The work done by a pump is related to the pressure difference it needs to overcome and the fluid properties. For an incompressible fluid, the specific work (work per unit mass) supplied to a pump is given by the integral of pressure with respect to specific volume: $w_p = \int_{P_1} ^{P_2} v dP$. Since water is assumed to be incompressible, its specific volume ($v$) is constant. Therefore, the integral simplifies to: $w_p = v \int_{P_1}^{P_2} dP = v (P_2 - P_1)$. Given: Initial pressure, $P_1 = 10 \text{ kPa} = 10 \times 10^3 \text{ Pa}$ Final pressure, $P_2 = 2 \text{ MPa} = 2 \times 10^6 \text{ Pa}$ Density of water, $\rho = 1000 \text{ kg/m}^3$. The specific volume is the reciprocal of density: $v = 1/\rho = 1/1000 \text{ m}^3/\text{kg}$. Now, substitute these values into the formula for pump work: $w_p = \frac{1}{1000} \text{ m} ^3/\text{kg} \times (2 \times 10^6 \text{ Pa} - 10 \times 10^3 \text{ Pa})$ $w_p = \frac{1}{1000} \text{ m}^3/\text{kg} \times (2000000 \text{ Pa} - 10000 \text{ Pa})$ $w_p = \frac{1}{1000} \text{ m}^3/\text{kg} \times 1990000 \text{ Pa}$ $w_p = 1990 \text{ Pa} \cdot \text{m}^3/\text{kg} $ Since $1 \text{ Pa} \cdot \text{m}^3 = 1 \text{ N/m}^2 \cdot \text{m}^3 = 1 \text{ N} \cdot \text{m} = 1 \text{ J}$. So, $w_p = 1990 \text{ J/ kg}$. To convert Joules to kilojoules, divide by 1000: $w_p = 1990 / 1000 \text{ kJ/kg} = 1.99 \text{ kJ/kg}$.
In a Rankine cycle, water is pumped from 10 kPa to 2 MPa. Assuming water to be incompressible, the specific work (in kJ/kg) supplied to the pump is given by (Assume water density = 1000 kg/m³) - If the question has multiple statements, format using ... - For Assertion/Reason, retain 'Statement-I' and 'Statement-II' clearly. - For matching pairs, use HTML
Asked in Assistant Eng. (Mechanical) Competitive Examination, 2023 · Topic: Heat and Thermodynamics (Zeroth and First Laws of Thermodynamics (Internal Energy, Work, Heat, Enthalpy))
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