Question 2.4: Pressure gage B is to measure the pressure at point A in a w...

• System sketch: The system is shown in Fig. E2.4.
• Assumptions: Hydrostatic fluids, no mixing, vertical “up” in Fig. E2.4.
• Approach: Sequential use of Eq. (2.14) to go from A to B.
Liquids:            p_2 – p_1 = -\gamma(z_2 – z_1)

or               z_1 – z_2 = \frac{p_2}{\gamma} – \frac{p_1}{\gamma}                         (2.14)

• Property values: From Table 2.1 or Table A.3:

^*In contact with air.
^†The viscosity–temperature variation of these liquids may be fitted to the empirical expression

with accuracy of ±6 percent in the range 0 ≤ T ≤ 100°C.

^‡Representative values. The SAE oil classifications allow a viscosity variation of up to ±50 percent, especially at lower temperatures.

• Solution steps: Proceed from A to B, “down” then “up,” jumping across at the left mercury meniscus:
p_A + \gamma_w |\Delta z|_w – \gamma_m |\Delta z_m| – \gamma_o |\Delta z|_o = p_B

or  p_A + (9790  N/m^3)(0.05  m) – (133,100  N/m^3)(0.07  m) – (8720  N/m^3)(0.06  m) = 87,000

or    p_A + 490 – 9317 – 523 = 87,000 Solve for p_A = 96,350  N/m^2 \approx 96.4  kPa

• Comments: Note that we abbreviated the units N/m^2 to pascals, or Pa. The intermediate five-figure result, p_A = 96,350  Pa, is unrealistic, since the data are known to only about three significant figures.