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3 years ago

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Vector ~A has a negative x-component 3.07 units in length and a positive y-component 3.17 units in length. When a vector ~B = b1

i + b2j is added to vector ~A the resulting vector has no x component and a negative y component of 4.43 units. Find the value of
a. b1
b. b2

1 answer:

luda_lava [24]3 years ago

4 0

Answer:

a. 3.07 b. 1.26

Explanation:

Given that A = -3.07i + 3.17j and B = b1i + b1j and C = A + B = 0i + 4.43j

Since A + B = -3.07i + 3.17j + b1i + b2j

= (-3.07 + b1)i + (3.17 + b2)j

So,(-3.07 + b1)i + (3.17 + b2)j = 0i + 4.43j

Comparing components,

-3.07 + b1 = 0 (1) and 3.17 + b2 = 4.43 (2)

a. From (1), b1 = 3.07

b. From(2) b2 = 4.43 - 3.17 = 1.26

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Answer:

$a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa$

Explanation:

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Specific volume, $s_2$ :

$\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg$

The pressures at $s_2 \ and\ s_3$ is:

$P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa$

.The thermal efficiency=> maximum temperature at $s_3$ can be obtained from the expansion work at constant pressure during $s_2-s_3$

$\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K$

b.Relative SV and enthalpy at $s_3$ are obtained for the given temperature with interpolation with data from A-17 : $a_r_3=4.553 \ and\ h_3=1909.62kJ/kg$

Relative SV at $s_4$ is

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Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

$n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563$

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

$MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa$

Hence, the mean effective relative pressure is 674.95kPa

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lora16 [44]

Answer:

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