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Anarel [89]
3 years ago
12

1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty per

cent of the flow is extracted at 1000 kPa to a feedwater heater and the remainder flows out at 200 kPa. Find the two exit temperatures and the turbine power output.
Engineering
1 answer:
KiRa [710]3 years ago
7 0

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

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Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2/s is being discharged by an 8-mm-diameter, 42-m-long horiz
sladkih [1.3K]

Answer:

The flow rate of oil through the pipe is 1.513E-7 m³/s.

Explanation:

Given

Density, ρ = 850 kg/m³

Kinematic viscosity, v = 0.00062 m²/s

Diameter, d = 8-mm = 0.008m

Length of horizontal pipe, L = 42-m

Height, h = 4-m.

We'll solve the flow rate of oil through the pipe by using Hagen-Poiseuille equation.

This is given as

∆P = (128μLQ)/πD⁴

Where ∆P = Rate of change of pressure

μ = Dynamic Viscosity

Q = Flow rate of oil through the pipe.

First, we need to determine the dynamic viscosity and the rate of change in pressure

Dynamic Viscosity, μ = Density (ρ) * Kinematic viscosity (v)

μ = 850 kg/m³ * 0.00062 m²/s

μ = 0.527kg/ms

Then, we calculate the rate of change of pressure.

Assuming that the velocity through the pipe is so small;

∆P = Pressure at the bottom of the tank

∆P = Density (ρ) * Acceleration of gravity (g) * Height (h)

Taking g = 9.8m/s²

∆P = 850kg/m³ x 9.8m/s² x 4m

∆P = 33320N/m²

Recall that Hagen-Poiseuille equation.

∆P = (128μLQ)/πD⁴ --- Make Q the subject of formula

Q = (πD⁴P)/(128μL)

By substituton;

Q = (π * 0.008⁴ * 33320)/(128 * 0.527 * 42)

Q = 0.00000015133693643099

Q = 1.513E-7 m³/s.

Hence, the flow rate of oil through the pipe is 1.513E-7 m³/s.

8 0
3 years ago
A pressure gage connected to a tank reads 50 psi at a location where the barometric reading is 29.1 inches Hg. Determine the abs
Effectus [21]

Answer:

Absolute pressure , P(abs)= 433.31 KPa

Explanation:

Given that

Gauge pressure P(gauge)=  50 psi

We know that barometer reads atmospheric pressure

Atmospheric pressure P(atm) = 29.1 inches of Hg

We know that

1 psi = 6.89 KPa

So 50 psi = 6.89 x 50 KPa

P(gauge)=  50 psi =344.72 KPa

We know that

1 inch = 0.0254 m

29.1 inches = 0.739 m

Atmospheric pressure P(atm) = 0.739 m of Hg

We know that density of Hg =13.6\times 10^3\ kg/m^3

P = ρ g h

P(atm) = 13.6 x 1000 x 9.81 x 0.739 Pa

P(atm) = 13.6  x 9.81 x 0.739 KPa

P(atm) =98.54 KPa

Now

Absolute pressure = Gauge pressure + Atmospheric pressure

P(abs)=P(gauge) + P(atm)

P(abs)= 344.72 KPa + 98.54 KPa

P(abs)= 433.31 KPa

3 0
3 years ago
A rigid, sealed tank initially contains 2000 kg of water at 30 °C and atmospheric pressure. Determine: a) the volume of the tank
Bad White [126]

Given:

mass of water, m = 2000 kg

temperature, T = 30^{\circ}C = 303 K

extacted mass of water = 100 kg

Atmospheric pressure, P = 101.325 kPa

Solution:

a) Using Ideal gas equation:

PV = m\bar{R}T                                        (1)

where,

V = volume

m = mass of water

P = atmospheric pressure

\bar{R} = \frac{R}{M}

R= Rydberg's constant = 8.314 KJ/K

M = molar mass of water = 18 g/ mol

Now, using eqn (1):

V = \frac{m\bar{R}T}{P}

V = \frac{2000\times \frac{8.314}{18}\times 303}{101.325}

V = 2762.44 m^{3}

Therefore, the volume of the tank is V = 2762.44 m^{3}

b) After extracting 100 kg of water, amount of water left, m' = m - 100

m' = 2000 - 100 = 1900 kg

The remaining water reaches thermal equilibrium with surrounding temperature at T' = 30^{\circ}C = 303 K

At equilibrium, volume remain same

So,

P'V = m'\bar{R}T'

P' = \frac{1900\times \frac{8.314}{18}\times 303}{2762.44}      

Therefore, the final pressure is P' = 96.258 kPa

4 0
3 years ago
7. Sockets internal designs come in what sizes?
MAXImum [283]

Answer:

These drive fittings come in four common sizes: 1⁄4 inch, 3⁄8 inch, 1⁄2 inch, and 3⁄4 inch (referred to as "drives", as in "3⁄8 drive").

3 0
3 years ago
Water vapor at 10 MPa, 600°C enters a turbine operating at steady state with a volumetric flow rate of 0.36 m3/s and exits at 0.
elena-s [515]

Answer:

Hook's law holds good up to. A elastic limit. B. plastic limit. C.yield point. D.Breaking point

8 0
3 years ago
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