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lina2011 [118]
3 years ago
13

Water is the working fluid in a Rankine cycle. Superheated vapor enters the turbine at 8 MPa, 560°C and the turbine exit pressur

e is 8 kPa. Saturated liquid enters the pump at 8 kPa. The heat transfer rate to the working fluid in the steam generator is 26 MW. The isentropic turbine efficiency is 88%, and the isentropic pump efficiency is 82%. Cooling water enters the condenser at 18°C and exits at 36°C with no significant change in pressure. Step 1 Determine the net power developed, in kW. Determine the thermal efficiency for the cycle. Type your answer here

Engineering
1 answer:
hoa [83]3 years ago
5 0

Answer:

1. The net power developed=9370.773KW

2. Thermal Efficiency= 0.058

Explanation

Check attachment

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LUNES MARTES MIÉRCOLES JUEVES VIERNES SÁBADO DOMINGO
scZoUnD [109]

Answer:

si

Explanation:

8 0
3 years ago
A motorist enters a freeway at 25 mi/h and accelerates uniformly to 65 mi/h. From the odometer in the car, the motorist knows th
Helga [31]

Answer:

a) 2.2 m/s² b) 8 s

Explanation:

a) Assuming that the acceleration is constant, we can use any of the kinematic equations to solve the question.

As we don´t know the time needed to accelerate, we can use the following equation:

vf2 – vo2 = 2*a*∆x

At first, we can convert the values of vf, vo and ∆x, to SI units, as follows:

vf = 65 mi/h* (1,605 m / 1mi) * (1h/3,600 sec) = 29 m/s

vo = 25 mi/h *(1,605 m / 1mi) * (1h/3,600 sec) = 11.2 m/s

∆x = 0.1 mi*(1,605 m / 1mi) = 160.5 m

Replacing these values in (1), and solving for a, we have:

a = (29 m/s – 11.2 m/s) / 321 m = 2.2 m/s2

b) In order to obtain the time needed to reach to 65 mi/h, we can rearrange the equation for the definition of acceleration, as follows:

vf = vo + at  

Replacing by the values already known for vo, vf and a, and solving for t, we get:

t = vf-vo /a = (29 m/s – 11.2 m/s) / 2.2 m/s = 8 sec

5 0
3 years ago
Six forces act on a beam that forms part of a building's
IrinaVladis [17]

Answer:

<h2> FA = 13 kN </h2><h2>FG = 15.3 kN</h2>

Explanation:

write each force in terms of magnitude and directions  

Fx = F sin Ф

Fy = F cos Ф

where Ф is to be measured from x axis.

∑F at y = o

FAy + FBy + FCy + FDy + FEy + FGy = 0

∑F at x = o

FAx + FBx + FCx + FDx + FEx + FGx = 0

Let  

FA = FA sin (110)   +   FA cos (110)

FB = 20 sin (270)  +  20 cos (270)

FC = 16 sin (140)    +  16 cos (140)

FD = 9 sin (40)       +  9 cos (40)

FE = 20 sin (270)    +  20 cos (270)

FG = FG sin (50)     +  FG cos (50)

add x and y forces:

FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

FA sin (110)  + 0  + 16 sin (140)  + 9 sin (40)  + 0   + FG sin (50) = 0

FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0

FA sin (110)  + 0  + 10.285  + 5.785  + 0   + FG sin (50) = 0

FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0

FA sin (110)  + 16.070 + FG sin (50) = 0        

FA cos (110) - 45.363 + FG cos (50) = 0

solving for FA, and FG

FA = 13 kN

FG = 15.3 kN

7 0
3 years ago
A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm (0.8 in.) on an edge is pulled intension with a load o
grigory [225]

Answer:

The elastic modulus of the steel is 139062.5 N/in^2

Explanation:

Elastic modulus = stress ÷ strain

Load = 89,000 N

Area of square cross section of the steel bar = (0.8 in)^2 = 0.64 in^2

Stress = load/area = 89,000/0.64 = 139.0625 N/in^2

Length of steel bar = 4 in

Extension = 4×10^-3 in

Strain = extension/length = 4×10^-3/4 = 1×10^-3

Elastic modulus = 139.0625 N/in^2 ÷ 1×10^-3 = 139062.5 N/in^2

7 0
3 years ago
Find the current Lx in the figure
AleksandrR [38]

Explanation:

\frac{1}{8}  +  \frac{1}{2}   \\ 1.6 + 1.4 = 3 \\  \frac{1}{3}  +  \frac{1}{9}   \\ 2.25 + 2 = 4.25 \: ohm

R total = 4.25 ohm

I total = Vt/Rt

I total= 17/4.25= 4 A

Ix= 600 mA

\frac{9}{9 + 3}  \times 4 = 3\\   \frac{2}{2 + 8} \times 3 = 0.6a \\  = 0.6 \: milli \: amper

6 0
3 years ago
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