PROBLEM 34:
A pressure gauge at elevation 8 m at the side
of a tank containing liquid reads 85 kPa .
Another pressure gage at elevation 3 m reads
120 kPa . Compute the specific gravity of the
liquid.

Ask by Fowler Graham. in the Philippines

Dec 10,2024

Question

PROBLEM 34:
A pressure gauge at elevation 8 m at the side
of a tank containing liquid reads 85 kPa .
Another pressure gage at elevation 3 m reads
120 kPa . Compute the specific gravity of the
liquid.

Ask by Fowler Graham. in the Philippines

Dec 10,2024

UpStudy AI · Accepted Answer

To find the specific gravity of the liquid, we can use the hydrostatic pressure equation, which relates the pressure difference to the height difference in a fluid column. The equation is given by:

$$
P = \rho g h
$$

Where:
- $$ P $$ is the pressure difference,
- $$ \rho $$ is the density of the liquid,
- $$ g $$ is the acceleration due to gravity (approximately $$ 9.81 \, \text{m/s}^2 $$),
- $$ h $$ is the height difference.

### Step 1: Calculate the pressure difference

The pressure readings at the two elevations are:
- At 8 m: $$ P_1 = 85 \, \text{kPa} = 85,000 \, \text{Pa} $$
- At 3 m: $$ P_2 = 120 \, \text{kPa} = 120,000 \, \text{Pa} $$

The pressure difference ($$ \Delta P $$) between the two gauges is:

$$
\Delta P = P_2 - P_1 = 120,000 \, \text{Pa} - 85,000 \, \text{Pa} = 35,000 \, \text{Pa}
$$

### Step 2: Calculate the height difference

The height difference ($$ h $$) between the two gauges is:

$$
h = 8 \, \text{m} - 3 \, \text{m} = 5 \, \text{m}
$$

### Step 3: Calculate the density of the liquid

Using the hydrostatic pressure equation, we can rearrange it to find the density ($$ \rho $$) of the liquid:

$$
\Delta P = \rho g h \implies \rho = \frac{\Delta P}{g h}
$$

Substituting the known values:

$$
\rho = \frac{35,000 \, \text{Pa}}{9.81 \, \text{m/s}^2 \times 5 \, \text{m}} = \frac{35,000}{49.05} \approx 712.5 \, \text{kg/m}^3
$$

### Step 4: Calculate the specific gravity

The specific gravity ($$ SG $$) of the liquid is defined as the ratio of the density of the liquid to the density of water (approximately $$ 1000 \, \text{kg/m}^3 $$):

$$
SG = \frac{\rho}{\rho_{\text{water}}} = \frac{712.5 \, \text{kg/m}^3}{1000 \, \text{kg/m}^3} \approx 0.7125
$$

### Conclusion

Rounding to two decimal places, the specific gravity of the liquid is approximately $$ 0.71 $$.

Thus, the answer is:

**d. 0.71**
The specific gravity of the liquid is approximately 0.71.

PROBLEM 34:
A pressure gauge at elevation 8 m at the side
of a tank containing liquid reads 85 kPa .
Another pressure gage at elevation 3 m reads
120 kPa . Compute the specific gravity of the
liquid.

Upstudy AI Solution

Answer

Solution

Step 1: Calculate the pressure difference

Step 2: Calculate the height difference

Step 3: Calculate the density of the liquid

Step 4: Calculate the specific gravity

Conclusion

Mind Expander

Related Questions

Latest Engineering Questions

PROBLEM 34: A pressure gauge at elevation 8 m at the side of a tank containing liquid reads 85 kPa . Another pressure gage at elevation 3 m reads 120 kPa . Compute the specific gravity of the liquid.

Upstudy AI Solution

Answer

Solution

Step 1: Calculate the pressure difference

Step 2: Calculate the height difference

Step 3: Calculate the density of the liquid

Step 4: Calculate the specific gravity

Conclusion

Mind Expander

Related Questions

Latest Engineering Questions

PROBLEM 34:
A pressure gauge at elevation 8 m at the side
of a tank containing liquid reads 85 kPa .
Another pressure gage at elevation 3 m reads
120 kPa . Compute the specific gravity of the
liquid.