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jek_recluse [69]
3 years ago
10

Consider Fe – 3.0 wt% C steel at a temperature just below the eutectic point. Determine the phase present in this alloy. Determi

ne the compositions of each phase. Determine the relative amount of each phase. Determine the relative amount of lamellar structure.

Engineering
1 answer:
Dmitry_Shevchenko [17]3 years ago
6 0

Answer

Please see attachment for the answer. I solved it in its simplest form.

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Refrigerant 134a at p1 = 30 lbf/in2, T1 = 40oF enters a compressor operating at steady state with a mass flow rate of 200 lb/h a
Mazyrski [523]

Answer:

a) \mathbf{Q_c = -3730.8684 \ Btu/hr}

b) \mathbf{\sigma _c = 4.3067 \ Btu/hr ^0R}

Explanation:

From the properties of Super-heated Refrigerant 134a Vapor at T_1 = 40^0 F, P_1 = 30  \ lbf/in^2 ; we obtain the following properties for specific enthalpy and specific entropy.

So; specific enthalpy h_1 = 109.12 \ Btu/lb

specific entropy s_1 = 0.2315 \ Btu/lb.^0R

Also; from the properties of saturated Refrigerant 134 a vapor (liquid - vapor). pressure table at P_2 = 160 \ lbf/in^2 ; we obtain the following properties:

h_2  = 115.91 \ Btu/lb\\\\ s_2 = 0.2157 \ Btu/lb.^0R

Given that the power input to the compressor is 2 hp;

Then converting to  Btu/hr ;we known that since 1 hp = 2544.4342 Btu/hr

2 hp = 2 × 2544.4342 Btu/hr

2 hp = 5088.8684 Btu/hr

The steady state energy for a compressor can be expressed by the formula:

0 = Q_c -W_c+m((h_1-h_e) + \dfrac{v_i^2-v_e^2}{2}+g(\bar \omega_i - \bar \omega_e)

By neglecting kinetic and potential energy effects; we have:

0 = Q_c -W_c+m(h_1-h_2) \\ \\ Q_c = -W_c+m(h_2-h_1)

Q_c = -5088.8684 \ Btu/hr +200 \ lb/hr( 115.91 -109.12) Btu/lb  \\ \\

\mathbf{Q_c = -3730.8684 \ Btu/hr}

b)  To determine the entropy generation; we employ the formula:

\dfrac{dS}{dt} =\dfrac{Qc}{T}+ m( s_1 -s_2) + \sigma _c

In a steady state condition \dfrac{dS}{dt} =0

Hence;

0=\dfrac{Qc}{T}+ m( s_1 -s_2) + \sigma _c

\sigma _c = m( s_1 -s_2)  - \dfrac{Qc}{T}

\sigma _c = [200 \ lb/hr (0.2157 -0.2315) \ Btu/lb .^0R  - \dfrac{(-3730.8684 \ Btu/hr)}{(40^0 + 459.67^0)^0R}]

\sigma _c = [(-3.16 ) \ Btu/hr .^0R  + (7.4667 ) Btu/hr ^0R}]

\mathbf{\sigma _c = 4.3067 \ Btu/hr ^0R}

5 0
4 years ago
What is a network? I'LL MARK BRAINLEST
Jobisdone [24]

Answer:

hsjeeieoj eu sou ku nahi u have UCC guide to buying it and I he was a temporary password for bees and u h ki tarah nahi to ye sab se jyada nahi hota nahi to kabhi bhi hai ki wo to sirf Tum nahi hota

7 0
3 years ago
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What information is needed to set up sales tax in QuickBooks Online for a client who only does business in their home state?
melisa1 [442]

Answer:

Their company address

When their last tax period started

How often they have to file a tax return

When they started collecting sales tax for the agency

Explanation:

For setup the sales tax information in Quickbooks online for a client who only does business in their home state, we need these information which are given below:

1. Their company addresses

2.  Last Tax Time period started

3.  How frequently they filed the tax return

3. when they begin to received sales tax

Therefore all the other options are not valid. Hence, ignored it

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3 years ago
Hãy trình bày sự hiểu biết của bạn về đo dòng điện
Sidana [21]

Answer:

-Khái niệm:

Đo dòng điện là sử dụng các dụng cụ như ôm kê, vôn kế, ampe kế, tần số kế… để xác định các đại lượng vật lý của dòng điện

-Đo lường điện để làm gì?

Phát hiện hư hỏng sự cố trong mạch điện và các thiết bị vi mạch

Xác định các giá trị cần đo

Đánh giá chất lượng của các thiết bị sau sản xuất

Xác định thông số kỹ thuật của thiết bị

-Phân loại dụng cụ đo điện: Hiên nay có 2 phương pháp phân loại chính

a. Theo nguyên lý làm việc

Dụng cụ đo kiểu điện từ

Dụng cụ đo kiểu điện động

Dụng cụ đo kiểu cảm ứng

Dụng cụ đo kiểu từ điện

b. Theo đai lượng, giá trị cần đo

+Đo điện năng: Ví dụ công tơ điện

+Đo điện áp: Ví dụ: Vôn kế

+Đo dòng điện: Ví dụ: Ampe kế

+Đo công suất: Oát kế

+Đo điện trở: Ôm kế

Sai số khi đo: Khi đo lường luôn xảy ra các sai số

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8 0
3 years ago
2 samples of water of equal volume are put into dishes and kept at room temp for several days. the water in the first dish is co
Over [174]

Answer:

Vaporization is the process by which a substance changes from its solid or liquid state to a gaseous state.

Since both liquids are of the same volume and are placed under the same temperature condition, for them to not to vaporize at the same time, they must have been in different containers.

For vaporization to take place, the volume of liquid, amount of air exposure and area of the surface must be considered.

Maybe the first liquid was in a dish which has a large opening, thereby exposing a large amount which can make water to evaporate faster, whereas the second liquid was somehow enclosed (in a deeper dish).

5 0
3 years ago
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