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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.: $\rho AV=\text{const}\rightarrow \rho_1A_1V_1=\rho_2A_2V_2$ $\rightarrow V_1=V_2\frac{\rho_2}{\rho_1}\frac{A_2}{A_1}$ Bernoulli Eq.: $p_1+\frac{1}{2}\rho_1V_1^2+\rho_1gy_1=p_2+\frac{1}{2}\rho_2V_2^2+\rho_2gy_2+W_{pump}$ 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: $p_1+\rho_1gy_1=p_2+\rho_2gy_2+W_{pump}$ Substituting for V2 and rearranging: $W_{pump}=(p_2-p_1)+(\rho_2-\rho_1)gy_2-\frac{1}{2}\rho_2V_1^2\left(\frac{\rho_1}{\rho_2}\frac{A_2}{A_1}\right)^2$ From the table, $\rho_1=999\text{ kg/m}^3, \rho_2=997\text{ kg/m}^3$ We don't know A1 and A2, but their ratio will cancel out: $W_{pump}=600\times10^3\text{ Pa}+997\times9.81\text{ m/s}^2\times105\text{ m}$ $-999\times9.81\text{ m/s}^2\times5\text{ m}-\frac{1}{2}\times997\times\left(\frac{10}{999}\frac{A_2}{A_1}\right)^2\text{ m}^2/\text{s}^2$ $=5.94\times10^6\text{ W}=\boxed{5.94\text{ MW}}$ Difficulty: Hard Subfield: Thermodynamics

Input

Output

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.:

$\rho AV=\text{const}\rightarrow \rho_1A_1V_1=\rho_2A_2V_2$

$\rightarrow V_1=V_2\frac{\rho_2}{\rho_1}\frac{A_2}{A_1}$

Bernoulli Eq.:

$p_1+\frac{1}{2}\rho_1V_1^2+\rho_1gy_1=p_2+\frac{1}{2}\rho_2V_2^2+\rho_2gy_2+W_{pump}$

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:

$p_1+\rho_1gy_1=p_2+\rho_2gy_2+W_{pump}$

Substituting for V2 and rearranging:

$W_{pump}=(p_2-p_1)+(\rho_2-\rho_1)gy_2-\frac{1}{2}\rho_2V_1^2\left(\frac{\rho_1}{\rho_2}\frac{A_2}{A_1}\right)^2$

From the table,

$\rho_1=999\text{ kg/m}^3, \rho_2=997\text{ kg/m}^3$

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

$W_{pump}=600\times10^3\text{ Pa}+997\times9.81\text{ m/s}^2\times105\text{ m}$

$-999\times9.81\text{ m/s}^2\times5\text{ m}-\frac{1}{2}\times997\times\left(\frac{10}{999}\frac{A_2}{A_1}\right)^2\text{ m}^2/\text{s}^2$

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

Difficulty: Hard

Subfield: Thermodynamics

Input

Input ID: bc891802-a40b-4c7f-8cd8-a58a1bf3c1e4
Created: February 15, 2024
Permission: Public