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abruzzese [7]
2 years ago
11

A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The

system is heated until the volume is 0.2 m3 and the pressure is 180 kPa. What is the work done by the air? External pressure is 100 kPa.
Engineering
1 answer:
AleksAgata [21]2 years ago
4 0

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = \frac{60+180}{2}  = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

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Water vapor at 100 psi, 500 F and a velocity of 100 ft./sec enters a nozzle operating at steady sate and expands adiabatically t
almond37 [142]

Answer:

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b) the amount of entropy produced is 0.006 Btu/Ibm.R

Explanation:

Given:

P1 = 100 psi

V1 = 100 ft./sec

T1 = 500f

P2 = 40 psi

n = 95% = 0.95

a) for nozzle:

Let's apply steady gas equation.

h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2}

h1 and h2 = inlet and exit enthalpy respectively.

At T1 = 500f and P1 = 100 psi,

h1 = 1278.8 Btu/Ibm

s1 = 1.708 Btu/Ibm.R

At P2 = 40psi and s1 = 1.708 Btu/Ibm.R

1193.5 Btu/Ibm

Let's find the actual h2 using the formula :

n = \frac{h_1 - h_2*}{h_1 - h_2}

n = \frac{1278.8 - h_2*}{1278.8 - 1193.5}

solving for h2, we have

h_2 = 1197.77 Btu/Ibm

Take Btu/Ibm = 25037 ft²/s²

Using the first equation, exit velocity of the steam =

(1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}

Solving for V2, we have

V2 = 2016.8 ft/s

b) The amount of entropy produced in BTU/ lbm R will be calculated using :

Δs = s2 - s1

Where s1 = 1.708 Btu/Ibm.R

At h2 = 1197.77 Btu/Ibm and P2 =40 psi,

S2 = 1.714 Btu/Ibm.R

Therefore, amount of entropy produced will be:

Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R

= 0.006 Btu/Ibm.R

3 0
3 years ago
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