The title block generally contains _______

A gas contained within a piston-cylinder undergoes the follow change in states: Process 1: Constant volume from p1 = 1 bar V1 =

laiz [17]

Answer:

Process 1:W=0

Process 2:W= -386.13 KJ

Process 3:W= -468 KJ

Explanation:

Process 1: $P_1=1 bar,V_1=2.6m^3$

Process 2: $P_2=2.7bar,V_2=2.6m^3$

Process 3: $V_3=1.5 m^3$

$V_4=0.5 m^3$

Process 1:

Work (W)=0 ,because it is constant volume process.

Process 2:

It is constant temperature process so PV=C

$P_2V_2=P_3V_3$

$P_3=\dfrac{P_2V_2}{V_3}$

$P_3=\dfrac{2.7\times 2.6}{1.5}$

$P_3=4.68$ bar

So work in constant temperature process

W= $P_2V_2\ ln\dfrac{V_3}{V_2}$

W= $270\times 2.6\ ln\dfrac{1.5}{2.6}$ (1 bar=100KPa)

W= -386.13 KJ

Negative sign means it is compression process.

Process 3:

It is a constant pressure.

So work W= $P_3(V_4-V_3)$

W=468(0.5-1.5) KJ

W= -468 KJ

Negative sign means it is compression process.

7 0

4 years ago

Vehicles arrive at a recreational park booth at a uniform deterministic rate of 5 veh/min. If uniform deterministic processing o

Natali5045456 [20]

Answer:

The right solution is "32 min".

Explanation:

The given values are:

Total delay,

= 3200 veh/min

Deterministic rate,

= 5 veh/min

For 1st arrival, time taken

= 20 minutes

Now,

To be clear the queue, the time take will be:

= $\frac{Total \ delay}{Deterministic \ rate\times (1st \ arrival \ time \ taken)}$

On substituting the values in the above formula, we get

= $\frac{3200}{5\times 20} \ min$

= $\frac{3200}{100}$

= $32 \ min$

8 0

3 years ago

Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa

Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air, R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes 2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0

4 years ago

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

trasher [3.6K]

Answer:

The publication of a parody for commercial gain does not fall within the protection afforded by Section 107, as it is used for commercial gain.

Explanation:

<h2>PLEASE MARK AS BRAINLIEST!!!!!</h2>

4 0

3 years ago

What is the composition of the two phases that form when a stream of 40% A, 39% B, and 21% C separates into two phases? Label th

irga5000 [103]

Answer:

vapor fraction = 0.4 and 0.08

Explanation:

At reasonably high temperatures, a mixture will exist in the form of a sub cooled liquid. Between these extremes, the mixture exists in a two phrase region where it is a vapor liquid equilibrium. From a vapor-liquid phase diagram, a mixture of 40% A, 39% B, and 21% C separates to give the vapor compositions of 0.4 and 0.08.

8 0

3 years ago

The title block generally contains ________.