What’s another name for a service overcurrent device?

A gas tank is known to have a thickness of 0.5 inches and an internal pressure of 2.2 ksi. Assuming that the maximum allowable s

sergiy2304 [10]

Answer:

$D_o=11.9inch$

Explanation:

From the question we are told that:

Thickness $T=0.5$

Internal Pressure $P=2.2Ksi$

Shear stress $\sigma=12ksi$

Elastic modulus $\gamma= 35000$

Generally the equation for shear stress is mathematically given by

$\sigma=\frac{P*r_1}{2*t}$

Where

r_i=internal Radius

Therefore

$12=\frac{2.2*r_1}{2*0.5}$

$r_i=5.45$

Generally

$r_o=r_1+t$

$r_o=5.45+0.5$

$r_o=5.95$

Generally the equation for outer diameter is mathematically given by

$D_o=2r_o$

$D_o=11.9inch$

Therefore

Assuming that the thin cylinder is subjected to integral Pressure

Outer Diameter is

$D_o=11.9inch$

7 0

3 years ago

Two different fuels are being considered for a 2.5 MW (net output) heat engine which can operate between the highest temperature

sveta [45]

Answer:

If the heat engine operates for one hour:

a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.

b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.

In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.

Explanation:

The Carnot efficiency is obtained as:

$\epsilon_{car}=1-\frac{T_c}{T_H}$

Where $T_c$ is the atmospheric temperature and $T_H$ is the maximum burn temperature.

For the case (B), the efficiency we will use is:

$\epsilon_{b}=0.4\epsilon_{car}$

The work done by the engine can be calculated as:

$W=\epsilon Q=\epsilon H_v\cdot m_{fuel}$ where Hv is the heat value.

If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.

$m=\frac{P\cdot t}{\epsilon H_v}$

If we want to calculate the total fuel cost, we only have to multiply the fuel mass with the cost per kilogram.

$TC=m\cdot c$

8 0

3 years ago

Suppose we have a database for an investment firm, consisting of the following attributes: B (broker), O (office of a broker), I

Snowcat [4.5K]

Answer:

Given, FDs are:

S -> D

I -> B

IS -> Q

B -> O

a)

"I" and "S" must be there in any candidate key because they do not appear on the right side of any functional dependency.

The only candidate key is: IS

IS -> ISBDQO

b)

Decomposition of R into 3NF: (I, B), (S, D), (B, O), (I, S, Q)

c)

Decomposition of R into BCNF:

Decompose R by I → B into R1 = (I, B) and R2 = (I, O, S, Q, D).

R1 is in BCNF

Decompose R2 by S → D into R21 = (S, D) and R22 = (O, I, S, Q).

R21is in BCNF

Decompose R22 by I → O into R221 = (I, O) and R222 = (I, S, Q).

R221 is in BCNF.

R222 is in BCNF.

The decomposition is: (I, B), (S, D), (I, O), (I, S, Q)

We can also write it as: (I, B), (S, D), (B, O), (I, S, Q)

Explanation:

The answer above is rendered in a very explanatory way.

8 0

3 years ago

Calculate the theoretical density of FCC iron (eg. austenitic stainless steel). The lattice parameter for FCC iron is 0.357 nm a

Ann [662]

Answer: 12.4 feet

Explanation:

If there is a smooth transition and there is no change in slopes, energy considerations can be used

The cube has a kinetic energy of

ke = mv^2/2 = 10 lbm * 20^2ft^2/s^2 / 2 = 2000 lbm-ft^2 / s^2

At the highest point when there is a gain in potential energy

pe = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2

If there is no loss in energies,

pe = ke

322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2

h = 2000 /322 = 6.211 (ft)

= h / sin(30) = 12.4 ft

8 0

3 years ago

Identify the branch of study each student should pursue.

Paha777 [63]

Answer:

Explanation:

i took the test and screenshot it

3 0

3 years ago