalpy is h_{i} = 295.17  \frac{KJ}{Kg}

Q = ( 16.8 × 210.48 - 5.91 × 210.49 )- 295.17 ( 16.8 - 5.91 )

Q = 2292.23 - 3214.4

Q_{a} = - 741.4 KJ

The heat transfer for the tank is given by

Q_{t} = m C (T_{2} - T_{1})

Q_{t} = 20 × 0.385 × ( 310-295 )

Q_{t} = + 115.5 KJ

Total heat transfer Q = Q_{a} + Q_{t}

Q = - 741.4 + 115.5

Q = - 625.9 KJ

This is the heat transfer to the surrounding from the tank.

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Write a C program that will update a bank balance. A user cannot withdraw an amount ofmoney that is more than the current balanc
GarryVolchara [31]

Answer:

Explanation:

Sample output:

BANK ACCOUT PROGRAM!

----------------------------------

Enter the old balance: 1234.50

Enter the transactions now.

Enter an F for the transaction type when you are finished.

Transaction Type (D=deposit, W=withdrawal, F=finished): D

Amount: 568.34

Transaction Type (D=deposit, W=withdrawal, F=finished): W

Amount: 25.68

Transaction Type (D=deposit, W=withdrawal, F=finished): W

Amount: 167.40

Transaction Type (D=deposit, W=withdrawal, F=finished): F

Your ending balance is $1609.76

Program is ending

Code to copy:

// include the necessary header files.

#include<stdio.h>

// Definition of the function

float withdraw(float account_balance, float withdraw_amount)

{

// Calculate the balace amount.

float balance_amount = account_balance - withdraw_amount;

// Check whether the withdraw amount

// is greater than 0 or not.

if (withdraw_amount > 0 && balance_amount >= 0)

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// Assign value.

account_balance = balance_amount;

}

// return account_balance

return account_balance;

}

// Definition of the function deposit.

float deposit(float account_balance, float deposit_amount)

{

// Check whether the deposit amount is greater than zero

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// Update account balance.

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// return account balance.

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{

// Declare the variables.

float account_balance;

float deposit_amount;

float withdrawl_amount;

char input;

// display the statement on console.

printf("BANK ACCOUT PROGRAM!\n");

printf("----------------------------------\n");

// prompt the user to enter the old balance.

printf("Enter the old balance: ");

// Input balance

scanf("%f", &account_balance);

// Display the statement on console.

printf("Enter the transactions now.\n");

printf("Enter an F for the transaction type when you are finished.\n");

// Start the do while loop

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// prompt the user to enter transaction type.

printf("Transaction Type (D=deposit, W=withdrawal, F=finished): ");

// Input type.

scanf(" %c", &input);

// Check if the input is D

if (input == 'D')

{

// Prompt the user to input amount.

printf("Amount: ");

// input amount.

scanf("%f", &deposit_amount);

// Call to the function.

account_balance=deposit(account_balance,deposit_amount);

}

// Check if the input is W

if (input == 'W')

{

printf("Amount: ");

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// Call to the function.

account_balance = withdraw(account_balance,withdrawl_amount);

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// Check if the input is F

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// Dispplay the amount.

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the picture uploaded below shows the program screenshot.

cheers, i hope this helps.

5 0
3 years ago
Design for human-fit strategies include:
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Answer:

B- extreme fit, close fit, adjustable fit

Explanation:

A human-fit design typically involves the process of manufacturing or producing products (tools) that are easy to use by the end users. Therefore, human-fit designs mainly deals with creating ideas that makes the use of a particular product comfortable and convenient for the end users.

The design for human-fit strategies include; extreme fit, close fit and adjustable fit.

Hence, when the aforementioned strategies are properly integrated into a design process, it helps to ensure the ease of use of products and guarantees comfort for the end users.

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3 years ago
Which statement about cathode ray tubes (CRTs) is true?
yuradex [85]

Answer:

CRTs consume less power than LCDs.

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2 years ago
A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th
SCORPION-xisa [38]

Answer:

F_{thrust} ≅ 111 KN

Explanation:

Given that;

A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8

mass = 85,000 kg

drag co-efficient (C) = 0.37

(velocity (v)= 230 m/s

density (ρ) = 1.0 kg/m³

To calculate the thrust; we need to determine the relation of the drag force; which is given as:

F_{drag} = \frac{1}{2} × CρAv²

where;

ρ = density of air wind.

C = drag co-efficient

A = Area of the jet

v = velocity of the jet

From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0

SO, F_{drag}-F_{thrust} = 0

We can as well say:

F_{drag}= F_{thrust}

We can now replace F_{thrust} with F_{drag} in the above equation.

Therefore, F_{thrust} = \frac{1}{2} × CρAv²

The A which stands as the area of the jet is given by the formula:

A=\frac{\pi d^2}{4}

We can now have a new equation after substituting our A into the previous equation as:

F_{thrust} = \frac{1}{2} × Cρ (\frac{\pi d^2}{4})v^2

Substituting our data from above; we have:

F_{thrust} = \frac{1}{2} × (0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2

F_{thrust} = \frac{1}{8}   (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2

F_{thrust} = 110,990N

F_{thrust}  in N (newton) to KN (kilo-newton) will be:

F_{thrust} = (110,990N)*\frac{1KN}{1,000N}

F_{thrust} = 110.990 KN

F_{thrust} ≅ 111 KN

In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.

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3 years ago
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dezoksy [38]

Answer:

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Explanation:

3 0
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