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bulgar [2K]
3 years ago
9

Determine the design stress for bolts in a cylinder cover where the load is fluctuating due to gas pressure. The maximum load on

the bolt is 50 kN and the minimum is 30 kN. The load is unpredictable and factor of safety is 3. The surface of the bolt is hot rolled and the surface finish factor is 0.9. During a simple tension test and rotating beam test on ductile materials (40 C 8 steel annealed), the following results were obtained : Diameter of specimen = 12.5 mm; Yield strength = 240 MPa; Ultimate strength = 450 MPa; Endurance limit = 180 MPa
​
Engineering
1 answer:
Svetach [21]3 years ago
3 0

Sorry I'm new and need points ty

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(a) Draw the Moore finite state machine (FSM) of an electronic combination lock with a RESET button, two number buttons (0 and 1
Dafna11 [192]

Answer:

Explanation:

There are a total of 6 states and 3 bits in this problem. Whenever the Reset button is pressed, RESET state is called otherwise the state according to the diagram is called. For the combination to be "01011", the input sequence has to be in the same order. If 0 is pressed instead of 1 in state "010", the last state of output ending with 0 will be called and likewise in all the states that follow.

7 0
3 years ago
A binary geothermal power plant uses geothermal water at 160°C as the heat source. The cycle operates on the simple Rankine cycl
bogdanovich [222]

A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined

Assumptions :

1.  Steady operating conditions exist.

2.  Kinetic and potential energy changes are negligible.

Properties:  The specific heat of geothermal water ( c_{geo}[) is taken to be 4.18 kJ/kg.ºC.  

Analysis (a) We need properties of isobutane, we can obtain the properties from EES.

a. Turbine

PP_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The  isentropic  efficiency, n_{T} = \frac{h_{3}-h_{4}  }{h_{3}- h_{4s} }

==\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788

b. Pump

h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} =  \frac{v_{1}(P_{2}-P_{1})   }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\          = 273.01+5.81\\           = 278.82 kJ/kg\\\\w_{T,out} = m^{.}  (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\

W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.}  w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.}  _{net} = W^{.} _{T, out} - W^{.}  _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\

c. The thermal efficiency of the cycle  n_{th}  =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%

7 0
4 years ago
Read 2 more answers
Write a SELECT statement that returns these column names and data from the Products table: product_name The product_name column
I am Lyosha [343]

Answer:

 SELECT  

   product_name, list_price, discount_percent,    

   ROUND(list_price * (discount_percent / 100), 2) AS discount_amount,

   ROUND(list_price - (discount_percent / 100 * list_price), 2)  AS discount_price  

FROM

Products

ORDER BY (list_price - (discount_percent / 100 * list_price)) DESC

LIMIT 5;

Explanation:

In the above SELECT statement is used to select columns from Products table.

The columns are  product_name,list_price and discount_percent

Next a column discount_amount is selected which is the calculated from previous two columns that are list_price and discount_percent. The discount amount is basically calculated by multiplying list_price with discount_percent/100 and the resultant column is named as discount_amount using Alias which is used to give a temporary name to set of columns or a table.

Next another column discount_price is obtained from previous three columns that are  list_price , discount_percent and discount_amount as: list_price - (discount_percent / 100 * list_price) This as a whole is given a column name discount_price.

FROM is used to refer to a table from which these columns are to be selected. The table name is Product.

The result set is sorted in descending order by discount_price so ORDER BY is used to order the resultant records and DESC is used to sort these records according to the discount_price in descending order.

LIMIT statement is used to extract the records from Product and limit the number of rows returned as a result based on a limit value which is 5 here.

5 0
4 years ago
An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at
Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft

G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%

With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

The station is given as

St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

Low point is given as

x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0
4 years ago
3. Which of these instruments is used to measure wind speed? A. anemometer C. wind sock B. thermometer D. wind vane It is an ins
siniylev [52]

Answer:

wind vane if it can be used to show wind speed and the other is a

Explanation:

please mark 5 star if im right and brainly when ya can

5 0
3 years ago
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