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Anni [7]
3 years ago
10

Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the cycle is modified with regeneration that

involves one open feedwater heater (select the correct statement per unit mass of steam flowing through the boiler),
(a) the turbine work output will decrease.

(b) the amount of heat rejected will increase.

(c) the cycle thermal efficiency will decrease.

(d) the quality of steam at turbine exit will decrease.

(e) the amount of heat input will increase.
Engineering
1 answer:
Elza [17]3 years ago
3 0

Answer:

The correct answer is OPTION A (the turbine work output will decrease.)

Explanation:

In open feedwater heaters, there is a mixing of cold feedwater and steam which raise the temperature of feed water causing the fluid average temperature to rise and thereby increasing the thermodynamic efficiency of the cycle.

In a simple ideal Rankine cycle, heating of the water is converted into work done by the cycle in potential and kinetic energy forms. If there is a fixed boiler flow rate, and the simple ideal Rankine cycle is modified with regeneration, turbine work output,  and heat supplied, will decrease.

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How are the particles moving and and arranged in a gas?
alexandr402 [8]

Answer:

The particles in gas do not have any particular arrangement and there are very, very weak forces between them. So, the particles in a gas can easily move around and fill the shape of the container they are in, meaning they have no fixed shape.

7 0
3 years ago
An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75
Inessa [10]

Answer:

(a). The value of temperature at the end of heat addition process            T_{3} = 2042.56 K

(b). The value of pressure at the end of heat addition process                    P_{3} = 1555.46 k pa

(c). The thermal efficiency of an Otto cycle   E_{otto} = 0.4478

(d). The value of mean effective pressure of the cycle P_{m} = 1506.41 \frac{k pa}{kg}

Explanation:

Compression ratio r_{p} = 8

Initial pressure P_{1} = 95 k pa

Initial temperature T_{1} = 278 °c = 551 K

Final pressure P_{2} = 8 × P_{1} = 8 × 95 = 760 k pa

Final temperature T_{2} = T_{1} × r_{p} ^{\frac{\gamma - 1}{\gamma} }

Final temperature T_{2} = 551 × 8 ^{\frac{1.4 - 1}{1.4} }

Final temperature T_{2} = 998 K

Heat transferred at constant volume Q = 750 \frac{KJ}{kg}

(a). We know that Heat transferred at constant volume Q_{S} = m C_{v} ( T_{3} - T_{2}  )

⇒ 1 × 0.718 × ( T_{3} - 998 ) = 750

⇒ T_{3} = 2042.56 K

This is the value of temperature at the end of heat addition process.

Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.

⇒ P ∝ T

⇒ \frac{P_{3} }{P_{2} } = \frac{T_{3} }{T_{2} }

⇒ P_{3} = \frac{2042.56}{998} × 760

⇒ P_{3} = 1555.46 k pa

This is the value of pressure at the end of heat addition process.

(b). Heat rejected from the cycle Q_{R} = m C_{v} ( T_{4} - T_{1}  )

For the compression and expansion process,

⇒ \frac{T_{3} }{T_{2} } = \frac{T_{4} }{T_{1} }

⇒ \frac{2042.56}{998} = \frac{T_{4} }{551}

⇒ T_{4} = 1127.7 K

Heat rejected Q_{R} = 1 × 0.718 × ( 1127.7 - 551)

⇒ Q_{R} = 414.07 \frac{KJ}{kg}

Net heat interaction from the cycle Q_{net} = Q_{S} - Q_{R}

Put the values of Q_{S} & Q_{R}  we get,

⇒ Q_{net} = 750 - 414.07

⇒ Q_{net} = 335.93 \frac{KJ}{kg}

We know that for a cyclic process net heat interaction is equal to net work transfer.

⇒ Q_{net} = W_{net}

⇒ W_{net} = 335.93 \frac{KJ}{kg}

This is the net work output from the cycle.

(c). Thermal efficiency of an Otto cycle is given by

E_{otto} = 1- \frac{T_{1} }{T_{2} }

Put the values of T_{1} & T_{2} in the above formula we get,

E_{otto} = 1- \frac{551 }{998 }

⇒ E_{otto} = 0.4478

This is the thermal efficiency of an Otto cycle.

(d). Mean effective pressure P_{m} :-

We know that mean effective pressure of  the Otto cycle is  given by

P_{m} = \frac{W_{net} }{V_{s} } ---------- (1)

where V_{s} is the swept volume.

V_{s} = V_{1}  - V_{2} ---------- ( 2 )

From ideal gas equation P_{1} V_{1} = m × R × T_{1}

Put all the values in above formula we get,

⇒ 95 × V_{1} = 1 × 0.287 × 551

⇒ V_{1} = 0.6 m^{3}

From the same ideal gas equation

P_{2} V_{2} = m × R × T_{2}

⇒ 760 × V_{2} = 1 × 0.287 × 998

⇒ V_{2} = 0.377 m^{3}

Thus swept volume V_{s} = 0.6 - 0.377

⇒ V_{s} = 0.223 m^{3}

Thus from equation 1 the mean effective pressure

⇒ P_{m} = \frac{335.93}{0.223}

⇒ P_{m} = 1506.41 \frac{k pa}{kg}

This is the value of mean effective pressure of the cycle.

4 0
3 years ago
Two concentric helical compression springs made of steel and having the same length when loaded and when unloaded are used to su
Andrews [41]

Answer:

see explaination for all the answers and full working.

Explanation:

deflection=8P*DN/Gd^4

G(for steel)=70Gpa=70*10^9N/m^2=70KN/mm^2

for outer spring,

deflection=8*3*50^3*5/(70*9^4)=32.66mm

for inner spring

deflection=8*3*30^3*10/(70*5^4)=148.11mm

max stress=k*8*P*C/(3.14*d^2)

for outer spring

c=50/9=5.55

k=(4c-1/4c-4)+.615/c=1.2768

max stress=1.2768*8*3*5.55/(3.14*9^2=.66KN.mm^2

for inner spring

c=6

k=1.2525

max stress=2.29KN/mm^2

6 0
4 years ago
Read 2 more answers
Using the tables for water determine the specified property data at the indicated state. For H2O at T = 140 °C and v = 0.2 m3/kg
Dennis_Churaev [7]

Answer:

h = 1429.74\,\frac{kJ}{kg}

Explanation:

The determination of any further properties requires the knowledge of two independent properties. (Temperature and specific volume in this case). The specific volumes for saturated liquid and vapor at 140 °C are, respectively:

\nu_{f} = 0.001080\,\frac{m^{3}}{kg}

\nu_{g} = 0.50850\,\frac{m^{3}}{kg}

Since \nu_{f} < \nu < \nu_{g}, it is a liquid-vapor mixture. The quality of the mixture is:

x = \frac{\nu-\nu_{f}}{\nu_{g}-\nu_{f}}

x = \frac{0.2\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }{0.50850\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }

x = 0.392

The specific enthalpies for saturated liquid and vapor at 140 °C are, respectively:

h_{f} = 589.16\,\frac{kJ}{kg}

h_{g} = 2733.5\,\frac{kJ}{kg}

The specific enthalpy is:

h = h_{f}+x\cdot (h_{g}-h_{f})

h = 589.16\,\frac{kJ}{kg}+0.392\cdot \left( 2733.5\,\frac{kJ}{kg} - 589.16\,\frac{kJ}{kg} \right)

h = 1429.74\,\frac{kJ}{kg}

6 0
4 years ago
Read 2 more answers
This is just so I can flip the picture
Pepsi [2]

Answer:

ok

Explanation:

thx for points

7 0
4 years ago
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