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Masja [62]
3 years ago
12

Is your mom the love of your life

Engineering
1 answer:
hodyreva [135]3 years ago
7 0
No but i love her
if she was the love of my life that would be weied
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A school is playing $0.XY per kWh for electric power. To reduce its power bill, the school installs a wind turbine with a rated
Semmy [17]

Answer: Your question has some missing figures so kindly plug in the values into the solution provided to get the exact amount of money saved

answer : Electric power generated = 216 * 10^6 kJ

             money saved = $0.XY * 60000 kwh

Explanation:

<u>Calculating  the amount of electric power generated by wind turbine</u>

power generated = ( 30 * 2000 ) kWh  = 60000 kWh

Electric energy generated = 60000 kWh * 3600 kJ = 216 * 10^6 kJ

<u>Calculate money saved by school per year </u>

$0.XY * 60000 kwh

5 0
3 years ago
You’ve experienced convection cooling if you’ve ever extended your hand out the window of a moving vehicle or into a flowing wat
Anna35 [415]

Answer:

Condition A

Heat flux is 1400 W/M^2

Condition B

Heat flux is 12800 w/m^2

Explanation:

Given that:

T_s is given as  30 degree celcius

condition A

Air temperature =  - 5 degree c

convection coefficient h = 40 w/m^2. k

heat\ flux = \frac{Q}{a}= h\Delta = 40{30 - (-5)} = 1400 w/m^2

condition A

water temperature  = 10 degree c

convection coefficient = 800 w/m^2.k

heat\ flux = \frac{Q}{A} = H(\Delta} = 800\times (30-14) = 12800w/m^2

7 0
3 years ago
Two Technicians are discussing ShopKey Pro. Technician
Leno4ka [110]

Answer:

Technician B

Explanation:

i took he test already

6 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
AVH
BigorU [14]

Answer: preventive maintenance

Explanation:

3 0
3 years ago
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