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Ber [7]
3 years ago
10

why is that most solidification does not begin until the temperature falls somewhat below the equilibrium melting temperature? p

lease give more details in explanation​
Engineering
1 answer:
Aleksandr [31]3 years ago
6 0

Answer:

Otherwise dies may fail under high operating pressure and temperature. ... The better description and understanding of the phase change processes can be ... occurs only when the temperature is dropped well below the equilibrium temperatures. ... The solidification starts at the bottom and the solidified volume grows more

Explanation:

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Write a SELECT statement that returns the same result set as this SELECT statement. Substitute a subquery in a WHERE clause for
sergiy2304 [10]

Answer:

SELECT distinct VendorName FROM Vendors

WHERE VendorID IN (

SELECT VendorID FROM Invoices

)

Explanation:

6 0
3 years ago
A 100-horsepower, three-phase squirrel-cage induction motor is connected to a 240-volt line. A dual-element time-delay fuse is t
antoniya [11.8K]

Based on the National Electrical Code (NEC), a 450-Ampere fuse should be used to protect this polyphase (three-phase) squirrel cage inductor motor.

<u>Given the following data:</u>

  • Power rating = 100-horsepower.
  • Voltage = 240 volt.

<h3>How to determine the correct fuse rating?</h3>

According to Table 430.52 of the National Electrical Code (NEC), a dual-element time delay fuse should be calculated at 175% (1.75) of the full-load current rating for an alternating current (AC) polyphase (three-phase) squirrel cage inductor motor.

In this scenario, the squirrel cage inductor motor didn't list a NEMA design code on its nameplate. Thus, we'll assume that the inductor motor is design B and its fuse rating is given by:

Fuse rating = 248 × 1.75 × 434

In conclusion, the nearest standard fuse size which is above the computed value listed in Section 240.6 of the National Electrical Code (NEC) is 450 amperes.

Read more on National Electrical Code here: brainly.com/question/10619436

#SPJ1

6 0
2 years ago
In a wheatstone bridge three out of four resistors have of 1K ohm each ,and the fourth resistor equals 1010 ohm. If the battery
Dima020 [189]

Answer:

  248.756 mV

  49.7265 µA

Explanation:

The Thevenin equivalent source at one terminal of the bridge is ...

  voltage: (100 V)(1000/(1000 +1000) = 50 V

  impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω

The Thevenin equivalent source at the other terminal of the bridge is ...

  voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V

  impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω

__

The open-circuit voltage is the difference between these terminal voltages:

  (50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage

__

The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:

  (50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA

8 0
3 years ago
For the pipe-fl ow-reducing section of Fig. P3.54, D 1 5 8 cm, D 2 5 5 cm, and p 2 5 1 atm. All fl uids are at 20 8 C. If V 1 5
bonufazy [111]

Answer:

The total force resisted by the flange bolts is  163.98 N

Explanation:

Solution

The first step is to find  the pipe cross section at the inlet section

Now,

A₁ = π /4 D₁²

D₁ =  diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of  the pipe at the inlet section

A₂ = π /4 D₂²

D₂ =  diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

= 1.963 * 10^⁻³ m²

Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

= 5.0265  * 5 = 1.963 * V₂

V₂ = 12.8 m/s

Note: Kindly find an attached copy of the part of the solution to the given question below

8 0
3 years ago
Oil enters a counterflow heat exchanger at 600 K with a mass flow rate of 10 kg/s and exits at 200 K. A separate stream of liqui
Elis [28]

Answer:

The minimum mass flow rate for the water is 14.39kg/s

Explanation:

In this question, we are asked to calculate the minimum mass flow rate for the water in kg/s.

Please check attachment for complete solution and step by step explanation

7 0
3 years ago
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