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azamat
2 years ago
10

There are 20 forging presses in the forge shop of a small company. The shop produces batches of forgings requiring a setup time

of 3.0 hours for each production batch. Average standard time for each part in a batch is 45 seconds, and there are an average of 600 parts in a batch. The plant workforce consists of two workers per press, two foremen, plus three clerical support staff. (a) Determine how many forged parts can be produced in 1 month, if there are 8 hours worked per day and an average of 21 days per month at one shift per day. (b) What is the labor productivity ratio of the forge shop, expressed as parts per worker-hour?
Engineering
1 answer:
Aleksandr-060686 [28]2 years ago
4 0

Answer:

Considering the guidelines of this exercise.

The pieces produced per month are 504 000

The productivity ratio is 75%

Explanation:

To understand this answer we need to analyze the problem. First of all, we can only produce 2 batches of production by the press because we require 3 hours to set it up. So if we rest those 6 hours from the 8 of the shift we get 6, leaving 2 for an incomplete bath. So multiplying 2 batches per day of production by press we obtain 40 batches per day. So, considering we work in this factory for 21 days per month well that makes 40 x 21  making 840 then we multiply the batches for the pieces 840 x 600 obtaining 504000 pieces produced per month. To obtain the productivity ratio we need to divide the standard labor hours meaning 6 by the amount of time worked meaning 8. Obtaining 75% efficiency.

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A steam power plant operates on an ideal reheat- regenerative Rankine cycle and has a net power output of 80 MW. Steam enters th
trasher [3.6K]

Answer:

flow(m) = 54.45 kg/s

thermal efficiency u = 44.48%

Explanation:

Given:

- P_1 = P_8 = 10 KPa

- P_2 = P_3 = P_6 = P_7 = 800 KPa

- P_4 = P_5 = 10,000 KPa

- T_5 = 550 C

- T_7 = 500 C

- Power Output P = 80 MW

Find:

-  The mass flow rate of steam through the boiler

-  The thermal efficiency of the cycle.

Solution:

State 1:

P_1 = 10 KPa , saturated liquid

h_1 = 192 KJ/kg

v_1 = 0.00101 m^3 / kg

State 2:

P_2 = 800 KPa , constant volume process work done:

h_2 = h_1 + v_1 * ( P_2 - P_1)

h_2 = 192 + 0.00101*(790) = 192.80 KJ/kg

State 3:

P_3 = 800 KPa , saturated liquid

h_3 = 721 KJ/kg

v_3 = 0.00111 m^3 / kg

State 4:

P_4 = 10,000 KPa , constant volume process work done:

h_4 = h_3 + v_3 * ( P_4 - P_3)

h_4 = 721 + 0.00111*(9200) = 731.21 KJ/kg

State 5:

P_5 = 10,000 KPa , T_5 = 550 C

h_5 = 3500 KJ/kg

s_5 = 6.760 KJ/kgK

State 6:

P_6 = 800 KPa , s_5 = s_6 = 6.760 KJ/kgK

h_6 = 2810 KJ/kg

State 7:

P_7 = 800 KPa , T_7 = 500 C

h_7 = 3480 KJ/kg

s_7 = 7.870 KJ/kgK

State 8:

P_8 = 10 KPa , s_8 = s_7 = 7.870 KJ/kgK

h_8 = 2490 KJ/kg

- Fraction of steam y = flow(m_6 / m_3).

- Use energy balance of steam bleed and cold feed-water:

                                        E_6 + E_2 = E_3

               flow(m_6)*h_6 + flow(m_2)*h_3 = flow(m_3)*h_3

                                    y*h_6 + (1-y)*h_3 = h_3

                                  y*2810 + (1-y)*192.8 = 721

Compute y:                          y = 0.2018

- Heat produced by the boiler q_b:

                             q_b = h_5 - h_4 +(1-y)*(h_7 - h_8)

                    q_b = 3500 -731.21 + ( 1 - 0.2018)*(3480 - 2810)

Compute q_b:               q_b = 3303.58 KJ/ kg

-Heat dissipated by the condenser q_c:

                                       q_c = (1-y)*(h_8 - h_1)

                                 q_c= ( 1 + 0.2018)*(2810 - 192)

Compute q_c:               q_c = 1834.26 KJ/ kg

- Net power output w_net:

                                     w_net = q_b - q_c

                                w_net = 3303.58 - 1834.26

                                    w_net = 1469.32 KJ/kg

- Given out put P = 80,000 KW

                                     flow(m) = P / w_net

compute flow(m)          flow(m) = 80,000 /1469.32 = 54.45 kg/s

- Thermal efficiency u:

                                     u = 1 - (q_c / q_b)

                                     u = 1 - (1834.26/3303.58)

                                     u = 44.48 %

5 0
2 years ago
5. Which statement regarding a finite state machine (FSM) is NOT true: (a) In a non-deterministic FSM, a string is invalid if th
valkas [14]

Answer:

The statement (a) In a non-deterministic FSM, a string is invalid if there is one path not leading to a final state is NOT true

Explanation:

A non-deterministic FSM, contrary to deterministic FSM which has only one possible thread of execution, has multiple threads and for the machine to be invalid, all threads should lead to a none accepting (final) state.

7 0
2 years ago
Use the predicate specifications(x, y): x beats yF (x): x is an (American) football teamQ(x, y): x is quarterback of yL(x,y): x
attashe74 [19]

Answer:

a) ∀y∃x(Q(x, y))

b) (B(Jayhawks, W ildcats)→¬∀y(L(Jayhawks, y)))

c) ∃x(B(Wildcats, x) ∧ B(x, Jayhawks))

Explanation:

a) The statement can be rewritten as "For all football teams, there exists a quarterback" which is written in logical symbols.

b) The statement is an implication and thus have a premise and a conclusion. The premise states "Jayhawks beat the Wildcats" which is translated using B(x, y). The conclusion can be rewritten as "It is not the case that Jayhawks lose to all football teams".

c) The statement is a simple conjunction which can be written as "There exists a team x such that the Wildcats beats x and x beats Jayhawks"

7 0
2 years ago
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
CHALLENGE ACTIVITY 1.4.1: Basic syntax errors. Type the statements. Then, correct the one syntax error in each statement. Hints:
Lunna [17]

Answer:

Lets check each statement for the errors.

Explanation:

System.out.println("Predictions are hard.");

This statement has no syntax errors. When this statement is executed the following line will be displayed:

Predictions are hard.

System.out.print("Especially ');

This statement is missing the closing quotation mark. A single quotation mark is placed instead of double quotation mark in the statement.

The following error message will be displayed when this program statement will be compiled:

Main.java:15: error: unclosed string literal

String literals use double quotes. So to correct this syntax error, the statement should be changed as follows:

System.out.print("Especially");

The output of this corrected line is as following:

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System.out.println("about the future.").

In this line a period . is placed at the end of the statement instead of a semicolon ; but according to the syntax rules statements should end in semicolons.

The error message displayed when this line is compiled is as following:

Main.java:15: error: ; expected

Main.java:15: error: not a statement

So in order to correct this syntax error the statement should be changed as following:

System.out.println("about the future.");    

The output of this corrected line is as following:

about the future

System.out.println("Num is: " - userName);

There is a syntax error in this statement because of - symbol used instead of +

+ symbol is used to join together a variable and a value a variable and another variable in a single print statement.

The error message displayed when this line is compiled is as following:

Main.java:13: error: bad operand types for binary operator '-'

So in order to correct this syntax error the statement should be changed as following:

System.out.println("Num is: " + userName);

This line will print two things one is the string Num is and the other is the value stored in userName variable.

So let userName= 5 then the output of this corrected line is as following:

Num is: 5

8 0
2 years ago
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