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Luchekv
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Homework Statement
- A well insulated rigid steel tank contains water (only)
- A resistance heating element (240V, 3.67Amp) designed to heat the water as well as inlet and outlet valves for - the water.
- At state-1, the water is in a saturated liquid-vapor mixture state at P(abs)=100kPa, measures 5kg and the portion of it in liquid form is 75%, the rest being vapor
- To reach state-2, more power is supplied to the heater, but just enough to vaporize all the remaining liquid. When all the liquid is vaporized, the heater is switched off.
1.) Calculate the (average) specific volume (units: m3 /kg) for the water at state-1.
2.) Calculate the total internal energy (units: kJ) for the water at state-1.
3.) Calculate the (average) specific volume (units: m3 /kg) for the water at state-2.
Homework Equations
yavg=yf+x(yfg)
ΔU=mΔu
The Attempt at a Solution
1.) [/B]I went to the tables and read of the vf and vg data @ 100kPa. For quality, seeing as 75% is liquid that would make it 0.35. I applied that data to the following equation: yavg=yf+x(yfg) and got:
vavg = 0.001043+0.25(1.6941-0.001043) = 0.4243 m^3/kg
2.) I did the same thing for this part I read the data for uf and ufg from the tables @ 100kPa and applied it to this equation: yavg=yf+x(yfg) and got:
uavg=417+0.25(2088.2) =939.65
Which I then plugged into ΔU=mΔu, seeing as its the same state I only needed uavg:
U= 5kg*1147.87 = 4695.25 kJ
3.) In the statement it says at state 2 all the water is vaporized which would make it a quality of 1 and it would be a saturated vapor..so all I did was read of the value for saturated vapor at 100kPa which is 1.6941m^3/kg
Wasn't 100% on a few of these, any feed back would be great :)
Thank you in advance
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