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2 years ago

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A gas-filled pneumatic strut in an automobile suspension system behaves like a piston-cylinder apparatus. At one instant when th

e piston is L=0:15 maway from the closed end of the cylinder, the gas density is uniform at rho=18 kg=m3 and the piston begins to move away from the closed end at V =12m=s. Assume as a simple model that the gas velocity is one-dimensional and proportional to distance from the closed end; it varies linearly from zero at the end to u=V at the piston. Find the rate of change of gas density at this instant. Obtain an expression for the average density as a function

1 answer:

natima [27]2 years ago

4 0

Answer: I will ask my dad

Explanation:

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3/63 A 2‐kg sphere S is being moved in a vertical plane by a robotic arm. When the angle θ is 30°, the angular velocity of the a

miss Akunina [59]

Answer:

Ps=19.62N

Explanation:

The detailed explanation of answer is given in attached files.

5 0

2 years ago

Twenty distinct cars park in the same parking lot every day. Ten of these cars are US-made, while the other ten are foreign-made

Zina [86]

Answer:

Total no. of ways to line up cars is 20! = 2.43 c 10^18

Probability that the cars alternate is 0.00001 or 0.001%

Explanation:

Since, the position of a car is random.Therefore, number ways in which cars can line up is given as:

<u>No. of ways = 20! = 2.43 x 10^18</u>

For the probability that cars alternate, two groups will be formed, one consisting of US-made 10 cars and other containing 10 foreign made. The number of favorable outcomes for this can be found out as the arrangements of 2! between these groups multiplied by the arrangements of 10! for each group, due to the arrangements among the groups themselves.

Favorable Outcomes = 2! x 10! x 10!

Thus the probability of event will be:

Probability = Favorable Outcomes/Total No. of Ways

Probability = (2! x 10! x 10!)/20!

<u>Probability = 0.00001 = 0.001%</u>

4 0

2 years ago

A suspension of Bacillus subtilis cells is filtered under constant pressure for recovery of protease. A pilot-scale filter is us

sattari [20]

The answer & explanation for this question is given in the attachment below.

7 0

2 years ago

2An oil pump is drawing 44 kW of electric power while pumping oil withrho=860kg/m3at a rate of 0.1m3/s.The inlet and outlet diam

Natasha2012 [34]

Answer:

$\eta = 91.7$ %

Explanation:

Determine the initial velocity

$v_1 = \frac{\dot v}{A_1}$

$= \frac{0.1}{\pi}{4} 0.08^2$

= 19.89 m/s

final velocity

$v_2 =\frac{\dot v}{A_2}$

$= \frac{0.1}{\frac{\pi}{4} 0.12^2}$

=8.84 m/s

total mechanical energy is given as

$E_{mech} = \dot m (P_2v_2 -P_1v_1) + \dot m \frac{v_2^2 - v_1^2}{2}$

$\dot v = \dot m v$ $( v =v_1 =v_2)$

$E_{mech} = \dot mv (P_2 -P_1) + \dot m \frac{v_2^2 - v_1^2}{2}$

$= mv\Delta P + \dot m \frac{v_2^2 -v_1^2}{2}$

$= \dot v \Delta P + \dot v \rho \frac{v_2^2 -v_1^2}{2}$

$= 0.1\times 500 + 0.1\times 860\frac{8.84^2 -19.89^2}{2}\times \frac{1}{1000}$

$E_{mech} = 36.34 W$

Shaft power

$W = \eta_[motar} W_{elec}$

$=0.9\times 44 =39.6$

mechanical efficiency

$\eta{pump} =\frac{ E_{mech}}{W}$

$=\frac{36.34}{39.6} = 0.917 = 91.7$ %

8 0

2 years ago

Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp

telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = $c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}$

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

$-0.40= 1.005(ln T_2 - 5.68697)- 0.5968$

Solving for T2 we have:

$T_2 = 5.8828$

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

$w_in = c_p(T_2 - T_1)+q_out$

$w_in = 1.005(358.8 - 295)+120$

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = $\frac{q_out}{T_1}$

$\frac{120kJ/kg.k}{295K}$

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

3 0

2 years ago