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3 years ago

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Water enters a leaky cylindrical tank (D = 1 ft) at a rate of 8 ft3/min. Water leaks out of the tank at a rate of 17% of the flo

w into the tank. At what rate will water rise in the tank (answer in ft/min)?

1 answer:

stellarik [79]3 years ago

3 0

Answer:

Water would rise in the tank at a rate of 8.45 ft/min

Explanation:

Diameter of leaky cylindrical tank (D) = 1 ft

Base area = πD^2/4 = 3.142×1^2/4 = 0.7855 ft^2

Volumetric flow rate at which water enters the tank = 8 ft^3/min

Volumetric flow rate at which water leaks out = 0.17 × 8 = 1.36 ft^3/min

Volumetric flow rate at which water rises = 8 - 1.36 = 6.64 ft^3/min

Rate at which water would rise in ft/min = volumetric flow rate at which water rises ÷ base area of the cylinder = 6.64 ft^3/min ÷ 0.7855 ft^2 = 8.45 ft/min

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A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

abruzzese [7]

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂ = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

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

Otto cycle T-S diagram

T₂ = 288.706* $6.25^{0.393}$ = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

$\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}$

T₄ = 2888.89 / $6.25^{0.393}$ = 1406.5 K

Work done, W = $c_v$ ×(T₃ - T₂) - $c_v$ ×(T₄ - T₁)

0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.

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True or false? if i were to hook up an ac voltage source to a resistor, the voltage drop across the resistor would be in phase w

hodyreva [135]

Answer: True

Explanation:

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2 years ago

A sheet of steel 3-mm thick has nitrogen atomospheres on both sides at 900 C and is permitted to achieve a steady-state di usion

kati45 [8]

Answer:

$X_B = 1.8 \times 10^{-3} m = 1.8 mm$

Explanation:

Given data:

Diffusion constant for nitrogen is $= 1.85\times 10^{-10} m^2/s$

Diffusion flux $= 1.0\times 10^{-7} kg/m^2-s$

concentration of nitrogen at high presuure = 2 kg/m^3

location on which nitrogen concentration is 0.5 kg/m^3 ......?

from fick's first law

$J = D \frac{C_A C_B}{X_A X_B}$

Take C_A as point on which nitrogen concentration is 2 kg/m^3

$x_B = X_A + D\frac{C_A -C_B}{J}$

Assume X_A is zero at the surface

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$X_B = 1.8 \times 10^{-3} m = 1.8 mm$

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3 years ago

Explain the working and performance of a centrifugal clutch with a sketch

FinnZ [79.3K]

Answer:

A centrifugal clutch works, as the name suggests, through centrifugal force.

One of the most common methods is by mounting the clutch onto the parallel or taper crank shaft of the engine.

When the crank shaft rotates the shaft of the clutch rotates at the same speed as the engine

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3 years ago

Read 2 more answers

The motor of an electric vehicle runs at an average of 50 hp for one hour and 25 minutes. Determine the total energy. Write the

Sunny_sXe [5.5K]

Answer:

The total energy of the motor of the electric vehicle is 1.902 × 10⁸ joules.

Explanation:

Power is the rate of change of work in time, since given input is average power, the total energy ( $\Delta E$ ) of the motor of the electric vehicle, measured in joules, is determined by this formula:

$\Delta E = \dot W \cdot \Delta t$

Where:

$\dot W$ - Average power, measured in watts.

$\Delta t$ - Time, measured in seconds.

Now, let convert average power and time into watts and seconds, respectively:

Average Power

$\dot W = (50\,hp)\times \frac{746\,W}{1\,hp}$

$\dot W = 3.730\times 10^{4}\,W$

Time

$\Delta t = (1\,h)\times \frac{3600\,s}{1\,h} + (25\,min)\times \frac{60\,s}{1\,min}$

$\Delta t = 5.100\times 10^{3}\,s$

Then, the total energy is:

$\Delta E = (3.730\times 10^{4}\,W)\cdot (5.100\times 10^{3}\,s)$

$\Delta E = 1.902\times 10^{8}\,J$

The total energy of the motor of the electric vehicle is 1.902 × 10⁸ joules.

6 0

3 years ago