Thermodynamics homework: Water heated in a sealed container

[erde]

Homework Statement

Hello, i have the following assignment due for monday and i have truly no idea how to tackle it, i have already tried it many times but all my results have been wrong. the problem is as follows.

inside a sealed container is an amount of water: mass=90.26 kg; pressure 9.1842 bar; x=0.00051 ;
h´=745.14[kJ/kg]; h= 746.17 [kJ/kg]; h´´=2773.6[kJ/kg];
v´=0.001[m^3/kg]; v= 0.0011[m^3/kg] ; v´´=0.21338[m^3/kg];
u´=744.22[kJ/kg] ; u=745.15[kJ/kg]; u´´=2580.12[kJ/kg]

the container is heated using an electric heat source with P=? (im looking for this)

the container has an entry and an exit valve, both closed.

1) the exit valve is opened; now calculate the heater power P=? in kW required for a constant exit vapor mass(dm/dt) of 0.1[kg/s] while maintaining h´´=2773.6[kJ/kg]

Homework Equations

(du/dt)=(du/dM) * (dM/dt)
take into account that the liquid water level inside the container sinks
only vapor h´´ leaves the container

The Attempt at a Solution

M=90,26 kg; m=0,1 kg/s
deltaU=-deltaM*h+P*t |d()/dt
dm/dt * u+ du/dt *m = -m*h+ P
dm/dt* u + du/dt*m= -m * h +P |m=const-> dm/dt=0
du/dt * m=-m*h + P |du/dt= du/dM * dM/dt
du/dM * dM/dt*m= -m*h + P
du * (m^2)/dM= -m*h +P |* dM

integrate (m^2 * du ](0 to-> u) =-h* Integrate (m*dm] (0 to -> M) + integrate (P dM)(0 to->M)
(m^2)*u = -m*M + M*P
m*u= -m +P
[kJ]= [kg/s] + [kJ/s] --> so wrong

 

[BvU]

Hello erde,

Pressure is high , so I won't pester you with guidelines and such, that'll come later. But making the 1. problem statement a bit more explicit would be helpful for you too (not just units, but also explaining the variables -- never mind, I can guess). And 2. the relevant equations need a few more relationships; so that all variables you need for 3. the attempt at solution actually appear. Plus the kind of physics law(s) you need to work this out.

No concrete help so far. But my first line of inquiry here (is this step 1 in a multi-part exercise?) would be if I can apply an enthalpy balance. Grossly 0.1 [kg/s] * 2773.6[kJ/kg] = 277.36 [kW] but I suppose the enthalpy of the vessel contents also changes a little bit.

On another note: I am surprised by the 1-digit accuracy of v and dm/dt appearing aside 5 or 6 digit others, but that's an issue for the exercise composer.

 

[Chestermiller]

It's hard to read your equations. Any chance of your using LaTex (tutorial provided in Help from INFO drop down menu). In your heat balance, you should have a -dm/dt in front of the h on the right hand side, rather than a -m. Also, can you please provide the exact statement of the problem. Also, what do the primes and double primes on the properties signify.

 

[erde]

the prime values are the values for liquid water at the given pressure and the double prime are the values for the vapor at the given pressure, x is the amount of vapor.

also for relevant equations, i tried to use U₂-U₁=-ΔM *h + P*t
ΔM being the amount of water that leaves the container in this case 0.1 kg/s, h being equal to h´´ as only vapor leaves the container, P the heating power and t the time
for U is the equation U=M*u; U₁ is M₁*u₁ which are the mass before anything happens 90.26kg and u for the given pressure and x, which is u=745.15[kJ/kg]
and U₂ would be the mass and the u sometime after.

i tried solving it again without integrals using the following:

U₂-U₁=-m*t*h´´ + P*t | m being the the per second amount 0.1kg/s as delta M is the difference between both masses M₁ and M₂
M₂*u₂ - M₁*u₁ = -m*t*h´´ + P*t |now with following changes: M₂=M₁-m*t
(M₁-m*t)*u₂ - M₁*u₁ = -m*t*h´´ + P*t |now for u₂: u₂= u´+x₂*(u´´-u´); x₂= (v₂-v´)/(v´´-v´) ; v₂=V/(M₂)=> v₂=V/(M₁-m*t) the container Volume V is given with 0.1m^3
(M₁-m*t)* [u´+ (V/(M₁-m*t)-v´)/(v´´-v´))*(u´´-u´)]- M₁*u₁ = -m*t*h´´ + P*t

this new equation should give M₂, u₂ and therefore U₂ together with ΔM values depending of the time t. the thing is P should be constant no matter how much time has passed since the vapor started leaving the container as we are looking for P for which m is constant so the value of P should NOT change for any values of t but when trying it out i get different P values for every different t values so this method is also wrong :(

 

[erde]

the exact statement of the question is: what value P has to have the electric heater H so that a constant vapor mass of 0.1kg/s leaves the container with h=2773.6 (tip du/dt=du/dm * dm/dt) take into account that during this exit process because of the continuous vapor formation the liquid water level sinks.
this question is the last one in a 45 points exercise and gives alone 18 out of 45 points(the exercise comes from an old exam for which 1 point is equal to 1 minute being a 90 points exam with a duration of 90 minutes)

i thought of doing the problem using P=m*(h´´-h´) with P the amount of energy needed to get 0.1kg/s of water from h´to h´´ but i don't see how the given help du/dt= du/dM * dM/dt helps here. also it isn't stated anywhere that the pressure inside the container remains constant, but the only pressure at which water has the h´´ value of 2773.6[kJ/kg] is the given one so the pressure has to remain constant in order for the water vapor to have that enthalpy. so the equation P=m*(h´´-h) would only take into account the energy need to generate the leaving vapor but it would not take into account the energy needed to increase u of the remaining water inside the container in order to maintain the pressure. so P > m*(h´´-h´)

 

[Chestermiller]

You are correct that the pressure and temperature don't change.

Let $m_L$ = mass of liquid in tank

$m_V$ = mass of vapor in tank

$v_V$ = specific volume of vapor in tank

$v_L$ = specific volume of liquid in tank

$\dot{m}$ = (constant) rate of mass exiting the tank

In terms of these parameters, what is the mass balance on the tank contents? In terms of these parameters, what is the volume V of the tank at any time? Is the volume V of the tank changing with time? From this information, what is $dm_V/dt$ in terms of $\dot{m}$, $v_V$, and $v_L$? In terms of these parameters, what is $dm_L/dt$ in terms of $\dot{m}$, $v_V$, and $v_L$?

 

[erde]

volumen of the tank is V=0.1m³=const
Mass at the start would be m_V,1=m₁*x; m_L,1=m₁*(1-x)
the total Mass in the tank would be m(t)=m₁- $\dot m$ * t
m_V(t)=m_V,1 - $\dot m$ * t + $\frac {P}{(h´´-h´)}$ *t
m_L(t)=m_L,1 -$\frac {P}{(h´´-h´)}$ *t
dm_V/dt>0
dm_L/dt<0
v_L=const
v_V=const
dm_V/dt=-$\dot m$ + $\frac {P}{(h´´-h´)}$
dm_L/dt=-$\frac {P}{(h´´-h´)}$
im really puzzled how to find dm/dt in terms of $\dot m$ v_V and v_L

m_L*v_L + m_V*v_V=0.1m³

 

[Chestermiller]

volumen of the tank is V=0.1m³=const
Mass at the start would be m_V,1=m₁*x; m_L,1=m₁*(1-x)
the total Mass in the tank would be m(t)=m₁- $\dot m$ * t
m_V(t)=m_V,1 - $\dot m$ * t + $\frac {P}{(h´´-h´)}$ *t
m_L(t)=m_L,1 -$\frac {P}{(h´´-h´)}$ *t
dm_V/dt>0
dm_L/dt<0
v_L=const
v_V=const
dm_V/dt=-$\dot m$ + $\frac {P}{(h´´-h´)}$
dm_L/dt=-$\frac {P}{(h´´-h´)}$
im really puzzled how to find dm/dt in terms of $\dot m$ v_V and v_L

m_L*v_L + m_V*v_V=0.1m³

First of all, the volume is not 0.1 m^3.

Mass Balance: $$\frac{dm_L}{dt}+\frac{m_V}{dt}=-\dot{m}\tag{1}$$

Tank Volume: $$m_Lv_L+m_Vv_V=V=const$$

So, $$v_L\frac{dm_L}{dt}+ v_V\frac{dm_V}{dt}=0\tag{2}$$
So, using only Eqns. 1 and 2, what do you get for $dm_L/dt$ and $dm_V/dt$ in terms of $\dot{m}$, $v_V$, and $v_L$? Using this same set of variables (and $u_L$, $u_V$, $h_V$, and P) what is the transient heat balance on the tank?

 

[erde]

1) dm_L/dt=- $\dot m$ -dm_v/dt
2) in 1) dm_L/dt=- $\dot m$ + v_L/v_V*dm_L/dt
$\dot m$= v_L/v_V*dm_L/dt - dm_L/dt
$\dot m$=[v_L/v_V-1] dm_L/dt

-> dm_L/dt=$\dot m$ / ([v_L/v_V]-1)

-> dm_V/dt=$\dot m$ / ([v_V/v_L] -1)

dm_L/dt* u_L + dm_V/dt * u_V = -$\dot m$ *h_V +P

$\frac {\dot m} { ( \frac {v_L}{v_V }-1) }*u_L + \frac {\dot m} { ( \frac {v_V}{v_L }-1) }*u_V = - \dot m *h_V + P$

 

[Chestermiller]

1) dm_L/dt=- $\dot m$ -dm_v/dt
2) in 1) dm_L/dt=- $\dot m$ + v_L/v_V*dm_L/dt
$\dot m$= v_L/v_V*dm_L/dt - dm_L/dt
$\dot m$=[v_L/v_V-1] dm_L/dt

-> dm_L/dt=$\dot m$ / ([v_L/v_V]-1)

-> dm_V/dt=$\dot m$ / ([v_V/v_L] -1)

dm_L/dt* u_L + dm_V/dt * u_V = -$\dot m$ *h_V +P

$\frac {\dot m} { ( \frac {v_L}{v_V }-1) }*u_L + \frac {\dot m} { ( \frac {v_V}{v_L }-1) }*u_V = - \dot m *h_V + P$

i tried solving this equation for P in my calculator and got P=18.48 kW which seems a bit too low as the energy required only for $\dot m$ is P=##\dot m *(h´_V - h_L)

Excellent! Now, solve for P, simplifying mathemtaically the resulting relationship as much as you can.

Incidentally, in LaTex, you don't need to use *'s to indicate multiplication.

 

[erde]

i tried solving this equation for P in my calculator and got $P=18.48 kW$ which seems a bit too low as the energy required only for $\dot m$ is $P=\dot m(h_V - h_L)=202.846 kW$ is that really the end result? also we didnt use the tip that $\frac {du} {dt}= \frac {du}{dm} \frac {dm}{dt}$

 

[Chestermiller]

i tried solving this equation for P in my calculator and got $P=18.48 kW$ which seems a bit too low as the energy required only for $\dot m$ is $P=\dot m(h_V - h_L)=202.846 kW$ is that really the end result? also we didnt use the tip that $\frac {du} {dt}= \frac {du}{dm} \frac {dm}{dt}$

I simplified the final result by using $$u_L=h_L-pv_L$$ and $$u_V=h_V-pv_V$$ to finally obtain: $$P=\dot{m}(h_V-h_L)\frac{v_V}{(v_V-v_L)}$$That's certainly a little greater than $P=\dot m(h_V - h_L)$.

Regarding the hint, I thought it was goofy, and didn't expect it to be useful. I never would have given such a hint.

 

[erde]

thank you a whole lot, i have tried again and got $203.8 kW$ which is indeed a little more. thank you a lot for your help!

 

[Chestermiller]

There is another (even simpler) method of solving this problem that doesn't even directly involve applying the energy balance. The rate at which liquid water is being vaporized in the tank is equal to $\left(-\frac{dm_L}{dt}\right)=\dot{m}\frac{v_V}{v_V-v_L}$. The electric heat source must be supplying the heat of vaporization. So, $$P=\left(-\frac{dm_L}{dt}\right)(h_V-h_L)=\dot{m}(h_V-h_L)\frac{v_V}{v_V-v_L}$$