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The main water line into a tall building has a pressure of 600 kPa at 5 m below ground level, as shown in<image 1>. A pump brings the pressure up so thatnthe water can be delivered at 200 kPa at the top floor 100 m above ground level. Assume a flow rate of 10 kg/s liquid water at 10°C and neglect any difference in kinetic energy and internal energy u. Find the pump work.


Google: Gemini Pro Vision 1.0
mmmu / Energy_and_PowerPublic

Continuity Eq.:

ρAV=const→ρ1A1V1=ρ2A2V2\rho AV=\text{const}\rightarrow \rho_1A_1V_1=\rho_2A_2V_2

→V1=V2ρ2ρ1A2A1\rightarrow V_1=V_2\frac{\rho_2}{\rho_1}\frac{A_2}{A_1}

Bernoulli Eq.:

p1+12ρ1V12+ρ1gy1=p2+12ρ2V22+ρ2gy2+Wpumpp_1+\frac{1}{2}\rho_1V_1^2+\rho_1gy_1=p_2+\frac{1}{2}\rho_2V_2^2+\rho_2gy_2+W_{pump}

Assuming the water main is large so that the flow velocity there is negligible, and using subscripts 1 for the main and 2 for the top floor, we get:

p1+ρ1gy1=p2+ρ2gy2+Wpumpp_1+\rho_1gy_1=p_2+\rho_2gy_2+W_{pump}

Substituting for V2 and rearranging:

Wpump=(p2−p1)+(ρ2−ρ1)gy2−12ρ2V12(ρ1ρ2A2A1)2W_{pump}=(p_2-p_1)+(\rho_2-\rho_1)gy_2-\frac{1}{2}\rho_2V_1^2\left(\frac{\rho_1}{\rho_2}\frac{A_2}{A_1}\right)^2

From the table,

ρ1=999 kg/m3,ρ2=997 kg/m3\rho_1=999\text{ kg/m}^3, \rho_2=997\text{ kg/m}^3

We don't know A1 and A2, but their ratio will cancel out:

Wpump=600×103 Pa+997×9.81 m/s2×105 mW_{pump}=600\times10^3\text{ Pa}+997\times9.81\text{ m/s}^2\times105\text{ m}

−999×9.81 m/s2×5 m−12×997×(10999A2A1)2 m2/s2-999\times9.81\text{ m/s}^2\times5\text{ m}-\frac{1}{2}\times997\times\left(\frac{10}{999}\frac{A_2}{A_1}\right)^2\text{ m}^2/\text{s}^2

=5.94×106 W=5.94 MW=5.94\times10^6\text{ W}=\boxed{5.94\text{ MW}}


Difficulty: Hard

Subfield: Thermodynamics

Evaluation

Evaluation ID
8580674a-3c69-4085-baab-8b21683eb6ff
Created
February 21, 2024
Raw Output
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        "content": " Continuity Eq.:\n\n$$\\rho AV=\\text{const}\\rightarrow \\rho_1A_1V_1=\\rho_2A_2V_2$$\n\n$$\\rightarrow V_1=V_2\\frac{\\rho_2}{\\rho_1}\\frac{A_2}{A_1}$$\n\nBernoulli Eq.:\n\n$$p_1+\\frac{1}{2}\\rho_1V_1^2+\\rho_1gy_1=p_2+\\frac{1}{2}\\rho_2V_2^2+\\rho_2gy_2+W_{pump}$$\n\nAssuming the water main is large so that the flow velocity there is negligible, and using subscripts 1 for the main and 2 for the top floor, we get:\n\n$$p_1+\\rho_1gy_1=p_2+\\rho_2gy_2+W_{pump}$$\n\nSubstituting for V2 and rearranging:\n\n$$W_{pump}=(p_2-p_1)+(\\rho_2-\\rho_1)gy_2-\\frac{1}{2}\\rho_2V_1^2\\left(\\frac{\\rho_1}{\\rho_2}\\frac{A_2}{A_1}\\right)^2$$\n\nFrom the table,\n\n$$\\rho_1=999\\text{ kg/m}^3, \\rho_2=997\\text{ kg/m}^3$$\n\n>>>>>>We don't know A1 and A2, but their ratio will cancel out:\n\n$$W_{pump}=600\\times10^3\\text{ Pa}+997\\times9.81\\text{ m/s}^2\\times105\\text{ m}$$\n\n$$-999\\times9.81\\text{ m/s}^2\\times5\\text{ m}-\\frac{1}{2}\\times997\\times\\left(\\frac{10}{999}\\frac{A_2}{A_1}\\right)^2\\text{ m}^2/\\text{s}^2$$\n\n$$=5.94\\times10^6\\text{ W}=\\boxed{5.94\\text{ MW}}$$"
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