Ask question

Ask question

All categories

3 years ago

13

Air is compressed adiabatically from p1 1 bar, T1 300 K to p2 15 bar, v2 0.1227 m3 /kg. The air is then cooled at constant volum

e to T3 300 K. Assuming ideal gas behavior, and ignoring kinetic and potential energy effects, calculate the work for the first process and the heat transfer for the second process, each in kJ per kg of air. Solve the problem each of two ways:

1 answer:

sashaice [31]3 years ago

7 0

Answer:

Work done for the adiabatic process = -247873.6 J/kg = - 247.9 KJ/kg

Heat transfer for the constant volume process = - 244.91 KJ/kg

Explanation:

For the first State,

P₁ = 1 bar = 10⁵ Pa

T₁ = 300 K

V₁ = ?

Second state

P₂ = 15 bar = 15 × 10⁵ Pa

T₂ = ?

V₂ = 0.1227 m³/kg

Third state

P₃ = ?

T₃ = 300 K

V₃ = ?

We require the workdone for step 1-2 (which is adiabatic)

And heat transferred for steps 2-3 (which is isochoric/constant volume)

Work done for an adiabatic process is given by

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

where γ = ratio of specific heats = 1.4 for air since air is mostly diatomic

K = PVʸ

Using state 2 to calculate for k

K = P₂V₂ʸ = (15 × 10⁵)(0.1227)¹•⁴ = 79519.5

We also need V₁

For an adiabatic process

P₁V₁ʸ = P₂V₂ʸ = K

P₁V₁ʸ = K

(10⁵) (V₁¹•⁴) = 79519.5

V₁ = 0.849 m³/kg

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

W = 79519.5 [(0.1227)⁻⁰•⁴ - (0.849)⁻⁰•⁴]/(1 - 1.4)

W = (79519.5 × 1.247)/(-0.4) = - 247873.6 J/kg = - 247.9 KJ/kg

To calculate the heat transferred for the constant volume process

Heat transferred = Cᵥ (ΔT)

where Cᵥ = specific heat capacity at constant volume for air = 0.718 KJ/kgK

ΔT = T₃ - T₂

We need to calculate for T₂

Assuming air is an ideal gas,

PV = mRT

T = PV/mR

At state 2,

V/m = 0.1227 m³/kg

P₂ = 15 bar = 15 × 10⁵ Pa

R = gas constant for air = 287.1 J/kgK

T₂ = 15 × 10⁵ × 0.1227/287.1 = 641.1 K

Q = 0.718 (300 - 641.1) = - 244.91 KJ/kg

You might be interested in

A completely reversible heat pump produces heat ata rate of 300 kW to warm a house maintained at 24°C. Theexterior air, which is

Triss [41]

Answer:

Change in entropy S = 0.061

Second law of thermodynamics is satisfied since there is an increase in entropy

Explanation:

Heat Q = 300 kW

T2 = 24°C = 297 K

T1 = 7°C = 280 K

Change in entropy =

S = Q(1/T1 - 1/T2)

= 300(1/280 - 1/297) = 0.061

There is a positive increase in entropy so the second law is satisfied.

6 0

3 years ago

Steep safety ramps are built beside mountain highway to enable vehichles with fedective brakes to stop safely. a truck enters a

Veronika [31]

Answer:

a. 6 seconds

b. 180 feet

Explanation:

Images attached to show working.

a. You have the position of the truck so you integrate twice. Use the formula and plug in the time t = 7 sec. Check out uniform acceleration. The time at which the truck's velocity is zero is when it stops.

b. Determine the initial speed. Plug in the time calculated in the previous step. From this we can observe that the truck comes to a stop before the end of the ramp.

7 0

3 years ago

Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu

Sonja [21]

Answer:

a

The rate of work developed is $\frac{\r W}{\r m}= 300kJ/kg$

b

The rate of entropy produced within the turbine is $\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

Explanation:

From the question we are told

The rate at which heat is transferred is $\frac{\r Q}{\r m } = - 30KJ/kg$

the negative sign because the heat is transferred from the turbine

The specific heat capacity of air is $c_p = 1.1KJ/kg \cdot K$

The inlet temperature is $T_1 = 970K$

The outlet temperature is $T_2 = 670K$

The pressure at the inlet of the turbine is $p_1 = 400 kPa$

The pressure at the exist of the turbine is $p_2 = 100kPa$

The temperature at outer surface is $T_s = 315K$

The individual gas constant of air R with a constant value $R = 0.287kJ/kg \cdot K$

The general equation for the turbine operating at steady state is \

$\r Q - \r W + \r m (h_1 - h_2) = 0$

h is the enthalpy of the turbine and it is mathematically represented as

$h = c_p T$

The above equation becomes

$\r Q - \r W + \r m c_p(T_1 - T_2) = 0$

$\frac{\r W}{\r m} = \frac{\r Q}{\r m} + c_p (T_1 -T_2)$

Where $\r Q$ is the heat transfer from the turbine

$\r W$ is the work output from the turbine

$\r m$ is the mass flow rate of air

$\frac{\r W}{\r m}$ is the rate of work developed

Substituting values

$\frac{\r W}{\r m} = (-30)+1.1(970-670)$

$\frac{\r W}{\r m}= 300kJ/kg$

The general balance equation for an entropy rate is represented mathematically as

$\frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma = 0$

=> $\frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)$

generally $(s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]$

substituting for $(s_1 -s_2)$

$\frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} + c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]$

Where $\frac{\sigma}{\r m}$ is the rate of entropy produced within the turbine

substituting values

$\frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]$

$\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

5 0

4 years ago

A clutch is chattering. Technician A says that uneven wear on the flywheel could be the cause. Technician B says that worn or br

svet-max [94.6K]

Answer:

Both Technician A and B are correct

Explanation:

Due to a contamination of the clutch disc friction surface, the clutch chatters

The contamination that causes clutch chattering includes

1) Worn or broken motor mounts

2) Bell housing bolts becoming loose

3) Clutch link damage

4) Warped flywheel, due to overheating, the flywheel can become warped such that the non uniform surface interfaces the clutch resulting in clutch chattering

5) Engine or transmission oil contaminating the disc.

4 0

3 years ago

Question 7.1: Two possible overhead valve combustion chambers are being considered – the first has two valves; the second has fo

AleksandrR [38]

Answer:

1) The adoption of the second design we can see that the total valve perimeter is increased by 60.8%

2) Increase in flow are : 29%

3) Additional benefits in using 4 valves per cylinder:

a)For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

Explanation:

1) Perimeter of the first possible overhead valve combustion chamber with two valves:

P₂ = πd = π × 23 = 72.26mm

Perimeter of the second possible overhead valve combustion chamber with four valves:

P₄ = π2d = π × 18.5 × 2 = 116.24 mm

If second design is adopted, percentage increase = ((P₄ - P₂)/P₂)×100

= ((116.24 - 72.26)/72.26)×100 = 0.6086 ×100 = 60.86%

Therefore, the total valve perimeter is shown to have increased by 60.8%

2) Formula for flow Area (A) = P × L = πkd²

Area of the first possible overhead valve combustion chamber with two valves: A₂ = πkd² = πk(23)² = 1662k mm²

Area of the first possible overhead valve combustion chamber with four valves: A₄ = πkd² = 2πk(18.5)² = 2150k mm²

The percentage increase in flow area: ((A₄ - A₂)/A₄)×100 = ((2150 - 1662)/2150)×100 = 29%

3) The additional benefits of using are:

a) For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

7 0

3 years ago