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3 years ago

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