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

Air enters a turbine with a stagnation pressure of 900 kPa and a stagnation temperature of 658K, and it is expanded to a stagnat

bezimeni [28]

Answer:

12.332 KW

The positive sign indicates work done by the system ( Turbine )

Explanation:

Stagnation pressure( P1 ) = 900 kPa

Stagnation temperature ( T1 ) = 658K

Expanded stagnation pressure ( P2 ) = 100 kPa

Expansion process is Isentropic, also assume steady state condition

mass flow rate ( m ) = 0.04 kg/s

<u>Calculate the Turbine power </u>

Assuming a steady state condition

( p1 / p2 )^(r-1/r) = ( T1 / T2 )

= (900 / 100)^(1.4-1/1.4) = ( 658 / T2 )

= ( 9 )^0.285 = 658 / T2

∴ T2 = 351.22 K

Finally Turbine Power / power developed can be calculated as

Wt = mCp ( T1 - T2 )

= 0.04 * 1.005 ( 658 - 351.22 )

= 12.332 KW

The positive sign indicates work done by the system ( Turbine )

6 0

3 years ago

Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp

telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = $c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}$

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

$-0.40= 1.005(ln T_2 - 5.68697)- 0.5968$

Solving for T2 we have:

$T_2 = 5.8828$

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

$w_in = c_p(T_2 - T_1)+q_out$

$w_in = 1.005(358.8 - 295)+120$

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = $\frac{q_out}{T_1}$

$\frac{120kJ/kg.k}{295K}$

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

3 0

3 years ago

Recommend the types of engineers needed to collaborate on a city project to build a skateboard park near protected wetlands.

4vir4ik [10]

Answer:

A civil engineer.

Explanation:

Civil engineering is the science that deals with the design, creation and maintenance of constructions for civil use on the earth's surface. Thus, this specialty seeks to adapt soils to the needs of life in society, creating buildings, bridges, and all other constructions adapted to civil life, while taking care of the correct use of soils and the correct distribution of spaces and resources. to be used for such constructions.

4 0

3 years ago

Design a 7.5-V zener regulator circuit using a 7.5-V zener specified at 10mA. The zener has an incremental resistance of rZ = 30

hram777 [196]

Answer:

The answer is given in the explanation.

Explanation:

The circuit is as indicated in the attached figure.

From the analytical description the zener voltage is given as

$V_z=V_z_o+I_zr_z$

Here

Vzo is the voltage at which the slope of 1/rz intersects the voltage axis it is equal to knee voltage.

The equivalent model is shown in the attached figure.

From the above equation, Vzo is calculated as

$V_z_o=V_z-I_zr_z$

Here Vz is given as 7.5 V

Iz is given as 10 mA

rz is given as 30 Ω

Thus the Vzo is given as

$V_z_o=V_z-I_zr_z\\V_z_o=7.4-30*10*10^{-3}\\V_z_o=7.5-0.3\\V_z_o=7.2 V$

The value of I_L is given as 5 mA

Now the expression of current is as

$I=I_z+I_L\\I=10mA+5mA\\I=15 mA$

Now the resistance is calculated as

$R=\dfrac{V-Vo}{I}\\R=\dfrac{10-7.2}{15*10^{-3}}\\R=186.66$

So the value of resistance is 186.66 Ω.

Considering the supply voltage is increased by 10%

V is 10-10%*10=10+1=11 so the

$R=\dfrac{V-Vo}{I}\\186.66=\dfrac{11-V_o}{15*10^{-3}}\\V_o=8.2 V$

Considering the supply voltage is decreased by 10%

V is 10-10%*10=10-1=9 so the

$R=\dfrac{V-Vo}{I}\\186.66=\dfrac{9-V_o}{15*10^{-3}}\\V_o=6.2 V$

Now if the supply voltage is 10% high and the value of Load is removed i.e I=Iz only which is 10mA

so

$R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{11-V_o}{10*10^{-3}}\\V_o=9.13 V$

Now the largest load current thus that the supply voltage is 10% low and the current of zener is knee current thus

$V_z_o=V_z-I_zr_z\\V_z_o=7.5-30*0.5*10^{-3}\\V_z_o=7.5-0.015\\V_z_o=7.485 V$

$R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{9-7.485}{I}\\I=10.71 mA$

The load voltage is 7.485 V

8 0

3 years ago

For a column that is pinned at both ends, the critical buckling load can be calculated as, Pcr = π2 E I /L^2 where E is Young's

gulaghasi [49]

When a slender member is subjected to an axial compressive load, it may fail by a ... Consider a column of length, L, cross-sectional Moment of Inertia, I, having Young's Modulus, E. Both ends are pinned, meaning they can freely rotate ... p2EI L2 ... scr, is the Euler Buckling Load divided by the columns cross-sectional area

6 0

4 years ago

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