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2 years ago

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9.63 Air enters the compressor of a regenerative air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 2

80 K. The compressor pressure ratio is 20, and the maximum cycle temperature is 2100 K. For the compressor, the isentropic efficiency is 92% and for the turbine the isentropic efficiency is 95%. For a regenerator effectiveness of 85%, determine (a) the net power developed, in MW. (b) the rate of heat addition in the combustor, in MW. (c) the thermal efficiency of the cycle. (d) Plot the quantities calculated in parts (a) through (c) for regenerator effectiveness values ranging from 0 to 100%. Discuss.

1 answer:

nika2105 [10]2 years ago

7 0

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

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Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is

Yuki888 [10]

Answer:

1113kN

Explanation:

The ouside diameter OD of the pipe is 61cm and the thickness T is 0.9cm, so the inside diameter ID will be:

Inside Diameter = Outside Diameter - Thickness

Inside Diameter = 61cm - 0.9cm = 60.1cm

Converting this diameter to meters, we have:

$60.1cm*\frac{1m}{100cm}=0.601m$

This inside diameter is useful to calculate the volume V of water inside the pipe, that is the volume of a cylinder:

$V_{water}=\pi r^{2}h$

$V_{water}=\pi (\frac{0.601m}{2})^{2}*120m$

$V_{water}=113.28m^{3}$

The problem gives you the water density d as 1.0kg/L, but we need to convert it to proper units, so:

$d_{water}=1.0\frac{Kg}{L}*\frac{1L}{1000cm^{3}}*(\frac{100cm}{1m})^{3}$

$d_{water}=1000\frac{Kg}{m^{3}}$

Now, water density is given by the equation $d=\frac{m}{V}$ , where m is the water mass and V is the water volume. Solving the equation for water mass and replacing the values we have:

$m_{water}=d_{water}.V_{water}$

$m_{water}=1000\frac{Kg}{mx^{3}}*113.28m^{3}$

$m_{water}=113280Kg$

With the water mass we can find the weight of water:

$w_{water}=m_{water} *g$

$w_{water}=113280kg*9.8\frac{m}{s^{2}}$

$w_{water}=1110144N$

Finally we find the total weight add up the weight of the water and the weight of the pipe,so:

$w_{total}=w_{water}+w_{pipe}$

$w_{total}=1110144N+2500N$

$w_{total}=1112644N$

Converting this total weight to kN, we have:

$1112644N*\frac{0.001kN}{1N}=1113kN$

7 0

2 years ago

What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.

yanalaym [24]

Answer:

$Re=\dfrac{\rho\ v\ l}{\mu }$

Explanation:

Reynolds number:

Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

$Re=\dfrac{F_i}{F_v}$

For plate can be given as

$Re=\dfrac{\rho\ v\ l}{\mu }$

Where ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.

Flow on plate is a external flow .The values of Reynolds number for different flow given as

$Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.$

$Reynolds\ number\is \$

7 0

2 years ago

Natural Convection from an Oven Wall. The oven wall in Example 4.7-1 is insulated so that the surface temperature is 366.5 K ins

lukranit [14]

Answer:

i) Heat transfer coefficient (h) = 7 w/m²k

ii) Heat transfer per meter width of wall

= h x L x 1 x (Ts - T₆₀)

= 7 x 0.3048 x (505.4 - 322) = 414.747 w/m

Explanation:

see attached image

8 0

2 years ago

Peter the postman became bored one night and, to break the monotony of the night shift, he carried out the following experiment

aleksandrvk [35]

Answer:

//This Program is written in C++

// Comments are used for explanatory purpose

#include <iostream>

using namespace std;

enum mailbox{open, close};

int box[149];

void closeAllBoxes();

void OpenClose();

void printAll();

int main()

{

closeAllBoxes();

OpenClose();

printAll();

return 0;

}

void closeAllBoxes()

{

for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150

{

box[i] = close; //Close all boxes

}

}

void OpenClose()

{

for(int i = 2; i < 150; i++) {

for(int j = i; j < 150; j += i) {

if (box[j] == close) //Open box if box is closed

box[j] = open;

else

box[j] = close; // Close box if box is opened

}

}

// At the end of this test, all boxes would be closed

}

void printAll()

{

for (int x = 0; x < 150; x++) //use this to test

{

if (box[x] = 1)

{

cout << "Mailbox #" << x+1 << " is closed" << endl;

// Print all close boxes

}

}

}

Explanation:

7 0

2 years ago

A heat pump operates on a Carnot heat pump cycle with a COP of 12.5. It keeps a space at 24°C by consuming 2.15 kW of power. Det

Vinil7 [7]

Answer:

a) $T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)$ , b) $\dot Q_{H} = 26.875\,kW$

Explanation:

a) The Coefficient of Performance of the Carnot Heat Pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

After some algebraic handling, the temperature of the cold reservoir is determined:

$T_{H}-T_{L} = \frac{T_{H}}{COP_{HP}}$

$T_{L} = T_{H}\cdot \left(1-\frac{1}{COP_{HP}} \right)$

$T_{L} = (297.15\,K)\cdot \left(1-\frac{1}{12.5}\right)$

$T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)$

b) The heating load provided by the heat pump is:

$\dot Q_{H} = COP_{HP}\cdot \dot W$

$\dot Q_{H} = (12.5)\cdot (2.15\,kW)$

$\dot Q_{H} = 26.875\,kW$

4 0

2 years ago