Check the level of motor oil in your engine by ?

Type the correct answer in the box. Spell all words correctly.

Rudik [331]

Answer:ii dant overstand

Explanation:

5 0

3 years ago

Zander worked at a pet shop in high school and during college began taking classes in veterinary medicine. To pay the bills duri

anastassius [24]

True.
To understand it better
First job : Pet shop
Second job : pizza place
The first job supports his career path he has experience.
The second job support life in making sure he gets to his career path/ does help financially for him to get there.
And it’s called career pathway.

3 0

2 years ago

Which of the following explains why material properties present challenges for engineers?

Maurinko [17]

Answer:

Explanation:

They are altered by variables such as temperature hence making materials challenging when dealing with them.

4 0

3 years ago

An industrial plant consists of several 60 Hz single-phase motors with low power factor. The plant absorbs 600 kW with a power f

Gelneren [198K]

Answer:

(a) $Q=332$ $kvar$ and $C=5.66$ $uF$

(b) $pf=0.90$ lagging

Explanation:

Given Data:

$P=600kW$

$V=12.47kV$

$f=60Hz$

$pf_{old} =0.75$

$pf_{new} =0.95$

(a) Find the required kVAR rating of a capacitor

$\alpha _{old}=cos^{-1}(0.75) =41.41$ °

$\alpha _{new}=cos^{-1}(0.95) =18.19$ °

The required compensation reactive power can be found by

$Q=P(tan(\alpha_{old}) - tan(\alpha_{new}))$

$Q=600(tan(41.41) - tan(18.19))$

$Q=332$ $kvar$

The corresponding capacitor value can be found by

$C=Q/2\pi fV^{2}$

$C=332/2*\pi *60*12.47^{2}$

$C=5.66$ $uF$

(b) calculate the resultant supply power factor

First convert the hp into kW

$P_{mech} =250*746=186.5$ $kW$

Find the electrical power (real power) of the motor

$P_{elec} =P_{mech}/n$

where $n$ is the efficiency of the motor

$P_{elec} =186.5/0.80=233.125$ $kW$

The current in the motor is

$I_{m} =(P/\*V*pf)$

The pf of motor is 0.85 Leading

Note that represents the angle in complex notation (polar form)

$I_{m} =(233.125/12.47*0.85)$

$I_{m}=18.694+11.586$ $j$ $A$

Now find the Load current

pf of load is 0.75 lagging (notice the minus sign)

$I_{load} =(600/12.47*0.75)$

$I_{load} =48.115-42.433$ $j$ $A$

Now the supply current is the current flowing in the load plus the current flowing in the motor

$I_{supply} =I_{m} + I_{load}$

$I_{supply}= (18.694+11.586)+(48.115-42.433)$

$I_{supply} =66.809-30.847j$ $A$

or in polar form

$I_{supply} =73.58$ °

Which means that the supply current lags the supply voltage by 24.78

therefore, the supply power factor is

$pf=cos(24.78)=0.90$ lagging

Which makes sense because original power factor was 0.75 then we installed synchronous motor which resulted in improved power factor of 0.90

8 0

3 years ago

Liquid ethanol is a flammable fluid and can release vapors that form explosive mixtures at temperatures above its flashpoint at

marta [7]

Answer:

The volume flow rate necessary to keep the temperature of the ethanol in the pipe below its flashpoint should be greater than 1.574m^3/s

Explanation:

Q = MCp(T2 - T1)

Q (quantity of heat) = Power (P) × time (t)

Density (D) = Mass (M)/Volume (V)

M = DV

Therefore, Pt = DVCp(T2 - T1)

V/t (volume flow rate) = P/DCp(T2 - T1)

P = 20kW = 20×1000W = 20,000W, D(rho) = 789kg/m^3, Cp = 2.44J/kgK, T2 = 16.6°C = 16.6+273K = 289.6K, T1 = 10°C = 10+273K = 283K

Volume flow rate = 20,000/789×2.44(289.6-283) = 20,000/789×2.44×6.6 = 1.574m^3/s (this is the volume flow rate at the flashpoint temperature)

The volume flow rate necessary to keep the ethanol below its flashpoint temperature should be greater than 1.574m^3/s

6 0

3 years ago