Are there any companies that you can get a job at as an air craft engeer after university

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

3 years ago

In a certain chemical plant, a closed tank contains ethyl alcohol to a depth of 71 ft. Air at a pressure of 17 psi fills the gap

Yuliya22 [10]

Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

Given that;

depth 1 = 71 ft

depth 2 = 10 ft

pressure p = 17 psi = 2448 lb/ft²

depth h = 71 ft - 10 ft = 61 ft

we know that;

p = P_air + yh

where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )

so we substitute;

p = 2448 + ( 49.3 × 61 )

= 2448 + 3007.3

= 5455.3 lb/ft³

= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

5 0

3 years ago

n the Spring of 2015, three utility companies in the Ukraine received email purporting to come from Ukraine's parliament, the Ra

Tresset [83]

Answer:

Trojan horse

Explanation:

A trojan horse attack is a type of malware that misleads users, as it appears unsuspicious at first, but actually presents a threat to the user. A common example is that of an email that contains a malicious attachment. Another common example is that of a fake advertisement. The name comes from the Greek story of the Trojan horse that led to the fall of the city of Troy.

5 0

3 years ago

Which term describes how a mineral looks when it breaks apart in an irregular way

attashe74 [19]

Answer:

Fracture.

Explanation:

Fracture describes how a mineral looks when it breaks apart in an irregular way.

8 0

2 years ago

Read 2 more answers

If the maximum allowable shear stress is 70 MPa, find the shaft diameter needed to transmit 40 kW when the shaft speed is 250 rp

victus00 [196]

Answer:

The diameter is 50mm

Explanation:

The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.

T=(P×60)/(2×pi×N)

T is the Torque

P is the the power to be transmitted by the shaft; 40kW or 40×10³W

pi=3.142

N is the speed of the shaft; 250rpm

T=(40×10³×60)/(2×3.142×250)

T=1527.689Nm

Diameter of a shaft can be obtained from the formula

T=(pi × SS ×d³)/16

Where

SS is the allowable shear stress; 70MPa or 70×10⁶Pa

d is the diameter of the shaft

Making d the subject of the formula

d= cubroot[(T×16)/(pi×SS)]

d=cubroot[(1527.689×16)/(3.142×70×10⁶)]

d=0.04808m or 48.1mm approx 50mm

7 0

3 years ago

Are there any companies that you can get a job at as an air craft engeer after university​

Are there any companies that you can get a job at as an air craft engeer after university