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natima [27]
3 years ago
10

A 600-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 40 percent. Determine the rate of hea

t transfer to the river water. Will the actual heat transfer rate be higher or lower than this value
Engineering
2 answers:
Serhud [2]3 years ago
5 0

Answer:

<em>A The heat transfer to the river would be 900 MW</em>

<em>B. The actual heat transfer would be lower than the theoretical heat transfer.</em>

Explanation:

Part A

For a steam power plant, the process is continuous and the law of thermodynamic is conserved. The steam power plant consists of the cold and hot reservoir and the rate of heat transfer in the steam power plant can be gotten with the expression below.

W = Q_{H} - Q_{c} ......................1

where W is the work output of the steam power plant

Q_{H}  is the heat transfer at the hot reservoir

Q_{c} is the heat transfer at the cold reservoir( heat transfer to the nearby river)

The efficiency of the steam power plant would be used to obtain the heat transfer at the hot reservoir (Q_{H}). The efficiency can be obtained with equation 2;

e = \frac{W}{Q_{H} }

e is the thermal efficiency of the steam power plant = 40 % = 0.4

W is the work output of the steam power plant = 600 MW

Q_{H}  is the heat transfer at the hot reservoir

Rearranging equation 2 to make  Q_{H}  the subject formula we have;

Q_{H}  = W/e

substituting the values we have;

Q_{H} = 600 MW/ 0.4

Q_{H} = 1500 MW

The heat transfer at the hot reservoir is 1500 MW and putting it into equation 1 to find the heat transfer to the river;

Q_{c} = Q_{H} - W

Q_{c} = 1500 MW - 600 MW

Q_{c} = 900 MW

Therefore the heat transfer to the river would be 900 MW

Part B

The actual heat transfer would be lower than the theoretical heat transfer. Losses dues to friction, head losses and heat loss by evaporation to the surroundings account for the lower value.

umka21 [38]3 years ago
3 0

Answer:

As thermal efficiency is of 40% hence rate of heat transfer is 40% of 600 MW

Rate of heat transfer=(40/100)*600 MW

Rate of heat transfer=240 MW and it is lower than 600 MW

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Answer:

(iv) second law of thermodynamics

Explanation:

The Clausius  inequality expresses the second law of thermodynamics it applies to the real engine cycle.It is defined as the cycle integral of change in entropy of a reversible system is zero. It is nothing but mathematical form of second law of thermodynamics . It also states that for irreversible process the cyclic integral of change in entropy is less than zero

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3 years ago
In RSA Digital Signature, Suppose Bob wants to send a signed message (x = 4) to Alice. The first steps are exactly t eps are exa
Luda [366]

Answer:

what r u on

Explanation:

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3 years ago
Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans
aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, T_{i} = 20^{\circ}C = 273 + 20 = 293 K

Temperature at the outlet, T_{o} = 200{\circ}C = 273 + 200 = 473 K

Pressure at inlet, P_{i} = 80 kPa = 80\times 10^{3} Pa

Pressure at outlet, P_{o} = 800 kPa = 800\times 10^{3} Pa

Speed at the outlet, v_{o} = 20 m/s

Diameter of the tube, D = 10 cm = 10\times 10^{- 2} m = 0.1 m

Input power, P_{i} = 400 kW = 400\times 10^{3} W

Now,

To calculate the heat transfer, Q, we make use of the steady flow eqn:

h_{i} + \frac{v_{i}^{2}}{2} + gH  + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}

where

h_{i} = specific enthalpy at inlet

h_{o} = specific enthalpy at outlet

v_{i} = air speed at inlet

p_{s} = specific power input

H and H' = Elevation of inlet and outlet

Now, if

v_{i} = 0 and H = H'

Then the above eqn reduces to:

h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}

Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}                (1)

Also,

p_{s} = \frac{P_{i}}{ mass, m}

Area of cross-section, A = \frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}

Specific Volume at outlet, V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s

From the eqn:

P_{o}V_{o} = mRT_{o}

m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s

Now,

p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg

Also,

\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg

Now, using these values in eqn (1):

Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW

Now, rate of heat transfer, q:

q = mQ = 0.925\times 813.33 = 752.33 kW

4 0
3 years ago
A particular motor rotates at 3000 revolutions per minute. What is its speed in rad/sec, and how many seconds does it takes to m
Leno4ka [110]

Answer:

ω=314.15 rad/s.

0.02 s.

Explanation:

Given that

Motor speed ,N= 3000 revolutions per minute

N= 3000 RPM

The speed of the motor in rad/s given as

\omega=\dfrac{2\pi N}{60}\ rad/s

Now by putting the values in the above equation

\omega=\dfrac{2\pi \times 3000}{60}\ rad/s

ω=314.15 rad/s

Therefore the speed in rad/s will be 314.15 rad/s.

The speed in rev/sec given as

\omega=\dfrac{ 3000}{60}\ rad/s

ω= 50 rev/s

It take 1 sec to cover 50 revolutions

That is why to cover 1 revolution it take

\dfrac{1}{50}=0.02\ s

4 0
3 years ago
A missile flying at high speed has a stagnation pressure and temperature of 5 atm and 598.59 °R respectively. What is the densit
alexdok [17]

Answer:

5.31\frac{kg}{m^3}

Explanation:

Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to \frac{m}{M}, where m is the mass and M the molar mass of the gas, and the density is \frac{m}{V}.

For air M=28.66*10^{-3}\frac{kg}{mol} and \frac{5}{9}R=K

So, 598.59 R*\frac{5}{9}=332.55K

pV=nRT\\pV=\frac{m}{M}RT\\\frac{m}{V}=\frac{pM}{RT}\\\rho=\frac{pM}{RT}\\\rho=\frac{(5atm)28.66*10^{-3}\frac{kg}{mol}}{(8.20*10^{-5}\frac{m^3*atm}{K*mol})332.55K}=5.31\frac{kg}{m^3}

7 0
3 years ago
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