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KonstantinChe [14]
3 years ago
15

6-What is the difference between the critical point and the triple point?

Engineering
2 answers:
hichkok12 [17]3 years ago
8 0

Answer:

The triple point represents the combination of pressure and temperature that facilitates all phases of matter at equilibrium. The critical point terminates the liquid/gas phase line .

Explanation:

Svetradugi [14.3K]3 years ago
8 0

Explanation:

<h2><u>Critical Point :</u></h2>
  • It's the end point of equilibrium curve .
  • It's typically higher than standard temperature.
  • pressure is also higher than standard pressure.

<h2><u>T</u><u>r</u><u>i</u><u>p</u><u>l</u><u>e</u><u> </u><u>Point</u><u> </u><u>:</u></h2>
  • It's point where the three equilibrium curves meet.
  • Temperature is lower than standard temperature.
  • pressure is lower than standard pressure.

<u>A</u><u>d</u><u>d</u><u>i</u><u>t</u><u>i</u><u>o</u><u>n</u><u>a</u><u>l</u><u> </u><u>points</u><u> </u><u>:</u>

  • The triple point represents combination of pressure .
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3. A 4-m × 5-m × 7-m room is heated by the radiator of a steam-heating system. The steam radiator transfers heat at a rate of 10
Natali [406]

Answer:

14.52 minutes

<u>OR</u>

14 minutes and 31 seconds

Explanation:

Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.

Specific heat at constant volume at 27°C = 0.718 kJ/kg*K

Initial temperature of room (in kelvin) = 283.15 K

Final temperature (required) of room = 293.15 K

Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg

Heat required at constant volume: 0.718 * (change in temp) * (mass of air)

Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ

Time taken for temperature rise: heat required / (rate of heat change)

Where rate of heat change = 10000 - 5000 = 5000 kJ/hr

Time taken = 1210.26 / 5000 = 0.24205 hours

Converted to minutes = 0.24205 * 60 = 14.52 minutes

4 0
3 years ago
Steam at a pressure of 100 bar and temperature of 600 °C enters an adiabatic nozzle with a velocity of 35 m/s. It leaves the noz
butalik [34]

Answer:

Exit velocity V_2=1472.2 m/s.

Explanation:

Given:

At inlet:

P_1=100 bar,T_=600°C,V_1=35m/s

Properties of steam at 100 bar and 600°C

        h_1=3624.7\frac{KJ}{Kg}

At exit:Lets take exit velocity V_2

We know that if we know only one property inside the dome  then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of nozzle  is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.

Properties of saturated steam at 80 bar

   h_f= 1316.61\frac{KJ}{Kg} ,h_g= 2757.8\frac{KJ}{Kg}

So the enthalpy of steam at the exit of turbine  

h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}

h_2=1316.61+0.85(2757.8-1316.61)\frac{KJ}{Kg}

 h_2=2541.62\frac{KJ}{Kg}

Now from first law for open system

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w

In the case of adiabatic nozzle Q=0,W=0

3624.7+\dfrac{35^2}{2000}+0=2541.62+\dfrac{(V_2)^2}{2000}+0

V_2=1472.2 m/s

So Exit velocity V_2=1472.2 m/s.

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3 years ago
A differential amplifier is very useful for removing common mode noise voltage that might be fed or induced in the signal cables
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I think will help
7 0
3 years ago
Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q whil
Karo-lina-s [1.5K]

Answer:

E = \frac{3Q}{2A\epsilon_0}

Explanation:

By Gauss Law for electric field:

E = \frac{\sigma}{2\epsilon_0}

Where \sigma is the charge density Q/A. Since we have 2 parallel  plates with different charge, the electric field at point P in the gap would be the sum of 2 field

E = E_1 + E_2

E = \frac{Q}{2A\epsilon_0} + \frac{2Q}{2A\epsilon_0}

E = \frac{3Q}{2A\epsilon_0}

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3 years ago
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