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3 years ago

How do Solar Engineers Help Humans? (2 or more sentences please)

1 answer:

8 0

Solar engineers work in any number of areas of engineering products that help harness energy from the sun. They may research, design, and develop new products, or they may work in testing, production, or maintenance. They may collect and manage data to help design solar systems.

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Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i

Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

$\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W$

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

$h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W$

∴ Power input to the pump 20.7088 kW

6 0

3 years ago

*3–32. The rubber block is subjected to an elongation of 0.03 in. along the x axis, and its vertical faces are given a tilt so t

riadik2000 [5.3K]

Rubber block is not shown. I have attached an image of it.

Answer:

A) ε_x = 0.0075

B) ε_y = 0.00375

C) γ_xy = 0.0122 rad

Explanation:

We are given;

δ = 0.03 in

L = 4 in

ν_r = 0.5

θ = 89.3° = 89.3π/180 rad

Let's calculate ε_x in the direction of axis x

Thus, ε_x = δ/L = 0.03/4 = 0.0075

Let's calculate ε_y in the direction of axis y;

ε_y = v•ε_x = 0.5 x 0.0075 = 0.00375

Now, shear strain is angle between π/2 rad surfaces at that point.

Thus,

γ_xy = π/2 - θ = π/2 - 89.3π/180

γ_xy = π(0.003889) = 0.0122 rad

3 0

2 years ago

Applying the Entropy Balance: Closed Systems Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston–

Mrrafil [7]

Answer:

a) the amount of energy produced in kJ/K is 0.73145 kJ/K

b) the amount of energy produced in kJ/K is 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

Explanation:

Draw the T-s diagram.

$C_p$ = 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280

R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar

Δs = $m[c_p ln(\frac{T_2}{T_1}) - Rln(\frac{P_2}{P_1})]$

Substitute all parameters in the equation

Δs = $5[(0.939) ln(\frac{520}{280}) - (\frac{8.314}{44.01})ln(\frac{20}{2})]$

Δs = 5 kg × 0.14629 kJ/kg.K

= 0.73145 kJ/K

Δs = $m[\frac{s^0(T_2) - s^0(T_1)}{M} - Rln(\frac{P_2}{P_1})]$

Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K

T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K

R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar

Δs = $5[\frac{236.575 - 211.376}{44.01} - (\frac{8.314}{44.01})ln(\frac{20}{2})]$

= 5 kg (0.13795 kJ/kg.K)

= 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

7 0

3 years ago

What is capillary action?

BARSIC [14]

Capillary action occurs When the adhesion to the walls stronger than dirt cohesive forces between a liquid molecules. the head towards Capillery action will take water in a uniform circular is limited by surface tension and, of course, gravity.

4 0

2 years ago

What is the core domain for Accenture’s Multi-party Systems practice?

12345 [234]

Answer:

Explanation:

digital identity is the answer

6 0

3 years ago