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irina1246 [14]
3 years ago
15

Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

ity of 150 m/s. The inlet area of the turbine is 60 cm2. If the power output of the turbine is 250 kW, determine the exit temperature of the argon.
Physics
1 answer:
Crazy boy [7]3 years ago
5 0

Answer:

Temperature at the exit = 267.3 C

Explanation:

For the steady energy flow through a control volume, the power output is given as

W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

Inlet area of the turbine = 60cm^{2}= 0.006m^{2}

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume v_{1}

as

v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg

m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec

for Ideal gasses, the enthalpy change can be calculated using the formula

h_{2}-h_{1}=C_{p}(T_{2}-T_{1})

hence we have

W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have T_{2}=267.3C

Hence, the temperature at the exit = 267.3 C

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Tems11 [23]

Answer:

2 N

Explanation:

From the question, it's given that

Mass m = 0.2 kg

Acceleration a = 10 m/s^2

The force a soccer goalie experience when stopping a ball will be equal to the force at which the ball is being kicked. This is

F = ma

Substitute all the parameters into the formula

F = 0.2 × 10

F = 2 Newton.

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3 years ago
A uniform rod of length 0.8 m and mass 1.4 kg, has two point masses at each end. The point mass on the left end has a mass 1.2 k
VladimirAG [237]

Answer:

Explanation:

1.2(0) + 3(0.8) + 1.4(0.8/2) / (1.2 + 3 + 1.4) = 0.5285714... ≈ 0.53 m

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2 years ago
4. Which of the following would NOT be a major process in the formation of sedimentary
sladkih [1.3K]

Answer:

b. melting

Explanation:

it is made of sediments and that is not necessary

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2 years ago
If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is u  =  7.73 \  m/s

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      s = h+  u t +  0.5 gt^2

Here s  is the distance covered by the bag at sea level which is zero

      0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>    0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>   u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}

=>   u  =  7.73 \  m/s

     

7 0
3 years ago
A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr
Molodets [167]

Answer:

Explanation:

Given

P_1=150 kPa

T_1=20^{\circ}C

V_1=0.5 m^3

T_2=140^{\circ}C

P_2=400 kPa

R for Helium R=2.076

c_v=3.115 kJ/kg-K

mass of gas m=\frac{P_1V_1}{RT_1}

m=\frac{150\times 0.5}{2.076\times 293}

m=0.123 kg

Similarly V_2 can be found

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V_2=0.264 m^3

Work done W=\int_{V_1}^{V_2}PdV

W=\frac{P_2V_2-P_1V_1}{n-1}

W=\frac{mR(T_2_T_1)}{n-1}

Since it is a polytropic Process

therefore PV^n=c

P_1V_1^n=P_2V_2^n

(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}

(\frac{0.5}{0.264})^n=\frac{400}{150}

n=\frac{\ln 2.66}{\ln 1.893}

n=1.533

W=\frac{0.123\times 2.076(140-20)}{1.533-1}

W=57.48 kJ    

From Energy balance

E_{in}-E_{out}=\Delta E_{system}

Neglecting kinetic and Potential Energy change

Q_{in}+W_{in}=change\ in\ Internal\ Energy

Change in Internal Energy \Delta U=u_2-u_1

\Delta U=mc_v(T_2-T_1)

\Delta U=0.123\times 3.115(140-20)

\Delta U=45.977 kJ

Q_{in}+57.48=45.977

Q_{in}=-11.50 kJ  

i.e. Heat is being removed

3 0
3 years ago
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