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Ann [662]
3 years ago
10

Consider air entering a heated duct at P1 = 1 atm and T1 = 288 K. Ignore the effect of friction. Calculate the amount of heat pe

r unit mass (in joules per kilogram) necessary to choke the flow at the exit of the duct, as well as the pressure and temperature at the duct exit, for an inlet Mach number of
(a) M1 = 2.0
(b) M1 = 0.2.
Engineering
2 answers:
Ne4ueva [31]3 years ago
5 0

Answer:

The solution for the given problem is done below.

Explanation:

M1 = 2.0

\frac{p1}{p*} = 0.3636

\frac{T1}{T*} = 0.5289

\frac{T01}{T0*} = 0.7934

Isentropic Flow Chart:  M1 = 2.0 , \frac{T01}{T1} = 1.8

T1 = \frac{1}{0.7934} (1.8)(288K) = 653.4 K.

In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.

At the inlet,

T02= \frac{T02}{T1}T1 = (1.8)(288K) = 518.4 K.

Q= Cp(T02-T01) = \frac{1.4(287 J / (Kg.K)}{1.4-1}(653.4-518.4)K = 135.7*10^{3} J/Kg.

QveST [7]3 years ago
4 0

Answer:

q=135.7×10^3 J/kg

T_0^*= 653.4K

Explanation:

q=135.7×10^3 J/kg

T_0^*= 653.4K

M_1=2.0 => p_1/p^* =0.3636 T_1/T^* =0.5289 T_01/T_0 =0.7934

Isentropic flow properties chart:

M_1=2.0 => T_01/T_1 =1.8

T_0^*= ‖(T_O^*)/T_01 ‖‖T_01/T_1 ‖ T_1=‖1/0.7934‖(1.8)(288K)= 653.4K

In order to choke the flow at the exit (M_2=1),the above T_0^*= T_02; p_2=p^*

T_0^* must be the stagnation temperature at the exit.

At the inlet

T_02=T_02/T_1 T_1=(1.8)(288K)=518.4K

q=c_p (T_02-T_01 )=γ^R/(γ-1) (T_02-T_01 )=1.4(287J/(kg.K))/(1.4-1) (653.4-518.4)K=135.7×10^3 J/kg

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If something is 50fficient, how many joules of wasted energy will there be if 750j of energy is put in?’
Sedbober [7]

Efficiency is the minimum use of energy to accomplish the task. The wasted energy will be 375 J when 750 J of energy is given.

<h3>What is wasted energy?</h3>

Wasted energy is energy that is not useful when the transformation in the system occurs.

Total energy = 750 J

The efficiency of the system = 50 %

Output work (OW) is calculated as:

Efficiency = output work ÷ input work × 100%

750 × 50 = 100 OW

OW = 375 J

Wasted energy = Total energy - output work

= 750 - 375

= 375 J

Therefore, the machine is 50 % inefficient and has wasted energy of 375 J.

Learn more about wasted energy here:

brainly.com/question/16177264

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8 0
2 years ago
I am trying to create a line of code to calculate distance between two points. (distance=[tex]\sqrt{ (x2-x1)^2+(y2-y1)^2}) My li
k0ka [10]

Answer:

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Explanation:

The distance formula is the difference of the x coordinates squared, plus the difference of the y coordinates squared, all square rooted.  For the general case, it appears you simply need to change how you have written the code.

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Note, by moving the 2 inside of the pow function, you have provided the second argument that it is requesting.

You were close with your initial attempt, you just had a parenthesis after x1 and y1 when you should not have.

Cheers.

6 0
3 years ago
P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diame
eimsori [14]

Answer:

a)  m=0.17kg/s

b)  Ma=0.89

Explanation:

From the question we are told that:

Pressure P=60kPa

Diameter d=3cm

Generally at sea level

T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4

Generally the Power series equation for Mach number is mathematically given by

\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}

\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}

Ma=0.89

Therefore

Mass flow rate

\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}

\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}

\rho=0.848kg/m^3

Generally the equation for Velocity at throat is mathematically given by

V=Ma(r*T_0\sqrt{T_e})

Where

T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}

T_e=248

Therefore

V=0.89(1.4*288\sqrt{248})\\\\V=284

Generally the equation for Mass flow rate is mathematically given by

m=\rho*A*V

m=0.84*\frac{\pi}{4}*3*10^{-2}*284

m=0.17kg/s

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Kobotan [32]

Answer:

Nuclear fission is almost 8,000 times more efficient than traditional fossil fuels at producing energy. That's a lot of energy packed into a small space. Nuclear energy is more efficient, which means it uses less fuel to power the plant and produces less waste.


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Disadvantages:
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Answer:

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Explanation:

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