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2 years ago

Tensile testing provides engineers with the ability to verify and establish material properties related to a specific material.

Engineering

1 answer:

Sedbober [7]2 years ago

7 0

Answer:

True

Explanation:

Tensile testing which is also referred to as tension testing is a process which materials are subjected to so as to know how well it can be stretched before it reaches breaking point. Hence, the statement in the question is true

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A BOD test is to be run on a sample of wastewater that has a five-day BOD of 230 mg/L. If the initial DO of a mix of distilled w

Alexxandr [17]

Answer:

Distribution factor P = =38.33

V = 7.826 ml

Explanation:

given details:

BOD =230 mg/l

DO inital = 8.0mg/l

DO final = 2.0mg/l

we know

BOD = [DO inital -DO final] * distribution factor

230 = [8 - 2] D.F

Distribution factor P $= \frac{230}{6}$

Distribution factor P = =38.33

THE RANGE OF WASTE WATER VOLUME IN 300 ml bottle is

distribution factor $= \frac{300}{V}$

$V = \frac{300}{38.33}$

V = 7.826 ml

6 0

2 years ago

Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

NemiM [27]

Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

(d) $X_{C} =49.37$ Ω

(e) $I_{cap} =9.72 A$ and $I_{line} =27.66 A$

Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

$pf_{1} =0.6$ $lagging$

$pf_{2}=0.8$ $leading$

$V=480$ $Volts$

(a) Draw the power triangle for each load and for the combined load.

$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

$S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA$

$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

$Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99$ ≅ $-6 kVAR$

The negative sign means that the load 2 is providing reactive power rather than consuming

Then the combined load will be

$P_{c} =P_{1} +P_{2} =15+8=23 kW$

$Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR$

(b) Determine the power factor of the combined load and state whether lagging or leading.

$S_{c} =P_{c} +jQ_{c} =23+14j$

or in the polar form

$S_{c} =26.92$ °

$pf_{C}=cos(31.32) =0.85$ $lagging$

The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

Current of the combined load can be found by

$I_{C} =S_{C}/\sqrt{3}*V$

$I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A$

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

$Q_{C} =3*V^2/X_{C}$

$X_{C} =3*V^2/Q_{C}$

$X_{C} =3*(480)^2/14*10^3$ Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is

$I_{cap} =V/X_{C} =480/49.37=9.72 A$

Line current flowing from the source is

$I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A$

8 0

2 years ago

What is flow energy? Do fluids at rest possess any flow energy?

anzhelika [568]

Answer:

Flow energy is defined as, flow energy is the energy needed to push fluids into control volume and it is the amount of work done required to push the entire fluid. It is also known as flow work. Flow energy is not the fundamental quantities like potential and kinetic energy.

Fluid at state of rest do not possess any flow energy. It is mostly converted into internal energy as, rising in the fluid temperature.

8 0

2 years ago

An induced-draft cooling tower cools 90,000 gallons per minute of water from 84 to 68oF. Air at 14.61 psia, 70oF dry bulb and 60

belka [17]

Answer:

a. V = 109.64 × 10⁵ ft/min

b. Mw = 654519.54 kg/hr

Explanation:

Given Parameters

mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s

inlet temperature of water, T1 = 84 F = 28.89 C

outlet temperature of water, T2 = 68 F = 20 C

specific heat capacity of water, c = 4.18kJ/kgK

rate of heat remover from water, Qw is given by

Qw = 6607.33[28.89 - 20] * 4.18

Qw = 245529.545kw

For air, inlet condition

DBT = 70 F hi = 43.43 kJ/kg

WBT = 60 F wi = 0.00874 kJ/kg

u1 = 0.8445 m/kg

oulet condition,

DBT = 70 F RH = 100.1

h1 = 83.504kJ/kg

Wo = 0.222kJ/kg

check the attached file for complete solution

3 0

2 years ago

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t

zepelin [54]

Answer:

attached below

Explanation:

4 0

3 years ago