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Liono4ka [1.6K]

3 years ago

10

An airplane flies with a constant speed of 720 km/hr. How long will it take to travel a distance of 1440 km? Please show your wo

rk

1 answer:

choli [55]3 years ago

6 0

Answer:

2 h

Explanation:

Velocity =Distance/time ⇒ time = distance/speed

= 1440/720

= 2 h

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4 years ago

An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2)

weqwewe [10]

Answer:

(1) The maximum air temperature is 1383.002 K

(2) The rate of heat addition is 215.5 kW

Explanation:

T₁ = 17 + 273.15 = 290.15

$\frac{T_2}{T_1} =r_v^{k - 1} =18^{0.4} =3.17767$

T₂ = 290.15 × 3.17767 = 922.00139

$\frac{T_3}{T_2} =\frac{v_3}{v_2} = r_c = 1.5$

Therefore,

T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K

The maximum air temperature = T₃ = 1383.002 K

(2)

$\frac{v_4}{v_3} =\frac{v_4}{v_2} \times \frac{v_2}{v_3} = \frac{v_1}{v_2} \times \frac{v_2}{v_3} = 18 \times \frac{1}{1.5} = 12$

$\frac{T_3}{T_4} =(\frac{v_4}{v_3} )^{k-1} = 12^{0.4} = 2.702$

Therefore;

$T_4 = \frac{1383.002}{2.702} =511.859 \ k$

$Q_1 = c_p(T_3-T_2)$

Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg

Heat rejected per kilogram is given by the following relation;

$c_v(T_4-T_1)$ = 0.718×(511.859 - 290.15) = 159.187 kJ/kg

The efficiency is given by the following relation;

$\eta = 1-\frac{\beta ^{k}-1}{\left (\beta -1 \right )r_{v}^{k-1}}$

Where:

β = Cut off ratio

Plugging in the values, we get;

$\eta = 1-\frac{1.5 ^{1.4}-1}{\left (1.5 -1 \right )18^{1.4-1}}= 0.5191$

Therefore;

$\eta = \frac{\sum Q}{Q_1}$

$\therefore 0.5191 = \frac{150}{Q_1}$

Heat supplied = $\frac{150}{0.5191} = 288.978 \ hp$

Therefore, heat supplied = 215491.064 W

Heat supplied ≈ 215.5 kW

The rate of heat addition = 215.5 kW.

7 0

4 years ago