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madreJ [45]
3 years ago
8

What is the focus of 7th grade civics

Engineering
2 answers:
Jobisdone [24]3 years ago
7 0
C. seems like the best answer. i may be wrong so don’t quote me on that
pishuonlain [190]3 years ago
7 0
I’m going to say C is the possible correct answer
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Which word from the passage best explains what the web in the passage symbolizes
frez [133]

Answer:

I cant see the passage

Explanation:

I cant see the passage

6 0
3 years ago
The distillation column in Figure 3 is set up for so-called boil-up (V) control. It has
11111nata11111 [884]

A distillation column is an essential item used in the distillation of liquid mixtures to separate the mixture into its component parts, or fractions, based on the differences in volatilities. Fractionating columns are used in small scale laboratory distillations as well as large scale industrial distillations.

3 0
3 years ago
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
I need help with my autos
oksano4ka [1.4K]

Answer:

what is wrong with it and what is the question

Explanation:

6 0
3 years ago
An uncharged capacitor is connected to a resistor and a battery. Choose what happens to current, potential difference and charge
Ilya [14]

Answer:

  • The charge on the plates will increase with time
  • The potential difference across the capacitor starts with zero and then increases gradually to a maximum value
  • The current through the circuit starts high and then drops exponentially

Explanation:

<u>Case : An uncharged capacitor is connected to a resistor and a battery in a closed circuit.</u>

  • The charge on the plates will increase with time

applying this equation : Q = Q_{0}  [ 1 - e^{\frac{-t}{RC} } ]  as the value of (t) increases the value of Q increases i.e. charge on the plates

  • The potential difference across the capacitor starts with zero and then increases gradually to a maximum value

applying this equation : V = V_{0}  [ 1 - e^{\frac{-t}{RC} } ]

  • The current through the circuit starts high and then drops exponentially

current : I = I_{0}  e^{\frac{-t}{RC} }

6 0
3 years ago
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