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m_a_m_a [10]
3 years ago
15

A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is he

ated, and the gage pressure at the nal state is 367 kPa. Determine the nal temperature, in °C. The local atmospheric pressure is 1 atm.
Engineering
1 answer:
sergejj [24]3 years ago
8 0

Answer:

T₂ =93.77  °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

We know that

Absolute pressure = Gauge pressure + Atmospheric pressure

Initial pressure ,P₁ = 300+1=301 kPa

Final pressure  ,P₂= 367+1 = 368  kPa

Lets take  temperature=T₂

We know that ,If the volume of the gas is constant ,then we can say that

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

{T_2}=T_1\times \dfrac{P_2}{P_1}

Now by putting the values in the above equation we get

{T_2}=300\times \dfrac{368}{301}\ K

The temperature in  °C

T₂ = 366.77 - 273  °C

T₂ =93.77  °C

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Answer:While heat cycles cause oil to darken, soot causes oil to turn black

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3 years ago
The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to
Sergeeva-Olga [200]

Answer:

hello below is missing piece of the complete question

minimum size = 0.3 cm

answer : 0.247 N/mm2

Explanation:

Given data :

section span : 10.9 and 13.4 cm

minimum load applied evenly to the top of span  : 13 N

maximum load for each member ; 4.5 N

lets take each member to be 4.2 cm

Determine the max value of P before truss fails

Taking average value of section span ≈ 12 cm

Given minimum load distributed evenly on top of section span = 13 N  

we will calculate the value of   by applying this formula

= \frac{Wl^2}{12}  =  (0.013 * 0.0144 )/ 12  =  1.56 * 10^-5

next we will consider section ; 4.2 cm * 0.3 cm

hence Z (section modulus ) = BD^2 / 6  

                                             = ( 0.042 * 0.003^2 ) / 6  = 6.3*10^-8

Finally the max value of P( stress ) before the truss fails

= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )  

          = 0.247 N/mm2

5 0
3 years ago
8. What are two ways SpaceX plans to change personal travel?
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2 years ago
Write down the types and tasks of the pressure control valves ?
Yuki888 [10]

Answer:

There are 6 types of pressure control valves and their function is to regulate the pressure below a threshold level within safe limits and to maintain and control  pressure of a particular circuit.

Explanation:

The six type of Pressure valve with their functions are given below:

a. Unloading Valve:

These type of pressure valve are used to pour fluid into the container at very low or no pressure.

b. Safety valve:

These are used when the pressure within the vessel is in excess as inside temperature is near about preset [point point then these valves are open to release the extra pressure and are closed once normal conditions are regained.

c. Pressure Reducing Valve:

These are basically used for the control of the pressure in downstream not exceeding the design limits.

d. Pressure Relief Valves:

These are basically used to limit and regulate the pressure of any system.

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These are used to develop pressure in the reverse direction at the actuator's return line in order to keep the load under control.

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8 0
3 years ago
An insulated, vertical piston-cylinder device initially contains 10kg of water, 6kg of which is in the vapor phase. The mass of
Alexeev081 [22]

Answer:

a)120C

b)29kg

Explanation:

Hello!

To solve this exercise follow the steps below

1. we will call 1 the initial state, 2 the steam that enters and 3 the final state

2. We find the quality of the initial state, dividing the mass of steam by the total mass.

q1=\frac{6kg}{10kg} =0.6

3 Find the internal energy in the three states using thermodynamic tables

note:Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

u1=IntEnergy(Water;x=0,6(quality);P=200kPa) =1719KJ/kg

u2=IntEnergy(Water;t=350;P=5000kPa) =2808KJ/kg

u3=IntEnergy(Water;x=1;P=200kPa) =2529KJ/kg

4. use the internal energy and pressure to find the temperature in state 3, using thermodynamic tables

T3=Temperature(Water;P=200kPa;u=u3=2529KJ/kg)=120C

5. Use the first law of thermodynamics in the system, it states that the initial energy in a system must be equal to the final

m1u1+m2u2=(m1+m2)u3

where

m1=inital mass=10kg

m2=the mass of the steam that has entered.

solve for m2

(m1)(u1-u3)=(m2)(u3)-(m2)(u2)

m2=m1\frac{u1-u3}{u3-u2} =10\frac{1719-2529}{2529-2808} =29kg

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3 years ago
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