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WARRIOR [948]
3 years ago
9

A __________ defines the area of land that will supply water to a stream.

Engineering
1 answer:
Vikentia [17]3 years ago
8 0

Answer:

drainage basin

Explanation:

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Design the necessary circuit using Logisim to implement the situation described above. Use Kmaps for simplification. Be VERY car
Phantasy [73]

Answer:

The circuit is shown in my attachments, please take a good look

3 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
A circuit contains a 40 ohm resistor and a 60 ohms resistor connected in parallel. If you test this circuit with a DMM you shoul
lorasvet [3.4K]
<h2>Answer:</h2>

24Ω

<h2>Explanation:</h2>

When resistors are connected in parallel, the reciprocal of their combined resistance, when read with a DMM (Digital Multimeter - for measuring various properties of a circuit or circuit element such as resistance...) is the sum of the reciprocals of their individual resistances.

For example if two resistors of resistances R₁ and R₂ are connected together in parallel, the reciprocal of their combined resistance Rₓ is given by;

\frac{1}{R_x} = \frac{1}{R_1} + \frac{1}{R_2}

Solving for Rₓ gives;

R_{x} = \frac{R_1 * R_2}{R_1 + R_2}          ------------------(i)

From the question;

Let

R₁ = resistance of first resistor = 40Ω

R₂ =  resistance of second resistor = 60Ω

Now,

To get their combined or total resistance, Rₓ, substitute these values into equation (i) as follows;  

R_{x} = \frac{40 * 60}{40 + 60}

R_{x} = \frac{2400}{100}

R_{x} = 24 Ω

Therefore, the total resistance is 24Ω

4 0
3 years ago
Schrodinger equation is a ........... Equation
Tanzania [10]

Answer:

linear partial differential

Explanation:

The Schrödinger equation is a linear partial differential equation that describes the wave function or state function of a quantum-mechanical system. It is a key result in quantum mechanics, and its discovery was a significant landmark in the development of the subject.

5 0
3 years ago
Relevance of engineering law and ethic in contemporary society​
baherus [9]

Answer:

Engineering law and ethics are the moral codes associated with the field in which engineers are meant to adhere strictly to.

These ethics help in the safety of the engineers as a result of wearing protective clothing being a part of the ethics . Accuracy should also be taken important in research and daily activities. Corrupt practices and activities which poses risk to the environment is also strictly prohibited.

7 0
4 years ago
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