Isothermal process also means adiabatic internal reversible process. a)-True b)-False

Answer:

The power input, in kW is -86.396 kW

Explanation:

Given;

initial pressure, P₁ = 1.05 bar

final pressure, P₂ = 12 bar

initial temperature, T₁ = 300 K

final temperature, T₂ = 400 K

Heat transfer, Q = 6.5 kW

volumetric flow rate, V = 39 m³/min = 0.65 m³/s

mass of air, m = 28.97 kg/mol

gas constant, R = 8.314 kJ/mol.k

R' = R/m

R' = 8.314 /28.97 = 0.28699 kJ/kg.K

Step 1:

Determine the specific volume:

p₁v₁ = RT₁

Step 2:

determine the mass flow rate; m' = V / v₁

mass flow rate, m' = 0.65 / 0.81997

mass flow rate, m' = 0.7927 kg/s

Step 3:

using steam table, we determine enthalpy change;

h₁ at T₁ = 300.19 kJ /kg

h₂ at T₂ = 400.98 kJ/kg

Δh = h₂ - h₁

Δh = 400.98 - 300.19

Δh = 100.79 kJ/kg

step 4:

determine work input;

W = Q - mΔh

Where;

Q is heat transfer = - 6.5 kW, because heat is lost to surrounding

W = (-6.5) - (0.7927 x 100.79)

W = -6.5 -79.896

W = -86.396 kW

Therefore, the power input, in kW is -86.396 kW

Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

Hello!

To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

3.

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

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

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

s1=7.234KJ/kg

4.

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

h2=Enthalpy(Water;s=s1;P=P2)

h2=3116KJ/kg

T2=Temperature(Water;P=P2;s=s1)

T2=330.8°C

5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

h3=2750KJ/kg

T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

6.

Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation

W=m1(h1-h2)+m3(h2-h3)

W=5(3457-3116)+4(3116-2750)=3168Kw

By Riemann sums, the boundarywork done by a gas during an expansionprocess based on the information given by the statement is approximately 0.243 joules.

How to determine the boundary work done by a gas during an expansion process

A process is a consecution of states of a system. The boundary work (W), in kilojoules, is the work done by the system on surroundings and in a P-Vdiagram this kind of work is equal to the area below the curve, which can be approximated by Riemann sums:

    (1)

Where:

• p - Pressure, in kilopascals.
• V - Volume, in cubic meters.

Now we proceed to calculate the boundary work:

W = 0.5 · [(300 kPa + 290 kPa) · (1.1 × 10⁻³ m³ - 1 × 10⁻³ m³) + (270 kPa + 290 kPa) · (1.2 × 10⁻³ m³ - 1.1 × 10⁻³ m³) + (250 kPa + 270 kPa) · (1.4 × 10⁻³ m³ - 1.2 × 10⁻³ m³) + (220 kPa + 250 kPa) · (1.7 × 10⁻³ m³ - 1.4 × 10⁻³ m³) + (200 kPa + 220 kPa) · (2 × 10⁻³ m³ - 1.7 × 10⁻³ m³)]

W = 0.243 kJ

By Riemann sums, the boundarywork done by a gas during an expansionprocess based on the information given by the statement is approximately 0.243 joules.

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Answer:

attached below

Explanation:

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Answer:

Option D

Material has uniform properties throughout.

Explanation:

A homogeneous material is a material that exhibits uniform properties throughout and these properties cannot be separated. These materials can be metals, ceramics or alloys that exhibit similar properties throughout. The similar properties may include evaporation point, density and other properties.