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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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Gray cast iron, with an ultimate tensile strength of 31 ksi and an ultimate compressive strength of 109 ksi, has the following s
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Using an appropriate failure theory, find the factor of safety in each case. State the name of the theory that you are using the theory is max stress theory.

<h3>Wat is the max stress theory?</h3>

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3 0
2 years ago
The flowrate through a rectangular channel is 20 cfs. The upstream width of the channel is 10 ft, and the depth of the water in
Liula [17]

To solve this problem we will use the Froude number that relates the Forces of Inertia with the Forces of Gravity. There will be jump in the downstream only if Froude Number (Fr) is greater than 1 at upstream. Our values are given as,

Q = 20cfs\\w= 10ft\\D= 1ft

Then the velocity would be:

V = \frac{Q}{wD}V = \frac{20}{10*1}V = 2ft/s

The number of Froude is given as,

Fr = \frac{V}{gD}^{1/2}

Where,

V = Velocity

g = Gravity

D = Diameter

Replacing we have that

Fr = \frac{2}{32.2}^{1/2}\\Fr = 0.352\\Fr

There will be no Jump, correct answer is B.

5 0
3 years ago
2. What is the charge, expressed in micro coulombs on two equally and similarly charges spheres placed in air with their centres
9966 [12]

Answer:

q = 0.1086 micro Coulombs

Explanation:

By Coulombs law, we have;

F =k \times  \dfrac{ q_1 \times q_2}{r^2}

Where;

F = The electric force = 120 mgm

q₁ and q₂ = Charge

r = The separating distance = 30 cm = 0.3 m

k = 8.9876×10⁹ kg·m³/(s²·C²)

Where, q₁ and q₂, we have;

F =k \times  \dfrac{ q^2}{r^2}

Whereby the force is the force of 120 milligram mass, we have;

0.00012 × 9.81 = 000011772 N

q = \sqrt{ \dfrac{ F\times r^2}{k}}

Substituting the values, we have;

q = \sqrt{ \dfrac{ 000011772 \times (0.3)^2}{8.9876 \times 10^9}} = 1.086 \times 10 ^{-7} \ Coulombs

q = 0.1086 micro Coulombs.

5 0
3 years ago
An engine operates on gasoline (LHV=44 MJ/kg) with a brake thermal efficiency of 37.9 % What is the brake specific fuel consumpt
scZoUnD [109]

Answer:

s =0.21\ kg/Kw.hr

Explanation:

Given that

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\eta =\dfrac{BP}{\dot{m_f}\times CV}

Where BP is the brake power

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\dfrac{BP}{\dot{m_f}}=16676

Brake specific fuel consumption (s)

s =\dfrac{\dot{m_f}}{BP}

s =\dfrac{3600\times \dot{m_f}}{BP}

s =\dfrac{3600}{16,676}\ kg/Kw.hr

s =0.21\ kg/Kw.hr

7 0
3 years ago
The technique of smoothing out joint compound on either side of a joint is known as which of the following
Black_prince [1.1K]

Answer:

a is the answer to your questions

3 0
3 years ago
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