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uysha [10]
3 years ago
8

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

Engineering
1 answer:
boyakko [2]3 years ago
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
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What are the characteristics of carbon fibre?
nexus9112 [7]

Answer:

high tensile strenth

high stiffness

Explanation:

5 0
3 years ago
What is the resistance of a circuit with three 1.5 v batteries and running at a current of 5A?
adoni [48]

Answer:

0.3 Ω

Explanation:

Resistance (R) = V/I

R = 1.5/5

R = 15/50

= 3/10

= 0.3Ω

hope it helps :)

3 0
3 years ago
FD=CD*((P*(V^2)*A)/2)<br><br>Please solve for V
REY [17]

Answer:

The answer is V = √2FD ÷ CD × PA

Explanation:

FD = CD × PV²A ÷ 2

V² = 2FD ÷ CD × PA

V = √2FD ÷ CD × PA

Thus, The value of V is V = √2FD ÷ CD × PA

 

<u>-TheUnknownScientist 72</u>

7 0
2 years ago
Describe the steps involved in accomplishing a life-cycle cost analysis (LCCA). I have listed the steps. just describe. a)-Opera
Tasya [4]

Explanation:

a). <u>Operational Performance</u>

     It is defined as the parameter that describes the percentage of accuracy of performing an operation or carrying out an activity.

b).  <u>Salvage externalities</u>

   Salvage in Life cycle cost analysis is a process of estimating the value of the remaining assets in the organisation,.

c). <u>Value Vs Risk</u>

   When we take risk in doing any activity we know the value of accomplising the activity. So value relates directly with risk. When the value of a certain task is high, the risk involve in it is also high.

d). <u>Initial expenditure</u>

   Initial expenditure is nothing but the cost involve in starting a particular acitvity or task at the starting phase.

e). <u>Maintenance implications</u>

   It lays emphasis in maintaining the cost of every possible parameters that are involve in the activity. It includes labour, machines, positions, energy, facilities, etc.

8 0
3 years ago
An electric water heater held at 120° F is kept in a 60°F room. When purchased, its insulation is equivalent to R-5. An owner pu
Trava [24]

Answer:

Total saving would be of 36.917 $\yr

Explanation:

Given Data:

T_{heater} = 120 Degree F

T_{room} = 60 Degree F

A = 30 ft^2

\eta = 100%

Heat loss before previous final value = \frac{A \Delta T}{R}

                                                              =\frac{30\times *(120-60)}{5}

                                                              = 360 Btu/hr

Heat loss after new value= \frac{30\times \times (120-60)}{15} = 120 Btu/hr

saving would be = 360 - 120 Btu/hr \times kw hr/ 3412 Btu\times 24 hr/day \times 365 day/year

                           = 616.1782 kw hr/yr

cost = 616.1782 \times 0.06$

         = 36.917 $\yr

6 0
4 years ago
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