Question: A compressor for deep sea diving gas is fed by two supply pipes of circular cross-section:
• the first, of 39.7 mm internal diameter, carries an oxygen/helium mix; at a density of 1.01 kg/m^3 at a mean velocity of 33.3 cm/s;
• the second carries 221.2 cubic centimetres per hour of water, at a density of 1,003 kg/m^3.
The cylinder outlet valve is currently delivering gas with density 2.966 kg/m^3 at the rate of 9.532 litres per minute.
Calculate the rate of increase in the mass of the cylinder (including contents) – quoting your answer in kg per hour to 4 significant figures.
1. **State the problem:** We need to find the rate of increase in the mass of the cylinder, given two supply pipes feeding gas and water, and an outlet valve delivering gas at a certain rate and density.
2. **Given data:**
- Pipe 1 diameter $d_1 = 39.7$ mm = $0.0397$ m
- Density of gas in pipe 1 $\rho_1 = 1.01$ kg/m$^3$
- Velocity in pipe 1 $v_1 = 33.3$ cm/s = $0.333$ m/s
- Volume flow rate in pipe 2 $Q_2 = 221.2$ cm$^3$/hour = $221.2 \times 10^{-6}$ m$^3$/hour
- Density of water in pipe 2 $\rho_2 = 1003$ kg/m$^3$
- Outlet gas density $\rho_o = 2.966$ kg/m$^3$
- Outlet volume flow rate $Q_o = 9.532$ litres/min = $9.532 \times 10^{-3}$ m$^3$/min
3. **Calculate volume flow rate in pipe 1:**
Cross-sectional area of pipe 1:
$$ A_1 = \pi \left(\frac{d_1}{2}\right)^2 = \pi \left(\frac{0.0397}{2}\right)^2 = \pi \times (0.01985)^2 = \pi \times 0.000394 = 0.001238 \, \text{m}^2 $$
Volume flow rate in pipe 1:
$$ Q_1 = A_1 \times v_1 = 0.001238 \times 0.333 = 0.000412 \text{ m}^3/\text{s} $$
Convert $Q_1$ to m$^3$/hour:
$$ Q_1 = 0.000412 \times 3600 = 1.483 \text{ m}^3/\text{hour} $$
4. **Calculate mass flow rate in pipe 1:**
$$ \dot{m}_1 = \rho_1 \times Q_1 = 1.01 \times 1.483 = 1.498 \text{ kg/hour} $$
5. **Calculate mass flow rate in pipe 2:**
$$ \dot{m}_2 = \rho_2 \times Q_2 = 1003 \times 221.2 \times 10^{-6} = 0.222 \text{ kg/hour} $$
6. **Calculate outlet volume flow rate in m$^3$/hour:**
$$ Q_o = 9.532 \times 10^{-3} \text{ m}^3/\text{min} = 9.532 \times 10^{-3} \times 60 = 0.5719 \text{ m}^3/\text{hour} $$
7. **Calculate outlet mass flow rate:**
$$ \dot{m}_o = \rho_o \times Q_o = 2.966 \times 0.5719 = 1.695 \text{ kg/hour} $$
8. **Calculate rate of increase in mass of cylinder:**
$$ \dot{m}_{\text{increase}} = (\dot{m}_1 + \dot{m}_2) - \dot{m}_o = (1.498 + 0.222) - 1.695 = 1.720 - 1.695 = 0.025 \text{ kg/hour} $$
9. **Final answer:**
The rate of increase in the mass of the cylinder is $0.02500$ kg/hour (to 4 significant figures).