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2 years ago

Select the properties and typical applications for the high carbon steels.

Engineering

1 answer:

yanalaym [24]2 years ago

3 0

Answer:

Option-(A)

Explanation:

Typical applications for the high carbon steels includes the following;

It is heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.

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Answer:

... spilling water or getting anything cascading onto the floor

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2 years ago

After replacing a vacuum booster, the brakes lock up on a road test. Technician A says there is air trapped inside the brake lin

vitfil [10]

Answer:

Technician B

Explanation:

The brakes can lockup due to the following reasons

1) Overheating break systems

2) Use of wrong brake fluid

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3 years ago

The specific volume of a system consisting of refrigerant-134a at 1.0 Mpa is 0.01 m /kg. The quality of the R-134a is: (a) 12.6%

Flura [38]

Answer:

option c is correct

47.2%

Explanation:

given data

consisting of refrigerant = 134 a

volume V = 0.01 m³/kg

pressure P = 1MPa = 1000 kPa

to find out

quality of the R 134a

solution

we will get here value of volume Vf and Vv from pressure table 60 kpa to 3 Mpa for 1 Mpa of R134 a

that is

Vf = 0.0008701 m³/kg

Vv = 0.0203 m³/kg

so we will apply here formula that is

quality = (V - Vf) / (Vv - Vf) ............1

put here value

quality = (0.01 - 0.0008701 ) / ( 0.0203 - 0.0008701 )

quality = 0.4698

so quality is 47 %

SO OPTION C IS CORRECT

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3 years ago

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Answer:

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2 years ago

Read 2 more answers

A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl

lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

let mass flow rate for cold and hot fluid = Mc and Mh respectively
let specific heat for cold and hot fluid = Cpc and Cph respectively
let heat capacity rate for cold and hot fluid = Cc and Ch respectively

Mc = 1.2 kg/s and Mh = 2 kg/s

Cpc = 4.18 kj/kg °c and Cph = 4.31 kj/kg °c

Using effectiveness-NUT method

First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate

Cc = Mc × Cpc = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c

Ch = Mh × Cph = 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= Cmin/Cmax

C = 5.016/8.62 = 0.582

.2 Second, we determine the maximum heat transfer rate, Qmax

Qmax = Cmin (Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Qmax = (5.016 kW/°c)(160 - 20) °c

Qmax = (5.016 kW/°c)(140) °c = 702.24 kW

.3 Third, we determine the actual heat transfer rate, Q

Q = Cmin (outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Qmax = (5.016 kW/°c)(60) °c = 303.66 kW

.4 Fourth, we determine Effectiveness of the heat exchanger, ε

ε = Q/Qmax

ε = 303.66 kW/702.24 kW

ε = 0.432

.5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger

NTU = $\\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )$

NTU = $\frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582 -1} )$

NTU = 0.661

.6 sixth, we determine Heat Exchanger surface area, As

From the question, the overall heat transfer coefficient U = 640 W/m²

As = $\frac{NTU C{min} }{U}$

As = $\frac{0.661 x 5016 W. °c }{640 W/m²}$

As = 5.18 m²

.7 Finally, we determine the length of the heat exchanger, L

L = $\frac{As}{\pi D}$

L = $\frac{5.18 m² }{\pi (0.015 m)}$

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0

3 years ago