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Tatiana [17]
2 years ago
12

Which of the following would most likely be operated by a sequential control system?

Engineering
2 answers:
Rudiy272 years ago
7 0

Answer:pizza oven

Explanation:

Serga [27]2 years ago
3 0

Answer:

D:    School bells that mark the beginning and end of classes

Explanation:

E d e n u i t y 2020-2021

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A large spherical tank contains gas at a pressure of 400 psi. The tank is constructed of high-strength steel having a yield stre
jok3333 [9.3K]

Answer:

t=2.025 inches

Explanation:

Given that

P = 400 Psi

Yield stress ,σ = 80 ksi

Diameter ,d= 45 ft

We know that

1 ft = 12 inches

d= 540 inches

Factor of safety ,K= 3

The required thickness given as

\dfrac {Pd}{4t}=\dfrac{\sigma}{K}

t=thickness

\dfrac {PdK}{4\sigma}=t

\dfrac {400\times 540\times 3}{4\times 80\times 1000}=t

t=2.025 inches

Therefore thickness will be 2.025 inches.

5 0
4 years ago
Define volume flow rate Q of air flowing in a duct of area A with average velocity V
Shalnov [3]

Answer:

The volume flow rate of air is Q=A\times V

Explanation:

A random duct is shown in the below attached figure

The volume flow rate is defined as the volume of fluid that passes a section in unit amount of time

Now by definition of velocity we can see that 'v' m/s means that in 1 second the flow occupies a length of 'v' meters

From the attached figure we can see that

The volume of the prism that the flow occupies in 1 second equals

Volume=Area\times V=A\times V

Hence the volume flow rate is Q=V\times A

3 0
3 years ago
Determine the total condensation rate of water vapor onto the front surface of a vertical plate that is 10 mm high and 1 m in th
castortr0y [4]

Answer:

Q =  63,827.5 W

Explanation:

Given:-

- The dimensions of plate A = ( 10 mm x 1 m )

- The fluid comes at T_sat , 1 atm.

- The surface temperature, T_s = 75°C  

Find:-

Determine the total condensation rate of water vapor onto the front surface of a vertical plate

Solution:-

- Assuming drop-wise condensation the heat transfer coefficient for water is given by Griffith's empirical relation for T_sat = 100°C.

                            h = 255,310 W /m^2.K

- The rate of condensation (Q) is given by Newton's cooling law:

                           Q = h*As*( T_sat - Ts )

                           Q = (255,310)*( 0.01*1)*( 100 - 75 )

                           Q =  63,827.5 W

8 0
3 years ago
Read 2 more answers
What is one of the most common ways workers get hurt around machines
-BARSIC- [3]

Answer:

if their body parts stuck in a machine,if machine expl

Explanation:

ode.

4 0
3 years ago
A 0.25in diameter steel rod BC is securely attached between two identical 1in diameter copper rods (AB and CD). Find the torque
Helen [10]

Answer:

Tmax= 46.0 lb-in

Explanation:

Given:

- The diameter of the steel rod BC d1 = 0.25 in

- The diameter of the copper rod AB and CD d2 = 1 in

- Allowable shear stress of steel τ_s = 15ksi

- Allowable shear stress of copper τ_c = 12ksi

Find:

Find the torque T_max

Solution:

- The relation of allowable shear stress is given by:

                             τ = 16*T / pi*d^3

                             T = τ*pi*d^3 / 16

- Design Torque T for Copper rod:

                             T_c = τ_c*pi*d_c^3 / 16

                             T_c = 12*1000*pi*1^3 / 16

                             T_c = 2356.2 lb.in

- Design Torque T for Steel rod:

                             T_s = τ_s*pi*d_s^3 / 16

                             T_s = 15*1000*pi*0.25^3 / 16

                             T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

                             T = min ( 2356.2 , 46.02 )

                             T = 46.02 lb-in

6 0
3 years ago
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