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3 years ago

What is the lowest Temperature in degrees C?, In degrees K? in degrees F? in degrees R

Engineering

1 answer:

Colt1911 [192]3 years ago

3 0

Answer:

-273.16 °C

-459.677 °F

0 °K

0 °R

Explanation:

The lowest temperature is the absolute zero.

Absolute zero is at 0 degrees Kelvin, or 0 degrees Rankine, because these are absolute scales that have their zero precisely at the absolute zero.

Celsius and Fahrenheit degrees are relative scales, these have their zeroes above the absolute zero.

Celsius scale has the same degree separation as the Kelvin scale, but the zero is separated by 273.16 degrees. Therefore the lowest temperature in the Celsius scale is -273.16 °C.

The Fahrenheit degrees have the same degree separation as the Rankine degrees, and the zero is 459.67 degrees. Therefore the lowest temperature in the Fahrenheit scale is -459.67 °F.

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What is shown in the above figure

Vikki [24]

Answer:

there is no figure .....

3 0

2 years ago

A sample of wastewater is diluted 10 times. The diluted solution has an ultimate biochemical oxygen demand (BOD), Lo, of 30 mg/L

zzz [600]

Answer:

474.59 mg/L

Explanation:

Given that

BOD = 30 mg/L

Original BOD = 30 mg/L × dilution factor

Original BOD = 30 mg/L × 10 = 300 mg/L

$L_o = \frac{BOD}{1-e^{-5t}}$

here $L_o$ is the ultimate BOD ; BOD is the biochemical oxygen demand ; t = 0.20 /day

$L_o = \frac{300}{1-e^{-5(0.20)}}$

$L_o = 474.59 \ mg/L$

3 0

3 years ago

2. One of the many methods used for drying air is to cool the air below the dew point so that condensation or freezing of the mo

REY [17]

Answer:

0.5°c

Explanation:

Humidity ratio by mass can be expressed as

the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air

Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.

Humidity ratio expressed by mass:

x = mw / ma (1)

where

x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)

mw = mass of water vapor (kg, lb)

ma = mass of dry air (kg, lb)

It can be as:

x = 0.005 (100) / [(100 - 100)]

x = 0.005 x 100 / (100 - 100)

x = 0.005 x 100 / 0

x = 0.5°c

So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c

6 0

3 years ago

What is the objective of phasing out an INDUCTION MOTOR before putting the machine into commission?

enyata [817]

The main objective of phasing out an INDUCTION MOTOR is to identify the ends of the stator coils.

<h3>What is an induction motor?</h3>

An induction motor is a device based on alternate electricity (AC) which is composed of three different stator coils.

An induction motor is a device also known as an asynchronous motor due to its irregular velocity.

In conclusion, the objective of phasing out an INDUCTION MOTOR is to identify the ends of the stator coils.

Learn more on induction motors here:

brainly.com/question/15721280

#SPJ1

8 0

2 years ago

Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb

guajiro [1.7K]

Answer:

$\eta_{turbine} = 0.603 = 60.3\%$

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in vapor state:

h₂ = $h_{g\ at\ 125KPa}$ = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than $s_g$ and greater than $s_f$ at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

$x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88$

Now, we will find $h_{2s}$ (enthalpy at the outlet for the isentropic process):

$h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg$

Now, the isentropic efficiency of the turbine can be given as follows:

$\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%$

3 0

3 years ago