Stress that acts in the plane of a cut section, rather than at right angles to the section is called:______

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

Which option identifies the type of power system Tommy will design in the following scenario?

Sedaia [141]

Answer:

diagram of an electrical curcuit

an sketch of an HVAC system

Also 3D image of a hydrualic piston

se

7 0

3 years ago

Used ______ must be hot drained for 12 hours or crushed before disposal.

WITCHER [35]

Answer:

A

Explanation:

The answer is towels because towels after a little bit of just sitting around have a chemical reaction that cam cause them to spontaneously combust

5 0

3 years ago

que sabemos de la revolución industrial y como ese proceso impulso el uso de los controles eléctricos en las industrias

Alex Ar [27]

Answer:

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▀█▌░░░▄░▀█▀░▀ ░░ Copy And Paste Him onto all of ur brainly answers

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░░░░░░░▀███▀█░▄░░ Over brainly

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Explanation:

8 0

3 years ago

The electron beam in a TV picture tube carries 1015 electrons per second. As a design engineer, determine the voltage needed to

leonid [27]

Answer:

The voltage needed to accelerate the electron beam is 2.46 x 10^16 Volts

Explanation:

The rate of electron flow is given as:

q = 1015 electrons per second

The total current is given by:

Total Current = (Rate of electron flow)(Charge on one electron)

Total Current = I = (1015 electrons/s)(1.6 x 10^-19 C/electron)

I = 1.624 x 10^-16 A

Now, we know that electric power is given as:

Electric Power = Current x Voltage

P = IV

V = P/I

V = 4 W/1.624 X 10^-16 A

<u>V = 2.46 x 10^16 Volts</u>

6 0

3 years ago

Stress that acts in the plane of a cut section, rather than at right angles to the section is called:_______