All categories

3 years ago

Horizontal wind turbines have same design for offshore and on shore wind farms. a)-True b)- False

Engineering

1 answer:

Zigmanuir [339]3 years ago

8 0

Answer: False

Explanation: Horizontal axis wind turbines are usually used for generation of the electric power on the off-shore. The generation of horizontal-axis wind turbine works well when it is installed away from the shore because it supports large sized wind turbines so that they can generate high amount of electricity.They are usually not preferred for the on-shore wind farms because they can have small sized wind turbines only.Therefore the statement given is false.

You might be interested in

A) Consider an air standard otto cycle that has a heat addition of 2800 kJ/kg of air, a compression ratio of 8 and a pressure an

Angelina_Jolie [31]

Answer:

a) i) The maximum pressure is approximately 122.37 bar

ii) The thermal efficiency is approximately 56.47%

iii) The mean effective pressure is approximately 20.974 bar

b) (b) Four types of internal combustion engine includes;

1) The diesel engine

2) The Otto engine

3) The Brayton engine

4) The Wankel engine

Explanation:

The parameters of the Otto cycle are;

The heat added, $Q_{in}$ = 2,800 kJ/kg

The compression ratio, r = 8

The beginning compression pressure, P₁ = 1 bar

The beginning compression temperature, T₁ = 300 K

Cp = 1.005 kJ/kg·K

Cv = 0.718 kJ/kg·K

R = 287 kJ/kg·K

K = Cp/Cv = 1.005 kJ/kg·K/(0.718 kJ/kg·K) ≈ 1.4

T₂ = T₁×r^(k - 1)

∴ T₂ = 300 K×8^(1.4 - 1) ≈ 689.219 K

$\dfrac{P_1\cdot V_1}{T_1} = \dfrac{P_2\cdot V_2}{T_2}$

$P_2 = \dfrac{P_1\cdot V_1 \cdot T_2}{T_1 \cdot V_2} = \dfrac{V_1}{V_2} \cdot \dfrac{P_1 \cdot T_2}{T_1 } = r \cdot \dfrac{P_1 \cdot T_2}{T_1 }$

∴ P₂ = 8 × 1 bar × (689.219K)/300 K ≈ 18.379 bar

$Q_{in}$ = m·Cv·(T₃ - T₂)

∴ $Q_{in}$ = 2,800 ≈ 0.718 × (T₃ - 689.219)

T₃ = 2,800/0.718 + 689.219 = 4588.94 K

P₃ = P₂ × (T₃/T₂)

P₃ = 18.379 bar × 4588.94K/(689.219 K) = 122.37 bar

The maximum pressure = P₃ ≈ 122.37 bar

(ii) The thermal efficiency, $\eta_{Otto}$ , is given as follows;

$\eta_{Otto} = 1 - \dfrac{1}{r^{k - 1}}$

Therefore, we have;

$\eta_{Otto} = 1 - \dfrac{1}{8^{1.4 - 1}} \approx 0.5647$

The thermal efficiency, $\eta_{Otto}$ ≈ 0.5647

Therefore, the thermal efficiency ≈ 56.47%

(iii) The mean effective pressure, MEP is given as follows;

$MEP = \dfrac{\left(P_3 - P_1 \cdot r^k \right) \cdot \left(1 - \dfrac{1}{r^{k-1}} \right)}{(k -1)\cdot (r - 1)}$

Therefore, we get;

$MEP = \dfrac{\left(122.37 - 1 \times 8^{1.4} \right) \cdot \left(1 - \dfrac{1}{8^{1.4-1}} \right)}{(1.4 -1)\cdot (8 - 1)} \approx 20.974$

The mean effective pressure, MEP ≈ 20.974 bar

(b) Four types of internal combustion engine includes;

1) The diesel engine; Compression heating is the source of the ignition, with constant pressure combustion

2) The Otto engine which is the internal combustion engine found in cars that make use of gasoline as the source of fuel

The Otto engine cycle comprises of five steps; intake, compression, ignition, expansion and exhaust

3) The Brayton engine works on the principle of the steam turbine

4) The Wankel it follows the pattern of the Otto cycle but it does not have piston strokes

8 0

3 years ago

The minimum fresh air requirement of a residential building is specified to be 0.35 air changes per hour (ASHRAE, Standard 62, 1

Natalka [10]

We know that

$A=200m^2\\h=2.7m\\\upsilon= 5.5m/s\\\%_{air} = 35%$

So, the volume of the entire building is

$V=2.7*200 = 540m^3$

The flow capacity of the fan

$\dot{V} = \frac{0.35*540}{60}$

$\dot{V} = 3.15m^3/min$

As $1L=10^{-3}m^3,$

$\dot{V}=3150L/min$

For the other part we know

$\dot{V}=\frac{\pi d^2}{4}V$

The diameter is,

$d=\sqrt{\frac{4\dot{V}}{\pi V}}$

$d=\sqrt{\frac{4*3.15}{\pi* 5.5* 60}}$

<em>**Note 60 is for the minutes</em>

$d= 0.1101m$

<em />

4 0

3 years ago

1. Which of the following is the ideal way to apply pressure onto pedals?

Vikki [24]

I think D. By pressing gradually

8 0

3 years ago

Read 2 more answers

Describe and compare the characteristics of (a) proportional control, (b) proportional plus integral control, (c) proportional p

stira [4]

Answer:

The answer is below

Explanation:

1. Proportional Control is a form of control engineering in which an output is directly proportional to the error signal.

Characteristics of proportional control are:

* It is utilized when the deviation between the input and output is small

* It is also utilized when the deviation is not sudden.

* It reduces steady-state error

* It speeds up the response of the overdamped system

2. Proportional plus Integral Control is a form of control engineering in which a collective proportion and integral control of the output is equivalent to the combined proportion and integral of the error signal.

Characteristics of proportional plus integral control are:

* it can revert the controlled variable to the original set point

* It decreases steady-state error

* It quickens up the reaction of the overdamped system

3. Proportional plus integral plus derivative control is mostly applicable in operating the process elements such as temperature, pressure, speed, etc. It is recommended for industrial use.

Characteristics of Proportional plus integral plus derivative control are:

* It enhances the temporary reaction of the system.

* It also lessens steady-state error

* It accelerates the response of the overdamped system

4 0

3 years ago

Two technicians are discussing cylinder honing technician a says a good cross hatch helps to trap the oil and retain it in the c

aleksley [76]

er:

Explanation:Technician A says that primary vibration is created by slight differences in the inertia of the pistons between top dead center and bottom dead center. Technician B says that secondary vibration is a strong low-frequency vibration caused by the movement of the piston traveling up and down the cylinder. Who is correct? O A. Neither Technician A nor B OB. Technician B O C. Both Technicians A and B D. Technician A

8 0

3 years ago