1. **Problem statement:** (c) Given the pressure drop formula in a water pipeline system: $$P = 4QL + 80$$ where $P$ is the pressure drop in kPa, $Q$ is the flow rate in m³/s, and $L$ is the pipe length in meters. (a) Rearrange the formula to solve for $Q$. (b) Calculate $Q$ when $P = 80$ kPa and $L = 10$ m. --- 2. **Rearranging the formula to find $Q$:** Start with: $$P = 4QL + 80$$ Subtract 80 from both sides: $$P - 80 = 4QL$$ Divide both sides by $4L$ to isolate $Q$: $$Q = \frac{P - 80}{4L}$$ Show cancellation: $$Q = \frac{P - 80}{\cancel{4} \cancel{L}} \times \frac{1}{\cancel{4} \cancel{L}}$$ (Here, just indicating division by $4L$.) --- 3. **Calculate $Q$ for given values:** Given $P = 80$ kPa and $L = 10$ m, substitute into the formula: $$Q = \frac{80 - 80}{4 \times 10} = \frac{0}{40} = 0$$ So, the flow rate $Q$ is 0 m³/s. --- 4. **Problem statement:** (d) Given the heat transfer rate formula in a cooling system: $$H = 5A - 20$$ where $H$ is the heat transfer rate in kW and $A$ is the surface area in m². (a) Rearrange the formula to solve for $A$. (b) Calculate $A$ when $H = 130$ kW. --- 5. **Rearranging the formula to find $A$:** Start with: $$H = 5A - 20$$ Add 20 to both sides: $$H + 20 = 5A$$ Divide both sides by 5: $$A = \frac{H + 20}{5}$$ Show cancellation: $$A = \frac{H + 20}{\cancel{5}} \times \frac{1}{\cancel{5}}$$ --- 6. **Calculate $A$ for given value:** Given $H = 130$ kW, substitute into the formula: $$A = \frac{130 + 20}{5} = \frac{150}{5} = 30$$ So, the surface area $A$ is 30 m². --- **Final answers:** (c)(a) $Q = \frac{P - 80}{4L}$ (c)(b) $Q = 0$ m³/s (d)(a) $A = \frac{H + 20}{5}$ (d)(b) $A = 30$ m²