Vital role of maritime english among seaferers

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

Greeley [361]

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.

6 0

3 years ago

18) Technician A says that adjustable wrenches should be used when

kykrilka [37]

Tryna boost my score for college stuff could you give me the brainiest and a thanks? Hope you find your answer your looking for!

4 0

3 years ago

a ten station assembly machine has ideal cycle time of 6 sec. the fraction defect rate at each station 0.005 and defect always j

MAXImum [283]

Answer:

$T_{P}=(2.6667)(10^{-3})h$

Explanation:

Let's write the equation of the production rate for the assembly machine :

$T_{P}=T_{C}+(n).(m).(p).(T_{D})$

Where $T_{P}$ is the production rate for the assembly machine.

Where $T_{C}$ is the ideal cycle time

Where n is the number of stations.

Where m is the number stations that get jam when the defect occurs.

Where p is the defect rate at each station.

And where $T_{D}$ is the average downtime per breakdown

We are looking for the hourly production rate ⇒

$1h=60min\\1min=60s$ ⇒

$1h=3600s$ ⇒

$6s=\frac{(6s)(1h)}{(3600s)}= \frac{1}{600}h$

$60min=1h$ ⇒

$1.2min=\frac{(1.2min)(1h)}{(60min)}=0.02h$

$T_{P}=\frac{1}{600}h+(10)(1.0)(0.005)(0.02h)=\frac{1}{375}h=(2.6667)(10^{-3})h$

m = 1.0 in the equation.

3 0

3 years ago

The device in Figure P6.27 is a sloshing bath used to wash vegetable produce. For the configuration shown, use the relative velo

Rufina [12.5K]

check photo (answer)

5 0

3 years ago

A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as

marysya [2.9K]

Answer:

1) 287760.4 Hp

2) 18410899.5 kPa

Explanation:

The parameters given are;

p₁ = 14.7 lbf/in² = 101325.9 Pa

v₁ = 0.0196 ft³ = 0.00055501 m³

T₁ = 80°F = 299.8167 K

k = 1.4

Assumptions;

1) Air standard conditions are appropriate

2) There are negligible potential and kinetic energy changes

3) The air behaves as an ideal gas and has constant specific heat capacities of temperature and pressure

1) Process 1 to 2

Isentropic compression

T₂/T₁ = (v₁/v₂)^(1.4 - 1) = 10^0.4

p₂/p₁ = (v₁/v₂)^(1.4)

p₂ = p₁×10^0.4 = 101325.9*10^0.4 = 254519.153 Pa

T₂ = 299.8167*10^0.4 = 753.106 K

p₃ = 1080 lbf/in² = 7,446,338 Pa

Stage 2 to 3 is a constant volume process

p₃/T₃ = p₂/T₂

7,446,338/T₃ = 254519.153/753.106

T₃ = 7,446,338/(254519.153/753.106) = 22033.24 K

T₃/T₄ = (v₁/v₂)^(1.4 - 1) = 10^0.4

T₄ = 22033.24/(10^0.4) = 8771.59 K

The heat supplied, Q₁ = cv(T₃ - T₂) = 0.718*(22033.24 -753.106) = 15279.14 kJ

The heat rejected = cv(T₄ - T₁) = 0.718*(8771.59 - 299.8167) = 6082.73 kJ

W(net) = The heat supplied - The heat rejected = (15279.14 - 6082.73) = 9196.41 kJ

The power = W(net) × RPM/2*1/60 = 9196.41 * 2800/2*1/60 = 214582.9 kW

The power by the engine = 214582.9 kW = 287760.4 Hp

2) The mean effective pressure, MEP = W(net)/(v₁ - v₂)

v₁ = 0.00055501 m³

v₁/v₂ = 10

v₂ = v₁/10 = 0.00055501/10 = 0.000055501

MEP = 9196.41/(0.00055501 - 0.000055501) = 18410899.5 kPa

4 0

3 years ago