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Sunny_sXe [5.5K]
3 years ago
5

Air enters an adiabatic turbine at 900 K and 1000 kPa. The air exits at 400 K and 100 kPa with a velocity of 30 m/s. Kinetic and

potential energy changes are negligible. If the power delivered by the turbine is 1000 kW.
Required:
a. Find the mass flow rate.
b. Find the diameter of the duct at the exit.
Engineering
1 answer:
Ivanshal [37]3 years ago
3 0

hooooooooooooooooooooooooooooooooooooooooooooooooooooooe    

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The term is hydraulics.
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What does carbon addition to iron do, what does it produce, how does it change properties, what are its reflections? Describe in
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A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given
viktelen [127]

Answer:

A.) P = 2bar, W = - 12kJ

B.) P = 0.8 bar, W = - 7.3 kJ

C.) P = 0.608 bar, W = - 6.4kJ

Explanation: Given that the relation between pressure and volume is

PV^n = constant.

That is, P1V1^n = P2V2^n

P1 = P2 × ( V2/V1 )^n

If the initial volume V1 = 0.1 m3,

the final volume V2 = 0.04 m3, and

the final pressure P2 = 2 bar. 

A.) When n = 0

Substitute all the parameters into the formula

(V2/V1)^0 = 1

Therefore, P2 = P1 = 2 bar

Work = ∫ PdV = constant × dV

Work = 2 × 10^5 × [ 0.04 - 0.1 ]

Work = 200000 × - 0.06

Work = - 12000J

Work = - 12 kJ

B.) When n = 1

P1 = 2 × (0.04/0.1)^1

P1 = 2 × 0.4 = 0.8 bar

Work = ∫ PdV = constant × ∫dV/V

Work = P1V1 × ln ( V2/V1 )

Work = 0.8 ×10^5 × 0.1 × ln 0.4

Work = - 7330.3J

Work = -7.33 kJ

C.) When n = 1.3

P1 = 2 × (0.04/0.1)^1.3

P1 = 0.6077 bar

Work = ∫ PdV

Work = (P2V2 - P1V1)/ ( 1 - 1.3 )

Work = (2×10^5×0.04) - (0.608 10^5×0.1)/ ( 1 - 1.3 )

Work = (8000 - 6080)/ -0.3

Work = -1920/0.3

Work = -6400 J

Work = -6.4 kJ

5 0
4 years ago
An insulated, vertical piston-cylinder device initially contains 10kg of water, 6kg of which is in the vapor phase. The mass of
Alexeev081 [22]

Answer:

a)120C

b)29kg

Explanation:

Hello!

To solve this exercise follow the steps below

1. we will call 1 the initial state, 2 the steam that enters and 3 the final state

2. We find the quality of the initial state, dividing the mass of steam by the total mass.

q1=\frac{6kg}{10kg} =0.6

3 Find the internal energy in the three states using thermodynamic tables

note:Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

u1=IntEnergy(Water;x=0,6(quality);P=200kPa) =1719KJ/kg

u2=IntEnergy(Water;t=350;P=5000kPa) =2808KJ/kg

u3=IntEnergy(Water;x=1;P=200kPa) =2529KJ/kg

4. use the internal energy and pressure to find the temperature in state 3, using thermodynamic tables

T3=Temperature(Water;P=200kPa;u=u3=2529KJ/kg)=120C

5. Use the first law of thermodynamics in the system, it states that the initial energy in a system must be equal to the final

m1u1+m2u2=(m1+m2)u3

where

m1=inital mass=10kg

m2=the mass of the steam that has entered.

solve for m2

(m1)(u1-u3)=(m2)(u3)-(m2)(u2)

m2=m1\frac{u1-u3}{u3-u2} =10\frac{1719-2529}{2529-2808} =29kg

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