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3 years ago

15

A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage

at 12 MPa, 5608C and expands to 1 MPa, where some of the steam is extracted and diverted to the open feedwater heater operating at 1 MPa. The remaining steam expands through the second turbine stage to the condenser pressure of 6 kPa. Saturated liquid exits the open feedwater heater at 1 MPa. The net power output for the cycle is 330 MW. For isentropic processes in the turbines and pumps.
Determine:
a. the cycle thermal efficiency.
b. the mass flow rate into the first turbine stage, in kg/s.
c. the rate of entropy production in the open feedwater heater, in kW/K.

1 answer:

faltersainse [42]3 years ago

4 0

Answer:

a) 0.489

b) 54.42 kg/s

c) 247.36 kW/s

Explanation:

Note that all the initial enthalpy and entropy values were gotten from the tables.

See the attachment for calculations

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Answer:

0.14% probability of observing more than 4 errors in the carpet

Explanation:

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In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8.

This means that $\mu = 0.8$

What is the probability of observing more than 4 errors in the carpet

Either we observe 4 or less errors, or we observe more than 4. The sum of the probabilities of these outcomes is 1. So

$P(X \leq 4) + P(X > 4) = 1$

We want P(X > 4). Then

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

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$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

$P(X = 1) = \frac{e^{-0.8}*(0.8)^{1}}{(1)!} = 0.3595$

$P(X = 2) = \frac{e^{-0.8}*(0.8)^{2}}{(2)!} = 0.1438$

$P(X = 3) = \frac{e^{-0.8}*(0.8)^{3}}{(3)!} = 0.0383$

$P(X = 4) = \frac{e^{-0.8}*(0.8)^{4}}{(4)!} = 0.0077$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.0077 = 0.9986$

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Answer:

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Flow and Pressure Drop of Gases in Packed Bed. Air at 394.3 K flows through a packed bed of cylinders having a diameter of 0.012

devlian [24]

The pressure drop of air in the bed is 14.5 kPa.

<u>Explanation:</u>

To calculate Re:

$R e=\frac{1}{1-\varepsilon} \frac{\rho q d_{p}}{\mu}$

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ρ = $\frac{2.2}{2.83 \times 10^{-3} \times 394.3}$

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ρ = $\frac{P}{RT}$

R = $\frac{R_{c} }{M}$ = 8.2 × $10^{-5}$ / 28.97× $10^{-3}$

R = 2.83 × $10^{-3}$ $m^{3}$ atm / K Kg

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$q=\frac{2.45}{1.97}$

q = 1.24 m/s

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Re = 2278

The Ergun equation is used when Re > 10,

$\frac{\Delta P}{L}=\frac{180 \mu}{d_{p}^{2}} \frac{(1-\varepsilon)^{2}}{\varepsilon^{3}} q+\frac{7}{4} \frac{\rho}{d_{p}} \frac{(1-\varepsilon)}{\varepsilon^{3}} q^{2}$

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= 4089.748 Pa/m

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