All categories

3 years ago

What is the t max for a carbon steel heater tube?

1 answer:

4 0

Overheating of Carbon Steel Heater Tubes. ... document and find the creep analysis in the appendix-J. Maximum temperature of the CS permitted by the API-581 is 410 Deg C after which you need to

You might be interested in

Where do greywater pipes generally feed into?

Mekhanik [1.2K]

Answer:

c Waste stack

Explanation:

7 0

3 years ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

4 years ago

Design circuits that demonstrate all of the principles listed below. Set up the circuits and take measurements to show that the

Nata [24]

<u>Explanation</u>:

For series

$\Delta V=V_{1}+V_{2}+\ldots+V_{n}=I R_{1}+I R_{2}+\ldots+I R_{n}(\text {voltages add to the batter } y)$

$$I=I_{1}=I_{2}=I_{n}$ (current is the same)$

$V=I R(\text {voltage is directly proportional to } R)$

$R_{e q}=R_{1}+R_{2}+\ldots+R_{n} \quad \text { (resistance increase) }$

For parallel

$\Delta V=\Delta V_{1}=\Delta V_{2}=\Delta V_{n} \quad(\text { same voltage })$

$I=I_{1}+I_{2}+\ldots+I_{n}(\text {current adds})$

$I=\frac{\Delta V}{R_{e q}} \quad(R \text { inversal } y \text { proportional to } I)$

$\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots+\frac{1}{R_{n}}$

3 0

3 years ago

Can someone draw a electric circuit with electrical symbols for this

erastovalidia [21]

Answer:

i cant draw it

i am sry

Explanation:

4 0

3 years ago

Read 2 more answers

A(n) ______ is used to measure fluid flow in engineering

Kazeer [188]

Explanation:

instrument of engineering and management skills that can be used to identify a device on a network of devices

6 0

3 years ago