An annulus of narrow clearance causes a large pressure drop and can be used to measure fluid viscosity accurately. A smooth annulus has: L=1.0 m,a=50 mm,b=49 mmL = 1.0\ \text{m}, \qquad a = 50\ \text{mm}, \qquad b = 49\ \text{mm}Oil flows through the annulus at Q=2.60×10−3 m3/s Q = 1.400\times10^{-3}\ \text{m}^3/\text{s} If the measured pressure drop is Δp=250 kPa\Delta p = 250\ \text{kPa} What is the oil viscosity μ\mu in kg/m.s?  

What is displayed by letter E? IMG_0052.jpeg  

The rate of heat loss through a window, Q˙loss\dot{Q}_{loss} ​loss​, depends on: Temperature difference ΔT\Delta T Surface area AA Thermal resistance RR (units: ft²·hr·°F/Btu) Using dimensional reasoning, express the relationship for heat loss. Select the correct answer: A. Q˙loss=Const A2RΔT\displaystyle \dot{Q}_{loss} = \text{Const}\,\frac{A^2 R}{\Delta T} B. Q˙loss=Const ΔTAR\displaystyle \dot{Q}_{loss} = \text{Const}\,\frac{\Delta T}{A R} C. Q˙loss=Const R2AΔT\displaystyle \dot{Q}_{loss} = \text{Const}\,\frac{R^2}{A \Delta T} D. Q˙loss=Const AΔTR\displaystyle \dot{Q}_{loss} = \text{Const}\,\frac{A \Delta T}{R}

Water flows upward in a 6 cm6\ \text{cm}-diameter pipe at a velocity of 4 m/s4\ \text{m/s}. The pipe length between points (1) and (2) is 5 m5\ \text{m}, and point (2) is 3 m3\ \text{m} higher than point (1). A mercury manometer connected between points (1) and (2) shows a reading of: h=120 mmh = 120\ \text{mm}with pressure at point (1) higher than at point (2). Given: γHg=133,100 N/m3,γw=9,790 N/m3\gamma_{\text{Hg}} = 133{,}100\ \text{N/m}^3, \quad \gamma_{w} = 9{,}790\ \text{N/m}^3 Determine the head loss hfh_f in meters. hint: use Bernoulli equation to calculate pressure drop then calculate the head losses.

  A viscous liquid forms a constant-thickness film of thickness hhh flowing in laminar motion down a plate inclined at an angle θ\thetaθ, as shown in the figure. The velocity profile is given by u(y)=Cy(2h−y),v=0,w=0u(y)=Cy(2h-y), \qquad v=0, \qquad w=0u(y)=Cy(2h−y),v=0,w=0 where CCC is a constant. Task Using the momentum equation for steady laminar flow down an incline, determine the constant CCC in terms of the fluid properties and the plate angle θ\thetaθ.     Select the correct answer: A. C=gsin⁡θ2\displaystyle C=\frac{g\sin\theta}{2} ​ B. C=ρsin⁡θμ\displaystyle C=\frac{\rho \sin\theta}{\mu}  ​ C. C=ρgsin⁡θ2μ\displaystyle C=\frac{\rho g\sin\theta}{2\mu} ​ D. C=2ρgsin⁡θμ2\displaystyle C=\frac{2\rho g\sin\theta}{\mu^2} 

SAE 50W oil at 20°C flows from one tank to another through a tube 175 cm long and 5 cm in diameter. Estimate the flow rate in m3/hr if z1 = 2 m and z2 = 0.8 m. 

When a tank of length LL and water depth hh is disturbed, the free surface oscillates with frequency Ω\Omega. The frequency is assumed to depend on: Tank length LL Water depth hh Fluid density ρ\rho Gravitational acceleration gg Using dimensional analysis, express this relationship in dimensionless form. Select the correct answer: A. ΩL2g=f ⁣(hL)\displaystyle \Omega \sqrt{\frac{L^2}{g}} = f\!\left(\frac{h}{L}\right) B. ΩLg=f ⁣(hL)\displaystyle \Omega \sqrt{\frac{L}{g}} = f\!\left(\frac{h}{L}\right) C. ΩL2g=f ⁣(hL)\displaystyle \Omega \frac{L^2}{g} = f\!\left(\frac{h}{L}\right) D. ΩLg=f ⁣(hL)\displaystyle \Omega \frac{L}{g} = f\!\left(\frac{h}{L}\right)

An 84% efficient pump delivers water at 20∘C20^\circ\text{C} from one reservoir to another reservoir 20 ft higher, as shown in the figure. The piping system consists of: 60 ft60\ \text{ft} of galvanized iron pipe Pipe diameter D=2 inD = 2\ \text{in} A reentrant entrance Two screwed 90∘90^\circ long-radius elbows A screwed-open gate valve A sharp exit The flow rate isQ=0.4 ft3/s For water at 20∘C20^\circ\text{C}, useρ=1.94 slug/ft3,μ=2.09×10−5 slug/(ft. s)\rho = 1.94\ \text{slug/ft}^3, \qquad \mu = 2.09\times10^{-5}\ \text{slug/(ft\cdot s)} For galvanized iron, takeε=0.0005 ft Assume the system is analyzed without the 6° well-designed conical expansion at the exit.   Determine the input power required in horsepower.   Some Information need to be extracted regard the minor losses (K value) use the table and figures provided.

Water flows from a reservoir through a cast iron pipe and a turbine, then exits to the atmosphere. Given: Pipe length: L=125 mL = 125\ \text{m} Diameter: D=5 cmD = 5\ \text{cm} Elevation difference: Δz= 31 m Flow rate: Q=0.004 m3/sQ = 0.004\ \text{m}^3/\text{s} Fluid properties (water at 20∘C20^\circ C: ρ=998 kg/m3,μ=0.001 kg/(m. s) Cast iron roughness: ε=0.26 mm,f≈0.0316\varepsilon = 0.26\ \text{mm}, \quad f \approx 0.0316 What is the power extracted by the turbine (W)? Please extract the required minor losses K value from the figure and tables to solve the problem.   

Water flows upward in a 6 cm-diameter pipe at a velocity of 4 m/s. The pipe length between points (1) and (2) is 5 m, and point (2) is 3 m higher than point (1). A mercury manometer connected between points (1) and (2) shows a reading of: h=225 mmwith pressure at point (1) higher than at point (2). Given:   γHg=133,100 N/m3,γw=9,790 N/m3 Determine the head loss hf in meters. hint: use Bernoulli equation to calculate pressure drop then calculate the head losses.