| 1B-5 : | Relationships between Different Types of Pressures | 5 pts |
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| Fill in the blank values in the table below. Assume Patm = 100 kPa and the density of liquid mercury (Hg) is 13,600 kg/m3. | ||||||||||||||
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| Read: | This problem requires
an understanding of the relationship between absolute and gage
pressure. It will also require the effective use of unit conversions. |
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| Given: | Patm | 100 | kPa | rH2O | 1000 | kg/m3 | ||||||||
| gc | 1 | kg-m/N-s2 | rHg | 13600 | kg/m3 | |||||||||
| a.) | Pgage | 17 | kPa | c.) | Pabs | 55 | mmHg | |||||||
| b.) | Pabs | 225 | kPa | d.) | Pgage | 32 | m H2O | |||||||
| Assumptions: | 1- Assume: | g | 9.8066 | m/s2 | ||||||||||
| Find: | Pgauge | ??? | kPa | Pabs | ??? | mmHg | ||||||||
| Pabs | ??? | kPa | Pgauge | ??? | m H2O | |||||||||
| Equations / Data / Solve: | ||||||||||||||
| There are two key relationships in the solution to this problem. | ||||||||||||||
| The first is the relationship between absolute and gage pressure : | ||||||||||||||
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Eqn 1 | |||||||||||||
| The second relationship is required in order to make sense of the units for pressure in the last two columns of the table in the problem statement. The 2nd relationship is the Manometer Equation. | ||||||||||||||
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Eqn 2 | |||||||||||||
| The reason we use the Manometer Equation is that when a pressure unit involves a length of a given fluid, as in the last two columns of the table given in this problem, it really means that this is the height that an open-ended manometer (for gage pressure) or a closed end manometer (for absolute pressure) would read if the given fluid were used as the manometer fluid. | ||||||||||||||
| Now, let's see how we use these 2 equations to complete the table. | ||||||||||||||
| Part a.) | In order to
fill in the 2nd column, we must solve Eqn 1 for the absolute pressure : |
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Eqn 3 | |||||||||||
| Therefore : | Pabs | 117 | kPa | |||||||||||
| To complete column 3, we must convert the units from our result for Pabs using Eqn 2. | ||||||||||||||
| In this case, Pin = Pabs and Pout = 0 (because it is a closed-end manometer). | ||||||||||||||
| Actually, Pout should be equal to the vapor pressure of the manometer fluid, but that is a topic for chapter 2. | ||||||||||||||
| Therefore, Eqn 2 becomes : | 
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Eqn 4 | ||||||||||||
| In this case, the answer for column 3 is actually the value of h, so we need to solve Eqn 4 for h : | ||||||||||||||
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Eqn 5 | |||||||||||||
| Be sure to convert kPa to Pa=N/m2 when plugging values into Eqn 5. | ||||||||||||||
| Pabs = h = | 0.877 | m Hg | Pabs | 877 | mm Hg | |||||||||
| Column 4 requires a gage pressure, so the open-end form of the Manometer Equation is used : | ||||||||||||||
| In this case, Pin = Pabs and Pout = Patm (because it is a closed-end manometer). | ||||||||||||||
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Eqn 6 | |||||||||||||
| Next, we solve Eqn 6 for h and make use of Eqn 1 if we want to use the given value of the gage pressure. | ||||||||||||||
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Eqn 7 | |||||||||||||
| Pgage = h | 1.7335 | m H2O | Pgage | 1.734 | m H2O | |||||||||
| Parts b-d) | The solution of the remaining parts of this problem involve the algebraic manipulation of Eqns 1, 3, 5 and 7, but does not involve any additional concepts, techniques or data. The final answers to parts b through d are provided in the table below. | |||||||||||||
| Answers: | Pgage (kPa) |
Pabs (kPa) |
Pabs (mmHg) |
Pgage (m H2O) |
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| a.) | 17 | 117 | 877 | 1.73 | ||||||||||
| b.) | 125 | 225 | 1690 | 22.9 | ||||||||||
| c.) | -92.7 | 7.34 | 55 | -0.909 | ||||||||||
| d.) | 314 | 414 | 3100 | 32 | ||||||||||
| Notice that I chose to use 3 significant figures in my answers. | ||||||||||||||
| This is somewhat arbitrary since the problem statement does not make it very clear how many significant figures exist in the given information. When in doubt, 3 significant figures is a reasonable choice. | ||||||||||||||
