What’s another name for wiring diagrams

Homes may be heated by pumping hot water through radiators. What mass of water (in g) will provide the same amount of heat when

Nitella [24]

Answer:

a mass of water required is mw= 1273.26 gr = 1.27376 Kg

Explanation:

Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:

Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L

where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation

therefore

mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )

replacing values

mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg

3 0

3 years ago

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

Rashid [163]

Answer:

Exit temperature = 32 °C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

Initial temperature;T1 = 27°C = 273 + 27K = 300 K

volume flow rate;V' = 15 m³/s

W = 130 Kw

Q = 80 Kw

Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

m' = PV'/RT

So at inlet,

m' = P1•V1'/(R•T1)

m' = (100 × 15)/(0.5 × 300)

m' = 10 kg/s

From steady flow energy equation, we know that;

m'•h1 + Q = m'h2 + W

Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

Cp•T1 + Q/m' = Cp•T2 + W/m'

Plugging in the relevant values, we have;

(1*300) - (80/10) = (1*T2) - (130/10)

Q and M negative because heat is being lost.

300 - 8 + 13 = T2

T2 = 305 K = 305 - 273 °C = 32 °C

13000 + 300 - 8000 = T2

6 0

3 years ago

A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th

Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588 = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ

b) $\Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)$

Where:

$x_A$ and $x_B$ are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, $x_A$ = 248 /(248 + 42.8) = 0.83

$x_B$ = 42.8/(248 + 42.8) = 0.1472

∴ $\Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472)$ = 1066.0279 J/K

5 0

3 years ago

Write a iterative function that finds the n-th integer of the Fibonacci sequence. Then build a minimal program (main function) t

Natasha2012 [34]

Answer:

Codes for each of the problems are explained below

Explanation:

PROBLEM 1 IN C++:

#include<iostream>

using namespace std;

//fib function that calculate nth integer of the fibonacci sequence.

void fib(int n){

// l and r inital fibonacci values for n=1 and n=2;

int l=1,r=1,c;

//if n==1 or n==2 then print 1.

if(n==1 || n==2){

cout << 1;

return;

}

//for loop runs n-2 times and calculates nth integer of fibonacci sequence.

for(int i=0;i<n-2;i++){

c=l+r;

l=r;

r=c;

cout << "(" << i << "," << c << ") ";

}

//prints nth integer of the fibonacci sequence stored in c.

cout << "\n" << c;

}

int main(){

int n; //declared variable n

cin >> n; //inputs n to find nth integer of the fibonacci sequence.

fib(n);//calls function fib to calculate and print fibonacci number.

}

PROBLEM 2 IN PYTHON:

def fib(n):

print("fib({})".format(n), end=' ')

if n <= 1:

return n

else:

return fib(n - 1) + fib(n - 2)

if __name__ == '__main__':

n = int(input())

result = fib(n)

print()

print(result)

7 0

3 years ago

Read 2 more answers

The way most recursive functions are written, they seem to be circular at first glance, defining the solution of a problem in te

EastWind [94]

Question Continuation

int factorial(int n) {

if(n == 0)

return 1;

else

return n * factorial(n - 1);

}

Provide a brief explanation why this recursive function works.

Show all steps involved in calculating factorial(3) using the function deﬁned.

Answer:

1. Brief explanation why this recursive function works.

First, the recursive method factorial is defined.

This is the means through with the machine identifies the method.

The method is defined as integer, the machine will regard it as integer.

When the factorial is called from anywhere that has access to it, which in this case is within the factorial class itself. This means you can call it from the main method, or you can call it from the factorial method itself. It's just a function call that, well, happens to call itself.

2. Steps to calculate factorial(3)

1 First, 3 is assigned to n.

2. At line 2, the machine checks if n equals 0

3. If yes, the machine prints 1

4. Else; it does the following from bottom to top

factorial(3):

return 3*factorial(2);

return 2*factorial(1):

return 1;

Which gives 3 * 2 * 1 = 6

5. Then it prints 6, which is the result of 3!

6 0

3 years ago