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My name is Ann [436]
3 years ago
11

Consider the control volume form of the basic laws. For the conservation of mass form for a control volume with mass flow into a

nd out of the control volume, the mass in the control volume is: a)- of a known magnitude. b)- always the same c)-dependent on the mass flows in and out d)- none of the above
Engineering
1 answer:
aleksandrvk [35]3 years ago
5 0

Answer:

C) Dependent on the mass flows in and out.

Explanation:

Lets take control volume(CV)

Take

          m_i =inlet mass flow rate

          m_e =exit mass flow rate

If we take unsteady flow process then inlet mass can not be equal to exit mass.Some mass can store if inlet mass flow rate is high and exit mass flow rate  is low.

So mass of control volume

    m_{cv}=m_i-m_e.

so above we can say that mass of control volume dependent on inlet and exit mass.

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Firefighters are holding a nozzle at the end of a hose while trying to extinguish a fire. The nozzle exit diameter is 8 cm, and
ivanzaharov [21]

Question

Determine the average water exit velocity

Answer:

53.05 m/s

Explanation:

Given information

Volume flow rate, Q=16 m^{3}/min

Diameter d= 8cm= 0.08 m

Assumptions

  • The flow is jet flow hence momentum-flux correction factor is unity
  • Gravitational force is not considered
  • The flow is steady, frictionless and incompressible
  • Water is discharged to the atmosphere hence pressure is ignored

We know that Q=AV and making v the subject then

V=\frac {Q}{A} where V is the exit velocity and A is area

Area, A=\frac {\pi d^{2}{4} where d is the diameter

By substitution

V=\frac {16\times 4}{\pi 0.08^{2}}=3183.098862 m/min

To convert v to m/s from m/s, we simply divide it by 60 hence

V=\frac {3183.098862  m/min}{60 s}=53.0516477 m/s\approx 53.05 m/s

3 0
3 years ago
The function below takes a single parameter, a list of numbers called number_list. Complete the function to return a string of t
makkiz [27]

Answer:

The solution code is written in Python:

  1. def convertCSV(number_list):
  2.    str_list = []
  3.    for num in number_list:
  4.        str_list.append(str(num))
  5.    
  6.    return ",".join(str_list)
  7. result = convertCSV([22,33,44])
  8. print(result)

Explanation:

Firstly, create a function "convertCSV" with one parameter "number_list". (Line 1)

Next, create an empty list and assign it to a new variable <em>str_list</em>. (Line 2)

Use for-loop to iterate through all the number in the <em>number_list</em>.(Line 4). Within the loop, each number is converted to a string using the Python built-in function <em>str() </em>and then use the list append method to add the string version of the number to <em>str_list</em>.

Use Python string<em> join() </em>method to join all the elements in the str_list as a single string. The "," is used as a separator between the elements (Line 7) . At the end return the string as an output.

We can test the function by calling the function and passing [22,33,34] as an argument and we shall see "22,33,44" is printed as an output. (Line 9 - 10)

6 0
3 years ago
Why do engineers need to be particularly aware of the impact of hubris?
miv72 [106K]

Answer:

A) Their creations change society.

7 0
3 years ago
Driving Distraction Brainstorming Session
Leto [7]

texting, phone calls, putting on makeup, brushing hair, movies playing in car, loud music, children, and that's pretty much all I could think of

please give <u>BRAINLIEST ANSWER └[T‸T]┘</u>

5 0
3 years ago
A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60
Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

                     T_2 = T_1 * r^(k-1/k)

                     T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

                     T_7 = T_6 * r^(-k-1/k)

                     T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

                     flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

                     flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

                     flow(m) = 7.941 lbm/s

- The heat input is as follows:

                     Q_in = c_p*(T_6 - T_5)

                     Q_in = 0.24*(1400 - 1022.84)

                     Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

                     Q_out = c_p*(T_10 - T_1)

                     Q_out = 0.24*(711.744 - 520)

                    Q_out = 56.01856 Btu/lbm

                                           

                     

5 0
3 years ago
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