A pump discharges 50 ton of water per hour at the height of 8 m. The overall efficiency of the pumping system being 69%. Calculate the energy consumed by the pump in 2 hours, expressed in MJ. A. 12.38 C. 12.68 B. 11.38 D. 11.68 Solution Click here to… Jhun Vert Thu, 07/04/2024 - 19:21 Solution Click here to expand or collapse this section For 1 hour: $\text{Energy} = Wh$ $\text{Energy} = \left( 50 ~ \text{ton} \times 1000 ~ \dfrac{\text{kg}}{\text{ton}} \times 9.81 ~ \text{m/sec}^2 \right) 8 ~ \text{m}$ $\text{Energy} = 3,924,000 ~ \text{N}\cdot\text{m or J}$ $\text{Energy} = 3.924 ~ \text{MJ}$ For 2 hours: $\text{Energy} = 2(3.924)$ $\text{Energy} = 7.848 ~ \text{MJ}$ Efficiency: $\text{Efficiency} = \dfrac{\text{Output}}{\text{Input}}$ $0.69 = \dfrac{7.848}{\text{Input}}$ $\text{Input} = 11.3739 ~ \text{MJ}$ Log in or register to post comments Log in or register to post comments
Solution Click here to… Jhun Vert Thu, 07/04/2024 - 19:21 Solution Click here to expand or collapse this section For 1 hour: $\text{Energy} = Wh$ $\text{Energy} = \left( 50 ~ \text{ton} \times 1000 ~ \dfrac{\text{kg}}{\text{ton}} \times 9.81 ~ \text{m/sec}^2 \right) 8 ~ \text{m}$ $\text{Energy} = 3,924,000 ~ \text{N}\cdot\text{m or J}$ $\text{Energy} = 3.924 ~ \text{MJ}$ For 2 hours: $\text{Energy} = 2(3.924)$ $\text{Energy} = 7.848 ~ \text{MJ}$ Efficiency: $\text{Efficiency} = \dfrac{\text{Output}}{\text{Input}}$ $0.69 = \dfrac{7.848}{\text{Input}}$ $\text{Input} = 11.3739 ~ \text{MJ}$ Log in or register to post comments
Solution Click here to…