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4 years ago

12

A water pump delivers 3 hp of shaft power when operating. If the pressure differential between the outlet and the inlet of the p

ump is measured to be 1.2 psi when the flow rate is 8 ft^3/s and the changes in velocity and elevation are negligible, determine the mechanical efficiency of this pump.

1 answer:

Natali [406]4 years ago

5 0

Answer:

Mechanical Efficiency = 83.51%

Explanation:

Given Data:

Pressure difference = ΔP=1.2 Psi

Flow rate = $V=8ft^3/s\\$

Power of Pump = 3 hp

Required:

Mechanical Efficiency

Solution:

We will first bring the change the units of given data into SI units.

$P=1.2*6.895 = 8.274KPa\\V=8*0.00283=0.226 m^3/s\\P=3*0.746=2.238KW$

Now we will find the change in energy.

Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.

Thus change in energy is

$=(Mass * change in P)/density\\= \frac{M*P}{p}\\\\$

As we know that Mass = Volume x density

substituting the value

Energy = Volume * density x ΔP / density

Change in energy = Volumetric flow x ΔP

Change in energy = 0.226 x 8.274 = 1.869 KW

Now mechanical efficiency = change in energy / work done by shaft

Efficiency = 1.869 / 2.238

Efficiency = 0.8351 = 83.51%

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Answer:

Conceptual or planning design phase.

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3 years ago

Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held

77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} }$

${T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} } = 280.15 \times \left (9 \right )^{\frac{1.333-1}{1.333} } = 485.03\ K$

The heat supplied = $\dot {m}_f$ × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = $\dot m$ · $c_p(T_3 - T_2)$

$\dot m$ = 20 kg/s

The heat supplied = 20* $c_p(T_3 - T_2)$ = 21,350 kJ/s

$c_p$ = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

$\dfrac{T_5}{T_4} = \dfrac{2}{1.333 + 1}$

T₅ = 1208.42*(2/2.333) = 1035.94 K

$C_j = \sqrt{\gamma \times R \times T_5}$ = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = $\dot m$ × $C_j$ = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

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3 years ago

Based on bonding theory, explain why coefficient of expansion increases when you consider ceramics, metals and polymers

Mrrafil [7]

Answer:

The ratio that a material expands in accordance with changes in temperature is called the coefficient of thermal expansion. Because Fine Ceramics possess low coefficients of thermal expansion, their distortion values, with respect to changes in temperature, are low. The coefficients of thermal expansion depend on the bond strength between the atoms that make up the materials.

Explanation:

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4 years ago

A certain solar energy collector produces a maximum temperature of 100°C. The energy is used in a cyclic heat engine that operat

Gwar [14]

Answer:

$\eta _{max} = 0.2413 = 24.13%$

$\eta' _{max} = 0.5061 = 50.61%$

Given:

$T_{1max} = 100^{\circ} = 273 + 100 = 373 K$

operating temperature of heat engine, $T_{2} = 10^{\circ} = 273 + 10 = 283 K$

$T_{3max} = 300^{\circ} = 273 + 300 = 573 K$

Solution:

For a reversible cycle, maximum efficiency, $\eta _{max}$ is given by:

$\eta _{max} = 1 - \frac{T_{2}}{T_{1max}}$

$\eta _{max} = 1 - \frac{283}{373} = 0.24$

$\eta _{max} = 0.2413 = 24.13%$

Now, on re designing collector, maximum temperature, $T_{3max}$ changes to $300^{\circ}$ , so, the new maximum efficiency, $\eta' _{max}$ is given by:

$\eta' _{max} = 1 - \frac{T_{2}}{T_{3max}}$

$\eta _{max} = 1 - \frac{283}{573} = 0.5061$

$\eta _{max} = 0.5061 = 50.61%$

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4 years ago