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3 years ago

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The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated

with an external surface finish. c)- both answers 1 and 2 d)- all of the above.

1 answer:

madam [21]3 years ago

8 0

Answer:

a)Are generally associated with factor.

Explanation:

We know that losses are two types

1.Major loss :Due to friction of pipe surface

2.Minor loss :Due to change in the direction of flow

As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and it produce losses in the energy.

Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as

$Losses=K\dfrac{V^2}{2g}$

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wariber [46]

Answer:

Approaching a problem and finding a solution within given guidelines.

Explanation:

To be a good engineer, you simply need to know how to create a solution to a problem with a given set of restraints or guidelines.

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As an engineer, your first thought should be, let's see what this "trivial task" is, and then after your observation, you should begin to pull upon your experience and resources to build a solution that can be low maintenance and under $40,000.

Put simply, when you approach a problem, consider all aspects of the problem and then build a solution that satisfies all requirements.

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2 years ago

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The penalty for littering 15 lb or less is _____.<br> A. $25<br> B. $50<br> C. $100<br> D. $150

Marina CMI [18]

Answer:

D the closest

Explanation:

i looked it up and 8t says 200

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3 years ago

Air at 38°C and 97% relative humidity is to be cooled to 14°C and fed into a plant area at a rate of 510m3/min. (a) Calculate th

Katarina [22]

To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.

The gas ideal law is given as,

$PV=mRT$

Where,

P = Pressure

V = Volume

m = mass

R = Gas Constant

T = Temperature

Our data are given by

$T_1 = 38\°C$

$T_2 = 14\°C$

$\eta = 97\%$

$\dot{v} = 510m^3/kg$

Note that the pressure to 38°C is 0.06626 bar

PART A) Using the ideal gas equation to calculate the mass flow,

$PV = mRT$

$\dot{m} = \frac{PV}{RT}$

$\dot{m} = \frac{0.6626*10^{5}*510}{287*311}$

$\dot{m} = 37.85kg/min$

Therfore the mass flow rate at which water condenses, then

$\eta = \frac{\dot{m_v}}{\dot{m}}$

Re-arrange to find $\dot{m_v}$

$\dot{m_v} = \eta*\dot{m}$

$\dot{m_v} = 0.97*37.85$

$\dot{m_v} = 36.72 kg/min$

PART B) Enthalpy is given by definition as,

$H= H_a +H_v$

Where,

$H_a$ = Enthalpy of dry air

$H_v$ = Enthalpy of water vapor

Replacing with our values we have that

$H=m*0.0291(38-25)+2500m_v$

$H = 37.85*0.0291(38-25)-2500*36.72$

$H = 91814.318kJ/min$

In the conversion system 1 ton is equal to 210kJ / min

$H = 91814.318kJ/min(\frac{1ton}{210kJ/min})$

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The cooling requeriment in tons of cooling is 437.2.

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3 years ago

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irakobra [83]

Answer:

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Explanation:

The air will lose heat and the oil will gain heat.

These heats will be equal in magnitude.

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They will be of different signs because one is entering iits system and the other is exiting.

The heat exchanged by oil is:

qo = Gp * Cpo * (tof - toi)

The heat exchanged by air is:

qa = Ga * Cpa * (taf - tai)

The specific heat capacity of air at constant pressure is:

Cpa = 0.24 BTU/(lbm*F)

Therefore:

Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)

Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))

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slava [35]

Answer:

The Employee

Explanation:

Because it is there responsibility

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3 years ago