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Simple question involving steam tables

  • #1
rock.freak667
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Homework Statement


Wet steam at 0.5Mpa, quality 0.84 is to be heated to dry saturated steam. Determine the heat required if it is heated at constant volume and at constant pressure.


Homework Equations



h=hf +xhfg

u=uf +xufg

s=sf +xsfg


v=vf +xvfg

The Attempt at a Solution




WET STEAM
Sat. pressure 500.0000 kPa
Sat. temperature 151.8438 C
Quality 0.8400
Moisture 0.1600
Enthalpy 2410.3480 kJ/kg
hf 640.1158 kJ/kg
hfg 2107.4194 kJ/kg
hg 2747.5351 kJ/kg

Internal energy 2252.8966 kJ/kg
uf 639.5693 kJ/kg
ufg 1920.6277 kJ/kg
ug 2560.1970 kJ/kg

Entropy 6.0258 kJ/kg/C
sf 1.8604 kJ/kg/C
sfg 4.9588 kJ/kg/C
sg 6.8192 kJ/kg/C

Specific volume 0.3149 m3/kg
vf 0.0011 m3/kg
vfg 0.3736 m3/kg
vg 0.3747 m3/kg


this is what I found so far from the tables


At constant volume: (W=0)

1st law: Q-W=U

therefore Q=U=m(u2-u1)

I can get the specific entropies, internal energies, volumes,enthalpies from the tables.

But how do I get the mass?
 

Answers and Replies

  • #2
Q_Goest
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Hi there rock.freak
But how do I get the mass?
I would assume the question is asking for heat per unit mass, not total heat.

therefore Q=U=m(u2-u1)
Note that this is for constant volume. Do you know how to reduce the first law for a constant pressure process? There's an easy way of remembering how to reduce the first law for these processes. If you're having trouble with this, I'd be glad to point it out.
 
  • #3
rock.freak667
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Hi there rock.freak

I would assume the question is asking for heat per unit mass, not total heat.
So then I'd just need to find the value of Q/m then?


Note that this is for constant volume. Do you know how to reduce the first law for a constant pressure process? There's an easy way of remembering how to reduce the first law for these processes. If you're having trouble with this, I'd be glad to point it out.
Well the first part is for a constant volume process.
So from the tables, I just need to get u2 and u1 right? u1 would be ug and u2 would be the value using u2=uf+ xufg right?

For constant pressure

Q=(u2-u1)+ P(v2-v1)

u2 and u1 would be the same for the constant volume as above.

v2 would need to be found using v2=vf+xvfg and v1=vf

Would this be correct?
 
  • #4
Q_Goest
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So then I'd just need to find the value of Q/m then?
Just to be clear, they're asking for energy per unit mass, so the equation you provided earlier:
therefore Q=U=m(u2-u1)
is correct for a constant volume process. The first law dU = m(u2-u1) = dQ then becomes
u2-u1 = Q/m

and all you need to do is look up u1 from initial conditions and u2 from the saturated gas condition.

For constant pressure

Q=(u2-u1)+ P(v2-v1)
Correct. If you're comfortable with that, continue on. If you'd like another way of looking at it consider this:

Draw a control volume and show Hin and Hout as mass entering and leaving the control volume. Then, rewrite the first law:
dU = dQ + dW + dH = (Qin-Qout) + (Win-Wout) + (Hin-Hout)

For a constant pressure process, such as a heat exchanger in which you have a flow in and a flow out and heat in or out. No work. No stored mass (dU = 0). So the first law as I've rewritten it can be reduced for a constant pressure process such as a heat exchanger:

dU = 0 = Hin - Hout + Qin
(no work done by or on the heat exchanger so dW = 0, energy flow is into exchanger only)

Now you can solve for energy per unit mass in a slightly different way:

Qin = Hout - Hin = m(hout-hin)

You have hin since you have initial conditions. You have hout since you know the steam is saturated gas and you know the pressure. Now all you do is solve for Q.

This method should give you the same result as the way you show, but personally I find it much easier.

Hope that's clear, I'm in a rush unfortunately but can respond later if there's a misunderstanding.
 

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