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Eduardwww [97]
3 years ago
15

A community plans to build a facility to convert solar radiation to electrical power. The community requires 2.80 MW of power, a

nd the system to be installed has an efficiency of 30.0% (that is, 30.0% of the solar energy incident on the surface is converted to useful energy that can power the community).
Assuming sunlight has a constant intensity of 1 180 W/m2, what must be the effective area of a perfectly absorbing surface used in such an installation?
Engineering
1 answer:
Greeley [361]3 years ago
6 0

Given Information:  

Output power required = Pout = 2.80 MW

Efficiency = η = 30%

Intensity = I = 1180 W/m²

Required Information:  

Effective area = A = ?

Answer:  

Effective area = A = 7.907x10³ m²

Step-by-step explanation:  

A community plans to build a facility to convert solar power into electrical power and this facility has an efficiency of 30%

As we know efficiency is given by

η = Pout/Pin

Where Pout is the output power and Pin is the input power.

Pin = Pout/η

Pin = 2.80x10⁶/0.30

Pin = 9.33x10⁶ W

The effective area of a perfectly absorbing surface used in such an installation can be found using

A = Pin/I

Where I is the in Intensity of the sunlight in W/m²

A = 9.33x10⁶/1180

A = 7.907x10³ m²

Therefore, the effective area of the absorbing surface would be 7.907x10³ m².

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An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
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Answer:

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2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

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a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

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m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

T_{2} =T_{sat @ 0.2MPa} = -10.09^{o}  C

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

w_{b, out}  = workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

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3 years ago
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