Thermodynamics - water vapour cycle

[michealyap]

Homework Statement

How to calculate the specific work done.

Homework Equations

Water vapor initially at 10 bar and 400 °C is contained within a piston-cylinder assembly. The water is cooled at constant volume until its temperature is 150 °C. The water is then condensed isothermally to saturated liquid.
a) Sketch the process in a T,v – diagram and determine the properties first.
b) For the water as a system, evaluate the specific work done on the system, in kJ/kg.
c) Determine the specific heat removed during condensation.

The Attempt at a Solution

first i got specific work done from steam table when the water vapor is 10 bar 400 c . the specific work done i got is 2937.6 kj/kg. After that i have no idea how to continue on it. Can anyone explain what is seeing the water as a system.

 

[BvU]

Relevant equations (yours are the problem statement !) ? Properties ? Sketch ?

first i got specific work done from steam table when the water vapor is 10 bar 400 c . the specific work done i got is 2937.6 kj/kg

five digits of what ? Is this H, perhaps ?

 

[michealyap]

sorry. it is internal energy 2957.3 kj/kg. first law of thermodynamics = change of internal energy = Q+W. then i have no idea what is going on.

 

[BvU]

it is internal energy 2957.3 kj/kg

Can't find that either. What steam tables do you use ?
What is the saturation T at 10 Bar ?

 

[michealyap]

saturation T at 10 bar is 400C .. i used moran, Shapiro.

 

[BvU]

My bad, I misread Cengel p 895, 2957.9 so the same. OK, that's $u$ at the start. Part a) asks for a sketch and the properties (most likely at all three states).
What's the situation after cooling down to 150 $^\circ$ at same volume ?

 

[BvU]

saturation T at 10 bar is 400C .. i used moran, Shapiro.

Strange. I find 179.91 $^\circ$C there !

 

[michealyap]

yeah.. the saturation T is 179.91C but the initially T is 400C so it is saturated vapour. and i got 2957.3

 

[BvU]

Your $u$ is correct, but it's really superheated water vapor as it says at the heading of the table A-4

 

[michealyap]

yeah.. i thought the situation is like that. initially the

 

[BvU]

What's on the axes ? And: what makes you think that ? Or are you just guessing ?
For the record: point 1 could be correct.

 

[michealyap]

Sorry, i think this is 1 should be correct. First the water is superheated then same volume to cool down. then same temperature.

 

[BvU]

Much better ! Now we have to underpin it to see if it's right.

What's the situation after cooling down to 150 $^\circ$ at same volume ?

 

[BvU]

For the record: point 1 could be correct, point 2 and 3 I'm not so convinced

 

[Chestermiller]

What is the specific volume of superheated steam at 10 bars and 400 C? At 150 C, would that same specific volume lie between that of the saturated vapor and the saturated liquid?

 

[BvU]

Good point ! Welcome in this thread !

 

[michealyap]

the specific volume is gradually decreased with the T decreased. mean that the graph for the 1st time is correct.

 

[BvU]

We have the given that the volume stays the same during cooling down in the first stage. So what does that mean for the situation (as I asked in #6) ? (note what Chet asks in #15 too)

 

[Chestermiller]

At 10 bars and 400 C, the specific volume of the superheated steam is 0.3066 m^3/kg. At 150 C, the specific volume of the saturated vapor is 0.393 m^3/kg and the specific volume of the saturated liquid is 0.001091 m^3/kg. What does this mean to you?

 

[michealyap]

the 1st stage, the specific volume is 0.3066m3/kg..since it is same volume and according to mass of conservation, so i think they will having the same speicific volume

 

[Chestermiller]

the 1st stage, the specific volume is 0.3066m3/kg..since it is same volume and according to mass of conservation, so i think they will having the same speicific volume

So did you take this into account in determining the internal energy change in the constant-volume step?

 

[michealyap]

yes.. as i sketched in the T-v diagram i sketched the line from point 1 to point 2 as a downward straight line.

 

[Chestermiller]

yes.. as i sketched in the T-v diagram i sketched the line from point 1 to point 2 as a downward straight line.

Point 2 is placed incorrectly in the diagram. It should be in the two-phase region. This is why I was asking.

 

[michealyap]

oh.. i understand it should be in the region of vapour-liquid region..

 

[michealyap]

ya. i just drew a new diagram, I sketched from point 1 and through the line to point 2.

 

[Chestermiller]

ya. i just drew a new diagram, I sketched from point 1 and through the line to point 2.

Much better. Now, what are the mass fractions of liquid and vapor at point 2? What is the specific internal energy at point 2?

 

[michealyap]

it seems unable to get the mass fraction as the formula is x= mg/total weight of liquid and gas. i do not have either mass of liquid of vapor . nor x.
I am quite confused already. the specific volume when the water vapor is superheated, it is 0.3066 m^3/kg since constant volume cooldown, so at point 2 it also should be 0.3066m^3/kg however when i check the steamtable, the saturated vapor of 0.3066m^3/kg does not fall into the 150 degree celcius.

 

[michealyap]

it seems unable to get the mass fraction as the formula is x= mg/total weight of liquid and gas. i do not have either mass of liquid of vapor . nor x.
I am quite confused already. the specific volume when the water vapor is superheated, it is 0.3066 m^3/kg since constant volume cooldown, so at point 2 it also should be 0.3066m^3/kg however when i check the steamtable, the saturated vapor of 0.3066m^3/kg does not fall into the 150 degree celcius.

 

[Chestermiller]

it seems unable to get the mass fraction as the formula is x= mg/total weight of liquid and gas. i do not have either mass of liquid of vapor . nor x.
I am quite confused already. the specific volume when the water vapor is superheated, it is 0.3066 m^3/kg since constant volume cooldown, so at point 2 it also should be 0.3066m^3/kg however when i check the steamtable, the saturated vapor of 0.3066m^3/kg does not fall into the 150 degree celcius.

At point 2, you have a combination of liquid and water vapor, with a weighted average specific volume of 0.3066. Let V be the volume of the container, and let M be the total mass of water in the container. Then $$0.3066M=V$$Now, if x is the mass fraction of water vapor and (1-x) is the mass fraction of liquid water, then the volume of water vapor must be $0.393Mx$ and the volume of liquid water must be $0.001091M(1-x)$. Therefore, the total volume is $$0.393Mx+0.001091(1-x)M=V=0.3066M$$So, $$0.393x+0.001091(1-x)=0.3066$$

 

[michealyap]

Hi, may i know how do you get the volume of water vapor must be 0.393Mx and liquid water must be 0.001091m. is it because to find the speicifc volume at the degree of 150 celcius?

 

[michealyap]

by solving the equation 0.393 x + 0.001091 ( 1 − x ) = 0.3066. i got x is 0.779 which is mass fraction of vapor then i got 0.221 for saturated liquid so for specific internal energy, i used interpolation to get the internal energy of saturated vapour and liquid the 150 degree celcius. I got 631.40075 kj/kg for saturated liquid, then i got 2559.49kj/kg for saturated vapor. so by multiplying mass fraction of respected specific internal energy, 631.4075*= specific internal energy for saturated liquid in the system at point 2, then 2559.49kj/kg/for saturated vapor in the system at point 2.To get Ufg = 2558.49-631.4065=1928.081 kj/kg. =using the fomula Uavg= Uf+ X*Ufg = 631.4075 +0.78 *1928.089= 2135.310kj/kg. specific internal energy is 2135.310kj/kg at point 2.

 

[Chestermiller]

by solving the equation 0.393 x + 0.001091 ( 1 − x ) = 0.3066. i got x is 0.779 which is mass fraction of vapor then i got 0.221 for saturated liquid so for specific internal energy, i used interpolation to get the internal energy of saturated vapour and liquid the 150 degree celcius. I got 631.40075 kj/kg for saturated liquid, then i got 2559.49kj/kg for saturated vapor. so by multiplying mass fraction of respected specific internal energy, 631.4075*= specific internal energy for saturated liquid in the system at point 2, then 2559.49kj/kg/for saturated vapor in the system at point 2.To get Ufg = 2558.49-631.4065=1928.081 kj/kg. =using the fomula Uavg= Uf+ X*Ufg = 631.4075 +0.78 *1928.089= 2135.310kj/kg. specific internal energy is 2135.310kj/kg at point 2.

Your approach looks correct. But there was actually no need to do any interpolation. In your set of steam tables, there should not only be a table for saturated conditions at specified pressures, but also a table for saturated conditions at 10 degree increments of temperature. For 150 C, that table gives the following values:

Equilibrium pressure = 4.758 bars
Specific volume of liquid = 0.001091 m^3/kg
Specific volume of vapor = 0.393 M^3/kg
Internal energy of liquid = 631.7 kj/kg
Internal energy of vapor = 2559 kj/kg
Enthalpy of saturated liquid = 632.2 kj/kg
Enthalpy of saturated vapor = 2746 kj/kg

 

[michealyap]

hi, after that i do not know how to proceed. 1st law of thermodynamics change of internal energy = Q- W. for the process of 1 to 2, constant volume so work done on this is equal to zero. therefore internal energy of procee 1 - process 2 = 2597.3 kj/kg - 2135.310 kj/kg = 462kj/kg. heat transfer out of the system. then how to find the work done for the system?

 

[Chestermiller]

hi, after that i do not know how to proceed. 1st law of thermodynamics change of internal energy = Q- W. for the process of 1 to 2, constant volume so work done on this is equal to zero. therefore internal energy of procee 1 - process 2 = 2597.3 kj/kg - 2135.310 kj/kg = 462kj/kg. heat transfer out of the system. then how to find the work done for the system?

Didn't you just say that the work is zero?

 

[michealyap]

Hello, yea i thought the question is asking the whole system first is constant volume , then is condensed isothermally. so the specific workdone is on the process of condensation.