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lys-0071 [83]
3 years ago
9

thermodynamics A nuclear power plant based on the Rankine cycle operates with a boiling-water reactor to develop net cycle power

of 3 MW. Steam exits the reactor core at 100 bar, 620°C and expands through the turbine to the condenser pressure of 1 bar. Saturated liquid exits the condenser and is pumped to the reactor pressure of 100 bar. Isentropic efficiencies of the turbine and pump are 87% and 78%, respectively. Cooling water enters the condenser at 15°C with a mass flow rate of 114.79 kg/s. Determine: (a) the percent thermal efficiency. (b) the temperature of the cooling water exiting the condenser, in °C.

Engineering
1 answer:
IrinaK [193]3 years ago
3 0

Answer:

(a) the percent thermal efficiency is 27.94%

(b) the temperature of the cooling water exiting the condenser is 31.118°C

Explanation:

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Explanation:

Sum of forces in the x direction:

∑Fx = ma

Rx − 250 N = 0

Rx = 250 N

Sum of forces in the y direction:

∑Fy = ma

Ry − 120 N − 300 N = 0

Ry = 420 N

Sum of forces in the z direction:

∑Fz = ma

Rz − 50 N = 0

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Sum of moments about the x axis:

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∑τy = Iα

My = 0 Nm

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2 years ago
The ice on the rear window of an automobile is defrosted by attaching a thin, transparent, film type heating element to its inne
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Answer:

A)Q = 1208.33 W/m²

B)K = 0.138 W/m.K

Explanation:

We are given;

inside air temperature;T_∞,i =25 °C = 25 + 273 = 298K

outside air temperature;T_∞,o = -10°C = - 10 + 273 = 263K

Inner surface temperature;T_s,i = 15 °C = 15 + 273 = 288K

Thickness, L = 4mm = 0.004m

convection heat transfer coefficient ; hi = 25 W/(m².K)

A) From an energy balance at the inner surface and the thermal circuit, the electric power required per unit window area is given as;

Q = [(T_s,i - T_∞,o)/((L/k) + (1/hi))] - [(T_∞,o - T_s,i)/(1/hi)]

Plugging in the relevant values with k for glass as 1.4 W/m.k, we have;

Q = [(288 - 263)/((0.004/1.4) + (1/25))] - [(263 - 288)/(1/25)]

Q = 583.33 + 625

Q = 1208.33 W/m²

B) The formula for thermal conductivity is;

K = (QL)/(AΔT)

Where;

K is the thermal conductivity in W/m.K

Q is the amount of heat transferred through the material

L is the distance between the two isothermal planes

A is the area of the surface in square meters

ΔT is the difference in temperature in Kelvin

ΔT = 298K - 263K = 35K

Now, since we have value of heat per unit area to be Q = 1208.33 W/m², let's rearrange the equation to reflect that; Thus ;

k = (Q/A) x (L/ΔT)

K = 1208.33 x (0.004/35)

K = 0.138 W/m.K

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