Question

Water (density = 1000 kg/m$^3$) and alcohol (specific gravity = 0.7) enter a Y-shaped channel at flow rates of 0.2 m$^3$/s and 0.3 m$^3$/s, respectively. Their mixture leaves through the other end of the channel, as shown in the figure. The average density (in kg/m$^3$) of the mixture is $________________$.

Hint:

To find the density of a mixture, always divide the \textbf{total mass flow rate} by the \textbf{total volume flow rate}. Remember: use mass = density $\times$ volume for each fluid first.

Answer 1

Correct Answer: 820

Step 1: Write given data.
Density of water: $\rho_w = 1000 \ \text{kg/m}^3$
Specific gravity of alcohol: $0.7 \Rightarrow \rho_a = 0.7 \times 1000 = 700 \ \text{kg/m}^3$
Flow rate of water: $Q_w = 0.2 \ \text{m}^3/\text{s}$
Flow rate of alcohol: $Q_a = 0.3 \ \text{m}^3/\text{s}$
Step 2: Compute mass flow rates.
Mass flow rate of water: \[ \dot{m}_w = \rho_w Q_w = 1000 \times 0.2 = 200 \ \text{kg/s} \] Mass flow rate of alcohol: \[ \dot{m}_a = \rho_a Q_a = 700 \times 0.3 = 210 \ \text{kg/s} \]
Step 3: Compute total volume flow rate.
\[ Q_{total} = Q_w + Q_a = 0.2 + 0.3 = 0.5 \ \text{m}^3/\text{s} \]
Step 4: Compute total mass flow rate.
\[ \dot{m}_{total} = \dot{m}_w + \dot{m}_a = 200 + 210 = 410 \ \text{kg/s} \]

Step 5: Average density of the mixture.
By definition: \[ \rho_{avg} = \frac{\dot{m}_{total}}{Q_{total}} \] Substitute values: \[ \rho_{avg} = \frac{410}{0.5} = 820 \ \text{kg/m}^3 \] Final Answer: \[ \boxed{820 \ \text{kg/m}^3} \]