Bernoulli's Equation — Question 7

Section 14-9 · Physics pages 2224–2225

Question 7 A pump adds energy while losses remove it Water at volume rate Q = 0.0020 m3/s passes through a pump and pipe system. From inlet to outlet, pressure rises by 20 kPa, speed rises from 1.0 to 3.0 m/s, and elevation rises by 5.0 m. Irreversible losses equal 2.0 m of water head. Use ρ = 1000 kg/m3 and g = 10 m/s2. 1 2 P A pump supplies mechanical energy; the specified losses must also be included in the energy balance. (a) Find the mechanical energy per unit mass that the pump must add, including the specified loss. (b) Find hydraulic power and required shaft power for pump efficiency 0.80. Explain why the unmodified Bernoulli equation cannot connect these endpoints.
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Question 7 - Solution A pump adds energy while losses remove it 1 2 P (a) Derivation and calculation Per unit mass, the extended mechanical-energy balance gives wp = p2 −p1 ρ + v2 2 −v2 1 2 + g(z2 −z1) + ghL. wp = 20000 1000 + 9 −1 2 + 10(5) + 10(2) = 94.0 J/kg . (b) Result and interpretation ˙m = ρQ = (1000)(.00200) = 2.00 kg/s. Phydraulic = ˙mwp = 188 W , Pshaft = Phydraulic/.80 = 235 W . Unmodified Bernoulli omits both the pump addition and dissipation; it cannot connect these endpoints. Physical check The pump supplies 74 J/kg of endpoint energy increase plus 20 J/kg of pipe loss.

Question 7

A pump adds energy while losses remove it

Water at volume rate Q=0.0020m3/sQ=0.0020\,\mathrm{m^3/s} passes through a pump and pipe system. From inlet to outlet, pressure rises by 20kPa20\,\mathrm{kPa}, speed rises from 1.01.0 to 3.0m/s3.0\,\mathrm{m/s}, and elevation rises by 5.0m5.0\,\mathrm{m}. Irreversible losses equal 2.0m2.0\,\mathrm{m} of water head. Use ρ=1000kg/m3\rho=1000\,\mathrm{kg/m^3} and g=10m/s2g=10\,\mathrm{m/s^2}.

Diagram shown in the original SVG worksheet.

  1. Find the mechanical energy per unit mass that the pump must add, including the specified loss.

  2. Find hydraulic power and required shaft power for pump efficiency 0.800.80. Explain why the unmodified Bernoulli equation cannot connect these endpoints.

Question 7 - Solution

A pump adds energy while losses remove it

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Per unit mass, the extended mechanical-energy balance giveswp=p2−p1ρ+v22−v122+g(z2−z1)+ghL.w_p=\frac{p_2-p_1}{\rho}+\frac{v_2^2-v_1^2}{2}+g(z_2-z_1)+gh_L.wp=200001000+9−12+10(5)+10(2)=94.0J/kg.\boxed{w_p=\frac{20000}{1000}+\frac{9-1}{2}+10(5)+10(2)=94.0\,\mathrm{J/kg}}.

(b) Result and interpretation

ṁ=ρQ=(1000)(.00200)=2.00kg/s.\dot m=\rho Q=(1000)(.00200)=2.00\,\mathrm{kg/s}.Phydraulic=ṁwp=188W,Pshaft=Phydraulic/.80=235W.\boxed{P_{\mathrm{hydraulic}}=\dot mw_p=188\,\mathrm W},\qquad\boxed{P_{\mathrm{shaft}}=P_{\mathrm{hydraulic}}/.80=235\,\mathrm W}.Unmodified Bernoulli omits both the pump addition and dissipation; it cannot connect these endpoints.

Physical check

The pump supplies 74 J/kg of endpoint energy increase plus 20 J/kg of pipe loss.