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3 years ago

1. Explain the term enginecompression

Engineering

2 answers:

GalinKa [24]3 years ago

5 0

Answer:

Compression, decrease in volume of any object or substance resulting from applied stress. Compression may be undergone by solids, liquids, and gases and by living systems

alexandr1967 [171]3 years ago

3 0

$\huge{\mathbb{QUESTION:}}$

1. Explain the term engine

compression

$\huge{\mathbb{ANSWER:}}$

compression confines and presses a mixture of air and fuel into a small volume within the area of the engine's cylinder. This process presses together all the molecules under very high pressure.

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Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of "84" m3/min and exits at 12

Nina [5.8K]

Answer:

W = - 184.8 kW

Explanation:

Given data:

$P_1 = 1.05 bar$

$T_1 = 300K$

$\dot V_1 = 84 m^3/min$

$P_2 = 12 bar$

$T_2 = 400 K$

We know that work is done as

$W = - [ Q + \dor m[h_2 - h_1]]$

for $P_1 = 1.05 bar, T_1 = 300K$

density of air is 1.22 kg/m^3 and $h_1 = 300 kJ/kg$

for $P_2 = 12 bar, T_2 = 400 K$

$h_2 = 400 kj/kg$

$\dot m = \rho \times \dor v_1 = 1.22 \frac{84}{60} =1.708 kg/s$

W = -[14 + 1.708[400-300]]

W = - 184.8 kW

8 0

3 years ago

A thick aluminum block initially at 26.5°C is subjected to constant heat flux of 4000 W/m2 by an electric resistance heater whos

Yanka [14]

Given Information:

Initial temperature of aluminum block = 26.5°C

Heat flux = 4000 w/m²

Time = 2112 seconds

Time = 30 minutes = 30*60 = 1800 seconds

Required Information:

Rise in surface temperature = ?

Answer:

Rise in surface temperature = 8.6 °C after 2112 seconds

Rise in surface temperature = 8 °C after 30 minutes

Explanation:

The surface temperature of the aluminum block is given by

$T_{surface} = T_{initial} + \frac{q}{k} \sqrt{\frac{4\alpha t}{\pi} }$

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.

After t = 2112 sec:

$T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (2112)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.51098)\\\\T_{surface} = 26.5 + 8.6\\\\T_{surface} = 35.1\\\\$

The rise in the surface temperature is

Rise = 35.1 - 26.5 = 8.6 °C

Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.

After t = 30 mins:

$T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (1800)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.4717)\\\\T_{surface} = 26.5 + 7.96\\\\T_{surface} = 34.5\\\\$