Thermodynamics of Steam Homework

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The discussion revolves around a homework problem involving the thermodynamics of steam, specifically calculating the heat required to convert 7.7 kg of water at 90°C to saturated steam at 1300 kPa and 71.6% dryness. The initial calculations provided by the student yield a heat requirement of 13,290.561 kJ, while the textbook states the answer is 14,248.567 kJ. The key correction suggested is that the student must consider the enthalpy of both the saturated liquid and vapor in the final state, rather than simply applying the percentage to the saturated vapor enthalpy. Additionally, a separate question about boiler efficiency and steam production from fuel is introduced, emphasizing the need to account for the efficiency in calculating the heat available for steam generation. Accurate calculations and unit checks are essential for both problems.
mvf
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Here is the question in my homework, but my response does not correspond with the answer given, even though I worked out the answer that was given backwards and it still does not work:

Determine the quantity of heat required, to raise 7.7 kg of water at 90°C, to saturated steam at 1300 kPa and 71.6% dry.

Here is my attempt at a solution:

First, I checked the steam table for water at 90°C. The (hf) value is 376.92. Then I checked the steam table for saturated steam (hg) at 1300 kPa and that is 2787.6.

Then I subtracted hf from hg (2787.6 - 376.92 = 2410.68. Then I multiplied that by 7.7.
(2410.68 X 7.7 = 18562.24). Then I took 71.6 percent of that number which is 13 290.561 KJ required to raise 7.7 kg of water to saturated steam.

The answer the book is giving is 14248.567 KJ required. Is this a typo or am I making a mistake. I would appreciate any help I can get, thank you.
 
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mvf said:
Here is the question in my homework, but my response does not correspond with the answer given, even though I worked out the answer that was given backwards and it still does not work:

Determine the quantity of heat required, to raise 7.7 kg of water at 90°C, to saturated steam at 1300 kPa and 71.6% dry.

Here is my attempt at a solution:

First, I checked the steam table for water at 90°C. The (hf) value is 376.92. Then I checked the steam table for saturated steam (hg) at 1300 kPa and that is 2787.6.

Then I subtracted hf from hg (2787.6 - 376.92 = 2410.68. Then I multiplied that by 7.7.
(2410.68 X 7.7 = 18562.24). Then I took 71.6 percent of that number which is 13 290.561 KJ required to raise 7.7 kg of water to saturated steam.

The answer the book is giving is 14248.567 KJ required. Is this a typo or am I making a mistake. I would appreciate any help I can get, thank you.
Hi mvf. Welcome to physics forums!

In the final state, you have a combination of saturated liquid and saturated vapor. So, you can't just use 71.6 percent of the saturated vapor enthalpy and call that the enthalpy of the mixture. You need to find the enthalpy of the saturated liquid at the final state, and determine the final enthalpy of the mixture by weighting their enthalpies in proportion to their amounts.

Chet
 
And put units into your calculations!
 
Sorry, I'm not sure how to calculate properly. Can you help me out? I'd appreciate it, thanks.
 
mvf said:
Sorry, I'm not sure how to calculate properly. Can you help me out? I'd appreciate it, thanks.
1. Write down the specific enthalpy of the saturated liquid
2. Write down the specific enthalpy of the saturated vapor
3. Multiply specific enthalpy of saturated liquid by (0.284)
4. Multiply specific enthalpy of saturated vapor by (0.716)
5. Add results of 3. and 4. to get specific enthalpy for final equilibrium state
6. Multiply 5. by 7.7 kg
7. Subtract enthalpy in initial state.

Chet
 
Here is my next question:

A boiler furnace releases 10800 KJ of heat for each kg of fuel burned and produces dry saturated steam at 10000 kPa from feedwater at 190 degrees C. How many kg of steam will be produced for each kg of fuel burned if the boiler had an efficiency of 70%.

Here is as far as got:
ms(h1 - h2)
Boiler efficiency = ------------------- X 70%
Heating value of the fuel

ms (2724.7 - 807.62)
= ------------------------ X 70%
Heating value of the fuel

I'm not sure where to go from there and I would appreciate any help I could get. Thanks.
 
mvf said:
Here is my next question:

A boiler furnace releases 10800 KJ of heat for each kg of fuel burned and produces dry saturated steam at 10000 kPa from feedwater at 190 degrees C. How many kg of steam will be produced for each kg of fuel burned if the boiler had an efficiency of 70%.

Here is as far as got:
ms(h1 - h2)
Boiler efficiency = ------------------- X 70%
Heating value of the fuel

ms (2724.7 - 807.62)
= ------------------------ X 70%
Heating value of the fuel

I'm not sure where to go from there and I would appreciate any help I could get. Thanks.
What they are implying is that 70% of the 10800 kJ is available for supplying heat to the water to make steam. If those h's are the enthalpy of the saturated steam at 10000kPa and the feedwater at 190 C, then the amount of steam produced is equal to the heat made available divided by the increase in enthalpy per kg. Just check the units, and they will tell you how to do many problems.

Chet
 

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