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Alex
3 years ago
12

An insulated, vertical piston-cylinder device initially contains 10kg of water, 6kg of which is in the vapor phase. The mass of

the piston is such that it maintains a constant pressure of 200kPa inside the cylinder. Now steam at .5MPa and 350C is allowed to enter the cylinder from a supply line until all the liquid in the cylinder has vaporized. Determine
(a) the final temperature in the cylinder and
(b) the mass of the steam that has entered.
Engineering
1 answer:
Alexeev081 [22]3 years ago
7 0

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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Answer:

The current through each lamp is 0.273 Amperes

Power dissipated in each lamp is 0.082W

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Battery v = 1.5 V

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The 5 lamps in series will therefore have total resistance, R = 5 * 1.1 = 5.5 Ohms

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3 years ago
The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse
sertanlavr [38]

Answer:

T = 5416.67 N

T = -2083.5 N

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Explanation:

Forward thrust has positive values and reverse thrust has negative values.

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The thrust force represents the horizontal or x-component of momentum equation:

T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (150 - 41.67)\\\\T = 5416.67 N

Answer: The thrust force T = 5416.67 N

part b

Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component, thus thrust equation is:

T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (0 - 41.67)\\\\T = -2083.5 N

Answer: The thrust force T = -2083.5 N reverse direction

part c

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Answer: T = 0 N

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3 years ago
A cylindrical 1045 steel bar is subjected to repeated compression-tension stress cycling along its axis. If the load amplitude i
Gre4nikov [31]

Answer:

13.4 mm

Explanation:

Given data :

Load amplitude ( F )  = 22,000 N

factor of safety ( N )= 2.0

Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa

<u>calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur</u>

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<em>attached below is a detailed solution</em>

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3 years ago
Consider a very long, slender rod. One end of the rod is attached to a base surface maintained at Tb, while the surface of the r
9966 [12]

Answer:

(a) Calculate the rod base temperature (°C). = 299.86°C

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see attached file below

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3 years ago
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mr_godi [17]

Answer:

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Q=-50.005 KJ

7 0
3 years ago
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