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Maurinko [17]
3 years ago
14

Which phase of project development provides engineers with an opportunity to determine the gaps or problems of the project on pa

per?
design

testing

requirements gathering

implementation: develop and build
Engineering
1 answer:
inysia [295]3 years ago
7 0
Testing or Requirements gathering
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Which of the following explains the difference between rangeland management specialists and conservation biologists?
tekilochka [14]

Answer:

b

Explanation:

8 0
3 years ago
Read 2 more answers
Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra
gladu [14]

Answer:

a) T_{2}=837.2K

b) e=91.3 %

Explanation:

A) First, let's write the energy balance:

W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})  (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 \frac{kJ}{kgK} And its specific R constant is 0.287 \frac{kJ}{kgK}.

The only unknown from the energy balance is T_{2}, so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K

B) The isentropic efficiency (e) is defined as:

e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}

Where {h_{2s} is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because h_{2}-h_{1} can be obtained from the energy balance  \frac{W}{m}=h_{2}-h_{1}

h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}

An entropy change for an ideal gas with  constant Cp is given by:

s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})

Applying logarithm properties:

ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}

Then,

T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K

So, now it is possible to calculate h_{2s}-h_{1}:

h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}

Finally, the efficiency can be calculated:

e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3 %

4 0
4 years ago
PWM input and output signals are often converted to analog voltage signals using low-pass filters. Design and simulate the follo
Temka [501]

Answer:

Attached below

Explanation:

PWM signal source has 1 KHz base frequency

Analog filter : with time constant = 0.01 s

low pass transfer function = \frac{1}{0.01s + 1 }

PWM duty cycle is a constant block

Attached below is the design and simulation into Simulink at 25% , 50% and 100% respectively

5 0
3 years ago
A discrete-time LTI system H has input x[n] and output y[n] related by the linear constant coefficient difference equation y[n]
Molodets [167]

Answer:

See explaination and attachment

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Have the solution as an attachment.

4 0
3 years ago
Consider the flow field given by V ! =xy2^i− 1 3 y3^j+xyk ^. Determine (a) the number of dimensions of the flow, (b) if it is a
Basile [38]

Answer:

a) The flow has three dimensions (3 coordinates).

b) ∇V = 0 it is a incompressible flow.

c) ap = (16/3) i + (32/3) j + (16/3) k

Explanation:

Given

V = xy² i − (1/3) y³ j + xy k

a) The flow has three dimensions (3 coordinates).

b) ∇V = 0

then

∇V = ∂(xy²)/∂x + ∂(− (1/3) y³)/∂y + ∂(xy)/∂z

⇒ ∇V = y² - y² + 0 = 0 it is a incompressible flow.

c) ap = xy²*∂(V)/∂x − (1/3) y³*∂(V)/∂y + xy*∂(V)/∂z

⇒ ap = xy²*(y² i + y k) - (1/3) y³*(2xy i − y² j + x k) + xy*(0)

⇒ ap = (xy⁴ - (2/3) xy⁴) i + (1/3) y⁵ j + (xy³ - (1/3) xy³) k

⇒ ap = (1/3) xy⁴ i + (1/3) y⁵ j + (2/3) xy³ k

At point (1, 2, 3)

⇒ ap = (1/3) (1*2⁴) i + (1/3) (2)⁵ j + (2/3) (1*2³) k

⇒ ap = (16/3) i + (32/3) j + (16/3) k

3 0
4 years ago
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