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professor190 [17]

4 years ago

11

A jetliner flying at an altitude of 10,000 m has a Mach number of 0.5. If the jetliner has to drop down to 1000 m but still main

tain the Mach number (0.5), how much faster must the jetliner fly?

1 answer:

andreev551 [17]4 years ago

6 0

Answer

Assuming

At 10000 m height temperature T = -55 C = 218 K

At 1000 m height temperature T = 0 C = 273 K

$\dfrac{V_1}{C_1} =\dfrac{V_2}{C_2} = 0.5$

R = 287 J/kg K

$C_1 = \sqrt{\gamma RT_1} = \sqrt{1.4\times 287\times 218} = 295 m/s$

$C_2 = \sqrt{\gamma RT_2} = \sqrt{1.4\times 287\times 273} = 331 m/s$

$V_2 = \dfrac{V_1}{C_1}\timesC_2$

V₂ = V₁ ×1.1222

V₁ = 0.5 × C₁ = 0.5 × 295 = 147.5 m/s

V₂ = 1.1222 × 147.5 = 165.49 m/s

so, the jetliner need to increase speed by ( V₂ -V₁ )

= 165.49 - 147.5

= 17.5 m/s

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Answer:

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b) the number-average molecular weight for a PTFE for which the degree of polymerization is 10,000 щₙ = 1000150 g/mole

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b)

To compute the number-average molecular weight for PTFE of which the degree of polymerization is 10,000

we say

DP = щₙ / MW

where щₙ is the number of average molecular weight,

MW is the repeat unit molecular weight give as 100.015 g/mole

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10,000 = щₙ / 100.015

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Heat transferred from the refrigerated space = 95.93 kJ/kg

Work required = 18.45 kJ/kg

Coefficient of performance = 3.61

Quality at the beginning of the heat addition cycle = 0.37

Explanation:

From figure

$Q_H$ is heat rejection process

$Q_L$ is heat transferred from the refrigerated space

$T_H$ is high temperature = 50 °C + 273 = 323 K

$T_L$ is low temperature = -20 °C + 273 = 253 K

$W_{net}$ is net work of the cycle (the difference between compressor's work and turbine's work)

Coefficient of performance of a Carnot refrigerator $(COP_{ref})$ is calculated as

$COP_{ref} = \frac{T_L}{T_H - T_L}$

$COP_{ref} = \frac{253 K}{323 K - 253 K}$

$COP_{ref} = 3.61$

From figure it can be seen that heat rejection is latent heat of vaporisation of R-12 at 50 °C. From table

$Q_H = 122.5 kJ/kg$

From coefficient of performance definition

$COP_{ref} = \frac{Q_L}{Q_H - Q_L}$

$Q_H \times COP_{ref} = (COP_{ref} + 1) \times Q_L$

$Q_L = \frac{Q_H \times COP_{ref}}{(COP_{ref} + 1)}$

$Q_L = \frac{122.5 kJ/kg \times 3.61}{(3.61 + 1)}$

$Q_L = 95.93 kJ/kg$

Energy balance gives

$W_{net} = Q_H - Q_L$

$W_{net} = 122.5 kJ/kg - 95.93 kJ/kg$

$W_{net} = 26.57 kJ/kg$

Vapor quality at the beginning of the heat addition cycle is calculated as (f and g refer to saturated liquid and saturated gas respectively)

$x = \frac{s_1 - s_f}{s_g - s_f}$

From figure

$s_1 = s_4 = 1.165 kJ/(K kg)$

Replacing with table values

$x = \frac{1.165 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}{1.571 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}$

$x = 0.37$

Quality can be computed by other properties, for example, specific enthalpy. Rearrenging quality equation we get

$h_1 = h_f + x \times (h_g - h_f)$

$h_1 = 181.6 kJ/kg + 0.37 \times 162.1 kJ/kg$

$h_1 = 241.58 kJ/kg$

By energy balance, $W_{t}$ turbine's work is

$W_{t} = |h_1 - h_4|$

$W_{t} = |241.58 kJ/kg - 249.7 kJ/kg|$

$W_{t} = 8.12 kJ/kg$

Finally, $W_{c}$ compressor's work is

$W_{c} = W_{net} + W_{t}$

$W_{c} = 26.57 kJ/kg + 8.12 kJ/kg$

$W_{c} = 34.69 kJ/kg$

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