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4 years ago

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For turbulent now the friction factor is function of (Reynolds number —surface roughness -both the Reynolds number and the surfa

ce roughness) of the pipe.

1 answer:

Ket [755]4 years ago

7 0

Answer:

Both Reynolds and surface roughness

Explanation:

For turbulent flow friction factor is a function of both Reynolds and surface roughness of the pipe.But on the other hand for laminar flow friction factor is a function of only Reynolds number.

Friction factor for turbulent flow:

1. For smooth pipe

$f=0.0032+\dfrac{0.221}{Re^{0.237}}$

$5\times 10^4$

2. For rough pipe

$\dfrac{1}{\sqrt f}=2\ log_{10}\frac{R}{K}+1.74$

Where R/K is relative roughness

Friction factor for laminar flow:

$f=\dfrac{64}{Re}$

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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.

5 0

3 years ago

Helium is used as the working fluid in a Brayton cycle with regeneration. The pressure ratio of the cycle is 8, the compressor i

Temka [501]

Answer:

Explanation:

Find the temperature at exit of compressor

$T_2=300 \times 8^{\frac{1.667-1}{1.667} }\\=689.3k$

Find the work done by the compressor

$\frac{W}{m} =c_p(T_2-T_1)\\\\=5.19(689.3-300)\\=2020.4kJ/kg$

Find the actual workdone by the compressor

$\frac{W}{m} =n_c(\frac{W}{m} )\\\\=1 \times 2020.4kJ/kg$

Find the temperature at exit of the turbine

$T_4=\frac{1800}{8^{\frac{1.667-1}{1.667} }} \\\\=787.3k$

Find the actual workdone by the turbine

$1 \times 5.19 (1800-783.3)\\=5276.6kJ/kg$

Find the temperature of the regeneration

$\epsilon = \frac{T_5-T_2}{T_4-T_2} \\\\0.75=\frac{T_5-689.3}{783.3-689.3} \\\\T_5=759.8k$

Find the heat supplied

$Q_i_n=c_p(T_3-T_5)\\\\=5.19(1800-759.8)\\\\=5388.2kJ/kg$

Find the thermal efficiency

$n_t_h=\frac{W_t-W_c}{Q_i_n} \\\\=\frac{5276.6-2020.4}{5388.2} \\\\n_t_h=60.4$

60.4%

Find the mass flow rate

$m=\frac{W_net}{P} \\\\\frac{60 \times 10^3}{5276.6-2020.4} \\\\=18.42$

Find the actual workdone by the compressor

$\frac{W_c}{m} =\frac{(\frac{W}{m} )}{n_c} \\\\=\frac{2020.4}{0.8} \\\\=2525.5kg$

Find the actual workdone by the turbine

$\frac{W_t}{m} =n_t(\frac{W}{m} )\\\\=0.8 \times5.19(1800-783.3)\\\\=4221.2kJ/kg$

Find the temperature of the compressor exit

$\frac{W_t}{m} =c_p(T_2_a-T_1)\\2525.5=5.18(T_2_a-300)\\T_2_a=787.5k$

Find the temperature at the turbine exit

$4221.2=5.18(1800-T_4_a)\\\\T_4_a=985k$

Find the temperature of regeneration

$\epsilon =\frac{T_5-T_2}{T_4-T_2}\\\\0.75=\frac{T_5-787.5}{985-787.5}\\\\T_5=935.5k$

6 0

3 years ago

Read 2 more answers

A gas is compressed from an initial volume of 0.42 m3 to a final volume of 0.12 m3. During the quasi-equilibrium process, the pr

Georgia [21]

Answer:

$W=-52 800\ \text{J}=-52.8\ \text{kJ}$

Explanation:

First I sketched the compression of the gas with the help of the given pressure change process relation. That is your pressure change due to change in volume.

To find the area underneath the curve (the same as saying to find the work done) you should integrate the given relation for pressure change:

$W=\int_{0.42}^{0.12}-1200V+500dV=-52.8\ \text{kJ}$

6 0

3 years ago

1. There are two categories (shapes) for the Virginia driver's license. The ________________________ shape license represents th

dolphi86 [110]

Answer:

Vertical; horizontal.

Explanation:

The Virginia Department of Motor Vehicles started issuing sets of newly designed driver's licenses to drivers in 2009. Although, the cards that were issued to drivers prior to the introduction of the new cards remained valid until they were expired.

There are two categories (shapes) for the Virginia driver's license. The vertical shape license represents the driver who is under the age of 21, and the horizontal shaped license represents the driver who is over the age of 21.

Additionally, the Virginia's driver license (vertical in shape) issued to drivers who are under the age of 21 has a background image of a dogwood flower while the horizontal shaped license issued to drivers who are over the age of 21 has a background image of the state capitol.

5 0

3 years ago

The minimum fresh air requirement of a residential building is specified to be 0.35 air changes per hour (ASHRAE, Standard 62, 1

Thepotemich [5.8K]

Answer:

a) The flow capacity of the fan is 3150 L/min

d) the minimum diameter is 0.11 m

Explanation:

given data:

A = area of residence = 200 m²

h = height of building = 2.7 m

Percentage of air that must be replaced by fresh air is 35%

v = velocity of air in the duct = 5.5 m/s

a) The volume of the entire building is:

$Volume=2.7*200=540m^{3}$

The flow capacity of the fan is equal to:

$Flow=\frac{0.35*540}{60} =3.15m^{3} /min$

$Flow=3.15\frac{m^{3} }{min} *\frac{1000L}{1m^{3} } =3150L/min$

b) The volume flow rate of fresh air is equal to:

$Flow=\frac{\pi *d^{2} }{4} V\\d=\sqrt{\frac{4*Flow}{\pi V} } =\sqrt{\frac{4*3.15}{\pi *5.5*60} } =0.11m$

6 0

4 years ago