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shusha [124]
3 years ago
8

When was solar power envold ​

Engineering
2 answers:
tatiyna3 years ago
6 0
In a year 1839 was envold
Lyrx [107]3 years ago
5 0

Answer:

1839

Explanation:

the first huge step in solar penal technology came when alxendare becquerel observed the photovolatic effect in 1839,which occurs when a material produces electric current when exposed to light.it was not untill 1888 that the first solar cell was actually built by aleksander stoletov.

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Water is the working fluid in an ideal Rankine cycle. The condenser pressure is 8 kPa, and saturated vapor enters the turbine at
sergeinik [125]

Explanation:

The obtained data from water properties tables are:

Point 1 (condenser exit) @ 8 KPa, saturated fluid

h_{f} = 173.358 \\h_{fg} = 2402.522

Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid

h_{2a} =  489.752\\h_{2b} =  313.2

Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam

h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876

Point 4 (Turbine exit) @ 8 KPa, mixed fluid

x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119

Calculate mass flow rates

Part a) @ 18 MPa

mass flow

\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a}  - h_{4a}) - (h_{2a}  - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26  - 2241.448938 ) - (489.752  - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}

Heat transfer rate through boiler

Q_{in}  = mass flow * (h_{3a} -  h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W

Heat transfer rate through condenser

Q_{out}  = mass flow * (h_{4a} -  h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W

Thermal Efficiency

n = \frac{W_{net}  }{Q_{in} } = \frac{100*10^3}{1542011.787}  \\\\n = 0.06485

Part b) @ 4 MPa

mass flow

\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b}  - h_{4b}) - (h_{2b}  - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14  - 2405.54119 ) - (313.12  - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}

Heat transfer rate through boiler

Q_{in}  = mass flow * (h_{3b} -  h_{2b})\\Q_{in} = (1125.65951)*(2634.14-313.12)\\\\Q_{in} = 2612678.236 W

Heat transfer rate through condenser

Q_{out}  = mass flow * (h_{4b} -  h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W

Thermal Efficiency

n = \frac{W_{net}  }{Q_{in} } = \frac{100*10^3}{1542011.787}  \\\\n = 0.038275

6 0
3 years ago
Hi. I would like to know why one side of an island can get more rain (more rain forms), while the other gets less.
adelina 88 [10]
The elevation might be higher causing more rain
6 0
4 years ago
Read 2 more answers
Which of the following is used as part of a four-wheel drive system? A. Front drive axles B. Transfer Case C. Front Drive Shaft
xxMikexx [17]

Answer:

D: All of the above.

Have a good day!

4 0
4 years ago
WHAT IS THIS PLSSSSSS HELP
alekssr [168]

Answer:

It looks like... A machine that reads electric pulse and surge... Not sure though.

Explanation:

8 0
3 years ago
Consider a steam turbine, with inflow at 500oC and 7.9 MPa. The machine has a total-to-static efficiency ofηts=0.91, and the pre
sergiy2304 [10]

Answer: \dot m_{in} = 23.942 \frac{kg}{s}, \dot H_{out} = 39632.62 kW

Explanation:

Since there is no information related to volume flow to and from turbine, let is assume that volume flow at inlet equals to \dot V = 1 \frac{m^{3}}{s}. Turbine is a steady-flow system modelled by using Principle of Mass Conservation and First Law of Thermodynamics:

Principle of Mass Conservation

\dot m_{in} - \dot m_{out} = 0

First Law of Thermodynamics

- \dot W_{out} + \eta\cdot (\dot m_{in} \dot h_{in} - \dot m_{out} \dot h_{out}) = 0

This 2 x 2 System can be reduced into one equation as follows:

-\dot W_{out} + \eta \cdot \dot m \cdot ( h_{in}- h_{out})=0

The water goes to the turbine as Superheated steam and goes out as saturated vapor or a liquid-vapor mix. Specific volume and specific enthalpy at inflow are required to determine specific enthalpy at outflow and mass flow rate, respectively. Property tables are a practical form to get information:

Inflow (Superheated Steam)

\nu_{in} = 0.041767 \frac{m^{3}}{kg} \\h_{in} = 3399.5 \frac{kJ}{kg}

The mass flow rate can be calculated by using this expression:

\dot m_{in} =\frac{\dot V_{in}}{\nu_{in}}

\dot m_{in} = 23.942 \frac{kg}{s}

Afterwards, the specific enthalpy at outflow is determined by isolating it from energy balance:

h_{out} =h_{in}-\frac{\dot W_{out}}{\eta \cdot \dot m}

h_{out} = 1655.36 \frac{kJ}{kg}

The enthalpy rate at outflow is:

\dot H_{out} = \dot m \cdot h_{out}

\dot H_{out} = 39632.62 kW

3 0
3 years ago
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