It tells the amount of materials to be purchased.

Air in a large tank at 300C and 400kPa, flows through a converging diverging nozzle with throat diameter 2cm. It exits smoothly

-Dominant- [34]

Answer:

The answer is " $3.74 \ cm\ \ and \ \ 0.186 \frac{kg}{s}$ "

Explanation:

Given data:

Initial temperature of tank $T_1 = 300^{\circ}\ C= 573 K$

Initial pressure of tank $P_1= 400 \ kPa$

Diameter of throat $d* = 2 \ cm$

Mach number at exit $M = 2.8$

In point a:

calculating the throat area:

$A*=\frac{\pi}{4} \times d^2$

$=\frac{\pi}{4} \times 2^2\\\\=\frac{\pi}{4} \times 4\\\\=3.14 \ cm^2$

Since, the Mach number at throat is approximately half the Mach number at exit.

Calculate the Mach number at throat.

$M*=\frac{M}{2}\\\\=\frac{2.8}{2}\\\\=1.4$

Calculate the exit area using isentropic flow equation.

$\frac{A}{A*}= (\frac{\gamma -1}{2})^{\frac{\gamma +1}{2(\gamma -1)}} (\frac{1+\frac{\gamma -1}{2} M*^2}{M*})^{\frac{\gamma +1}{2(\gamma -1)}}$

Here: $\gamma$ is the specific heat ratio. Substitute the values in above equation.

$\frac{A}{3.14}= (\frac{1.4-1}{2})^{-\frac{1.4+1}{2(1.4 -1)}} (\frac{1+\frac{1.4-1}{2} (1.4)^2}{1.4})^{\frac{1.4+1}{2(1.4-1)}} \\\\A=\frac{\pi}{4}d^2 \\\\10.99=\frac{\pi}{4}d^2 \\\\d = 3.74 \ cm$

exit diameter is 3.74 cm

In point b:

Calculate the temperature at throat.

$\frac{T*}{T}=(1+\frac{\Gamma-1}{2} M*^2)^{-1}\\\\\frac{T*}{573}=(1+\frac{1.4-1}{2} (1.4)^2)^{-1}\\\\T*=411.41 \ K$

Calculate the velocity at exit.

$V*=M*\sqrt{ \gamma R T*}$

Here: R is the gas constant.

$V*=1.4 \times \sqrt{1.4 \times 287 \times 411.41}\\\\=569.21 \ \frac{m}{s}$

Calculate the density of air at inlet

$\rho_1 =\frac{P_1}{RT_1}\\\\=\frac{400}{ 0.287 \times 573}\\\\=2.43\ \frac{kg}{m^3}$

Calculate the density of air at throat using isentropic flow equation.

$\frac{\rho}{\rho_1}=(1+\frac{\Gamma -1}{2} M*^2)^{-\frac{1}{\Gamma -1}} \\\\\frac{\rho *}{2.43}=(1+\frac{1.4-1}{2} (1.4)*^2)^{-\frac{1}{1.4-1}} \\\\\rho*= 1.045 \ \frac{kg}{m^3}$

Calculate the mass flow rate.

$m= \rho* \times A* \times V*\\\\= 1.045 \times 3.14 times 10^{-4} \times 569.21\\\\= 0.186 \frac{kg}{s}$

5 0

2 years ago

PLEASE QUICK!!!!!!!!!!! Overtime people have given names to groups of stars called_____

kirza4 [7]

Answer:

Constellations

Explanation:

Constellations describes a group of stars that appear to form a pattern or a picture. Example of common patterns are : Leo the Lion or Orion the Great Hunter. It is easy to recognize constellations and most people will tend to orient with these patterns. There are 88 patterns currently known.

6 0

3 years ago

.. You should

Darya [45]

Answer:

C.

Explanation:

You want to make sure it still works. You don't want to move it periodically though in case of an emergency.

5 0

3 years ago

Read 2 more answers

Consider a junction that connects three pipes A, B and C. What can we say about the mass flow rates in each pipe for steady flow

Elis [28]

Answer:

The statement regarding the mass rate of flow is mathematically represented as follows $\Rightarrow \rho \times Q_{3}=\rho \times Q_{1}+\rho \times Q_{2}$

Explanation:

A junction of 3 pipes with indicated mass rates of flow is indicated in the attached figure

As a basic sense of intuition we know that the mass of the water that is in the pipe junction at any instant of time is conserved as the junction does not accumulate any mass.

The above statement can be mathematically written as

$Mass_{Junction}=Constant\\\\\Rightarrow Mass_{in}=Mass_{out}$

this is known as equation of conservation of mass / Equation of continuity.

Now we know that in a time 't' the volume that enter's the Junction 'O' is

1) From pipe 1 = $V_{1}=Q_{1}\times t$

1) From pipe 2 = $V_{2}=Q_{2}\times t$

Mass leaving the junction 'O' in the same time equals

From pipe 3 = $V_{3}=Q_{3}\times t$

From the basic relation of density, volume and mass we have

$\rho =\frac{mass}{Volume}$

Using the above relations in our basic equation of continuity we obtain

$\rho \times V_{3}=\rho \times V_{1}+\rho \times V_{2}\\\\Q_{3}\times t=Q_{1}\times t+Q_{2}\times t\\\\\Rightarrow Q_{3}=Q_{1}+Q_{2}$

Thus the mass flow rate equation becomes $\Rightarrow \rho \times Q_{3}=\rho \times Q_{1}+\rho \times Q_{2}$

6 0

3 years ago

9.21 LAB: Sorting TV Shows (dictionaries and lists) Write a program that first reads in the name of an input file and then reads

natka813 [3]

The following code or the program will be used

<u>Explanation:</u>

def readFile(filename):

dict = {}

with open(filename, 'r') as infile:

lines = infile.readlines()

for index in range(0, len(lines) - 1, 2):

if lines[index].strip()=='':continue

count = int(lines[index].strip())

name = lines[index + 1].strip()

if count in dict.keys():

name_list = dict.get(count)

name_list.append(name)

name_list.sort()

else:

dict[count] = [name]

print(count,name)

return dict

def output_keys(dict, filename):

with open(filename,'w+') as outfile:

for key in sorted(dict.keys()):

outfile.write('{}: {}\n'.format(key,';'.join(dict.get(key))))

print('{}: {}\n'.format(key,';'.join(dict.get(key))))

def output_titles(dict, filename):

titles = []

for title in dict.values():

titles.extend(title)

with open(filename,'w+') as outfile:

for title in sorted(titles):

outfile.write('{}\n'.format(title))

print(title)

def main():

filename = input('Enter input file name: ')

dict = readFile(filename)

if dict is None:

print('Error: Invalid file name provided: {}'.format(filename))

return

print(dict)

output_filename_1 ='output_keys.txt'

output_filename_2 ='output_titles.txt'

output_keys(dict,output_filename_1)

output_titles(dict,output_filename_2)

main()

8 0

3 years ago