Question

An oil has a density of 50 lbm/ft³ at Standard Temperature and Pressure (STP) conditions. The density of the oil (in ^oAPI) is ________. [Given: water density is 62.4 lbm/ft³ at STP]

Hint:

Find the specific gravity first, then apply the API gravity formula 141.5 divided by SG minus 131.5.

Answer 1

Correct Answer: 45.1

Step 1: Calculate the specific gravity of the oil:
Specific gravity (SG) of the oil relative to water is defined as \[ SG = \frac{\rho_{oil}}{\rho_{water}} \] Substituting the given densities, \[ SG = \frac{50 \ \text{lbm/ft}^3}{62.4 \ \text{lbm/ft}^3} = 0.8013 \]
Step 2: Recall the API gravity formula:
The API gravity scale, defined by the American Petroleum Institute, relates specific gravity to API degrees using \[ {}^{\circ}API = \frac{141.5}{SG} - 131.5 \]
Step 3: Substitute the specific gravity value:
\[ {}^{\circ}API = \frac{141.5}{0.8013} - 131.5 \] Calculating the division term, \[ \frac{141.5}{0.8013} = 176.59 \]
Step 4: Complete the subtraction to get the final API value:
\[ {}^{\circ}API = 176.59 - 131.5 = 45.09 \approx 45.1 \]
Step 5: Sanity check the result:
An oil with a density less than water, 50 lbm/ft³ versus 62.4 lbm/ft³, should have an API gravity above 10, since 10 API corresponds to SG equal to 1.0, the same density as water. An API value of about 45.1 corresponds to a light crude oil, which is consistent with the oil being noticeably less dense than water.
Final Answer:
\[ \boxed{45.1 \ {}^{\circ}API} \]