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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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Free ideas free points. You will be reported for answering "no" or I don't know
KengaRu [80]

Answer:

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-Zero fuel aircraft

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These are some things I've been working on for a few years lol, maybe you will have more luck

5 0
3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Assoli18 [71]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}  

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

8 0
3 years ago
The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,
babunello [35]

Answer:

Qx = 9.109.10^5 \times 10^{-6} m³/s  

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × 10^6 Pa

viscosity n = 100 Pas

angle = 18°

so  Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA   ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 \times 10^{-3} )²× 9  \times 10^{-3}  × sin18 × cos18

Qd = 94.305 × 10^{-6} m³/s

and

Qb = p × π × D × dc³ × sin²A ÷  12  × n × L    ............2

Qb = 11 × 10^{6} × π × 85 \times 10^{-3}  × ( 9  \times 10^{-3} )³ × sin²18 ÷  12  × 100 × 2

Qb = 85.2 × 10^{-6} m³/s

so here

volume flow rate Qx = Qd - Qb   ..............3

Qx =  94.305 × 10^{-6}  - 85.2 × 10^{-6}  

Qx = 9.109.10^5 \times 10^{-6} m³/s  

8 0
3 years ago
State the four advantages of levers
dezoksy [38]

Answer:

Here are 2 sense i cant find 4

Explanation:

Levers are used to multiply force, In other words, using a lever gives you greater force or power than the effort you put in.

In a lever, if the distance from the effort to the fulcrum is longer than the distance from the load to the fulcrum, this gives a greater mechanical advantage.

3 0
2 years ago
provides steady-state operating data for a solar power plant that operates on a Rankine cycle with Refrigerant 134a as its worki
Vaselesa [24]

Answer:

hello some parts of your question is missing attached below is the missing part ( the required fig and table )

answer : The solar collector surface area = 7133 m^2

Explanation:

Given data :

Rate of energy input to the collectors from solar radiation = 0.3 kW/m^2

percentage of solar power absorbed by refrigerant = 60%

Determine the solar collector surface area

The solar collector surface area = 7133 m^2

attached below is a detailed solution of the problem

8 0
3 years ago
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