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Ahat [919]
3 years ago
9

Can someone pls give me the answer to this?

Engineering
1 answer:
Zolol [24]3 years ago
7 0
During World War ll, the United States and the Soviet Union fought together as the allies against the Axis power. However, the relationship between the two nations was a tense one. Americans had long been wary of Soviet communism and concerned about Russian leader Joseph Stalin’s tyrannical rule of his own country
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A steam power plant operating on a simple ideal Rankine cycle maintains the boiler at 6000 kPa, the turbine inlet at 600 °C, and
ohaa [14]

Answer:

ηa=0.349

ηb=0.345

Explanation:

The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:

h_{3}=3658.8kJ/kg\\ s_{3}=7.1693kJ/kg

The quality at state 4 is determined from the condition  s_{4} =s_{3} and the entropies of the components at the condenser pressure taken from table:

 q_{4} =\frac{s_{4}-s_{liq50}  }{s_{evap,50} } \\=\frac{7.1693-1.0912}{6.5019}=0.935

The enthalpy at state 4 then is:  

h_{4} =h_{liq50} +q_{4} h_{evap,50}\\ (340.54+0.935*2304.7)kJ/kg\\=2495.43kJ/kg\\

Part A

In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:  

h_{1}=340.54kJ/kg\\ \alpha _{1}=0.00103m^3/kg

The enthalpy at state 2 is determined from an energy balance on the pump:

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

    =346.67 kJ/kg

The thermal efficiency is then determined from the heat input and output in the cycle:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.349

Part B  

In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from  and the saturated liquid values:  

h_{1}=293.7kJ/kg\\ \alpha _{1}=0.001023m^3/kg

The enthalpy at state 2 is then determined from an energy balance on the pump:  

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

    =299.79 kJ/kg  

The thermal efficiency in this case then is:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.345

8 0
3 years ago
Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−
sergeinik [125]

Answer:

Power output, P_{out} = 178.56 kW

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, T = 350^{\circ}C

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, \dot{m} = 0.1 kg.s^{- 1}

Diameter of exhaust pipe, d_{h} = 15 cm = 0.15 m

Pressure at exhaust, P' = 50 kPa

temperature, T' =  100^{\circ}C

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

\dot{m} = \frac{Av_{i}}{V_{1}}

0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}

0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}

v_{i} = 3.986 m/s

Now, eqn for compressible fluid:

\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}

Now,

\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}

\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}

\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}

v_{e} = 19.33 m/s

Now, the power output can be calculated from the energy balance eqn:

P_{out} = -\dot{m}W_{s}

P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}

P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW

4 0
3 years ago
Which of the following is not one of the systems required to ensure the safe and correct operation of an engine?
velikii [3]
Brake system

Explanation: the engine doesn’t need to be running to make the brake system work the brake system it’s independent
6 0
3 years ago
If a signal is transmitted at a power of 250 mWatts (mW) and the noise in the channel is 10 uWatts (uW), if the signal BW is 20M
Bess [88]

Answer:

C = 292 Mbps

Explanation:

Given:

- Signal Transmitted Power P = 250mW

- The noise in channel N = 10 uW

- The signal bandwidth W = 20 MHz

Find:

what is the maximum capacity of the channel?

Solution:

-The capacity of the channel is given by Shannon's Formula:

                            C = W*log_2 ( 1 + P/N)

- Plug the values in:

                            C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)

                            C = (20*10^6)*log_2 (25001)

                            C = (20*10^6)*14.6096

                           C = 292 Mbps

3 0
3 years ago
IM JI Suneou uo mm
Oksi-84 [34.3K]

Answer: g

Explanation:

5 0
3 years ago
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