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3 years ago

6

A flat plate is subjected to a load of 5KN as shown in figure. The plate material is grey cast iron FFG 200 and the factor of sa

fety is 2.5. Determine the thickness of the plate

1 answer:

lianna [129]3 years ago

3 0

Answer:

57.14 N/mm2

Explanation:

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A mass of air occupying a volume of 0.15m^3 at 3.5 bar and 150 °C is allowed [13] to expand isentropically to 1.05 bar. Its enth

e-lub [12.9K]

Answer:

Total work: -5.25 kJ

Total Heat: 52 kJ

Explanation:

V0 = 0.15

P0 = 350 kPa

t0 = 150 C = 423 K

P1 = 105 kPa (isentropical transformation)

Δh1-2 = 52 kJ (at constant pressure)

Ideal gas equation:

P * V = m * R * T

m = (R * T) / (P * V)

R is 0.287 kJ/kg for air

m = (0.287 * 423) / (350 * 0.15) = 2.25 kg

The specifiv volume is

v0 = V0/m = 0.15 / 2.25 = 0.067 m^3/kg

Now we calculate the parameters at point 1

T1/T0 = (P1/P0)^((k-1)/k)

k for air is 1.4

T1 = T0 * (P1/P0)^((k-1)/k)

T1 = 423 * (105/350)^((1.4-1)/1.4) = 300 K

The ideal gas equation:

P0 * v0 / T0 = P1 * v1 / T1

v1 = P0 * v0 * T1 / (T0 * P1)

v1 = 350 * 0.067 * 300 / (423 * 105) = 0.16 m^3/kg

V1 = m * v1 = 2.25 * 0.16 = 0.36 m^3

The work of this transformation is:

L1 = P1*V1 - P0*V0

L1 = 105*0.36 - 350*0.15 = -14.7 kJ/kg

Q1 = 0 because it is an isentropic process.

Then the second transformation. It is at constant pressure.

P2 = P1 = 105 kPa

The enthalpy is raised in 52 kJ

Cv * T1 + P1*v1 = Cv * T2 + P2*v2 + Δh

And the idal gas equation is:

P1 * v1 / T1 = P2 * v2 / T2

T2 = T1 * P2 * v2 / (P1 * v1)

Replacing:

Cv * T1 + P1*v1 + Δh = Cv * T1 * P2 * v2 / (P1 * v1) + P2*v2

Cv * T1 + P1*v1 + Δh = v2 * (Cv * T1 * P2 / (P1 * v1) + P2)

v2 = (Cv * T1 + P1*v1 + Δh) / (Cv * T1 * P2 / (P1 * v1) + P2)

The Cv of air is 0.7 kJ/kg

v2 = (0.7 * 300 + 105*0.16 + 52) / (0.7 * 300 * 105 / (105 * 0.16) + 105) = 0.2 m^3/kg

V2 = 2.25 * 0.2 = 0.45 m^3

T2 = 300 * 105 * 0.2 / (105 * 0.16) = 375 K

The heat exchanged is Q = Δh = 52 kJ

The work is:

L2 = P2*V2 - P1*V1

L2 = 105 * 0.45 - 105 * 0.36 = 9.45 kJ

The total work is

L = L1 + L2

L = -14.7 + 9.45 = -5.25 kJ

8 0

3 years ago

A large class with 1,000 students took a quiz consisting of ten questions. To get an A, students needed to get 9 or 10 questions

VMariaS [17]

Answer:

a. 0.11

b. 110 students

c. 50 students

d. 0.46

e. 460 students

f. 540 students

g. 0.96

Explanation:

(See attachment below)

a. Probability that a student got an A

To get an A, the student needs to get 9 or 10 questions right.

That means we want P(X≥9);

P(X>9) = P(9)+P(10)

= 0.06+0.05=0.11

b. How many students got an A on the quiz

Total students = 1000

Probability of getting A = 0.11 ---- Calculated from (a)

Number of students = 0.11 * 1000

Number of students = 110 students

So,the number of students that got A is 110

c. How many students did not miss a single question

For a student not to miss a single question, then that student scores a total of 10 out of possible 10

P(10) = 0.05

Total Students = 1000

Number of Students = 0.05 * 1000

Number of Students = 50 students

We see that 5

d. Probability that a student pass the quiz

To pass, a student needed to get at least 6 questions right.

So we want P(X>=6);

P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)

=0.08+0.12+0.15+0.06+0.05=0.46

So, the probability of a student passing the quiz is 0.46

e. Number of students that pass the quiz

Total students = 1000

Probability of passing the quiz = 0.46 ----- Calculated from (d)

Number of students = 0.46 * 1000

Number of students = 460 students

So,the number of students that passed the test is 460

f. Number of students that failed the quiz

Total students = 1000

Total students that passed = 460 ----- Calculated from (e)

Number of students that failed = 1000 - 460

Number of students that failed = 540

So,the number of students that failed is 540

g. Probability that a student got at least one question right

This means that we want to solve for P(X>=1)

Using the complement rule,

P(X>=1) = 1 - P(X<1)

P(X>=1) = 1 - P(X=0)

P(X>=1) = 1 - 0.04

P(X>=1) = 0.96

7 0

3 years ago

a centrifugal pump delivers water at the rate of 50kg/is . the inlet and outlet pressure are 2 bar and 6.2 bar respectively. the

Monica [59]

Answer:

21.6 kw

Explanation:

Given data:

m = 50 kg/s

Inlet pressure (p1) = 2 bar

outlet pressure(p2) = 6.2 bar

suction ( h1 ) = -2.2 m

delivery ( h2 ) = 8.5 m

d1 = 200 mm = 0.2 m

d2 = 100 mm = 0.1 m

Vs of water = 0.001 m^3/kg

next we have to determine the Q value

Q = V*A

Q = 0.001 * 50 = 0.05 m^3/s

next we have to calculate the various V's

V1 = Q/A1 = $\frac{0.05*4}{\pi *0.2^2}$ = 1.59 m/s

V2 = Q/A2 = $\frac{0.05*4}{\pi *0.1^2}$ = 6.37 m/s

Determine the capacity of the electric motor

attached below is the detailed solution

7 0

3 years ago

Which of the following statements about cylinder placement are true?

Alenkinab [10]

What is the following?

3 0

3 years ago

if this light stays on after the initial startup of the vehicle it signals a warning that there is something wrong with the serv

NemiM [27]

Service brake system indicator is the warning that there is something wrong with the service brake system. Hence option f is correct.

<h3>What is indicator?</h3>

Amber-colored indicator lights can be found at the front, back, and occasionally on the left and right sides of the vehicle. Whether you're turning left, right, or into oncoming traffic, you use your indicators to signal your planned change of direction.

When this light turns on, one of two things will happen. Either the parking brake is engaged or the hydraulic fluid (brake fluid) in the master cylinder is low. Your brakes are made up of a system of hydraulic oil-filled tubes called brake lines.

Thus, service brake system indicator is the warning that there is something wrong with the service brake system. Hence option f is correct.

To learn more about indicator, refer to the link below:

brainly.com/question/28093573

#SPJ1

7 0

1 year ago