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seropon [69]
3 years ago
11

Write down a transfer function of a stable system for which pure proportional feedback could drive the system unstable.

Engineering
1 answer:
gtnhenbr [62]3 years ago
6 0

Answer:

Gs = 1 / (s + 1)^3

Explanation:

A system is known to be a stable system if the poles of the system lie on the negative real axis of the s plane.

A system that could be driven unstable by a P-controller would be:

Gs = 1 / (s + 1)^3

As the value of K increases, the closed loop poles move into the right half s plane

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Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid-vapor mixture region using 0.96 kg of
Eduardwww [97]

Answer:

P_m_i_n= 356.9 KPa

Explanation:

Coefficient of performance (COP) of refrigeration cycle is given by:

COP=\frac{1}{\frac{T_H}{T_L}-1 }

We are given:

T_H=1.2T_L

COP=\frac{1}{1.2-1 }

COP= 5

We can also write Coefficient of performance (COP) of refrigeration cycle  as:

COP_R=\frac{Q_L}{W_i_n}

Amount of heat absorbed by low temperature reservoir can be found as:

Q_L=COP_R * W_i_n

Q_L=5 * 22 KJ

Q_L=110 KJ

According to first law of thermodynamics amount of heat rejected by hot reservoir is given by:

Q_H=Q_L + W_i_n

Q_H=110 KJ + 22 KJ

Q_H=132 KJ

We are given the mass of 0.96 kg. So,

q_H=\frac{Q_H}{m}

q_H=\frac{132 KJ}{0.96Kg}

q_H=137.5 KJ/Kg

Since it is a saturated liquid-vapour mixture q_H=h_f_g.

q_H=h_f_g=137.5 KJ/Kg

From Refrigerant 134-a tables T_H at h_f_g=137.5 KJ/Kg is 61.3 C. (We calculated this by interpolation)

Converting T_H from Celsius to Kelvin:

61.3^{o} C+273 = 334.3^{o} K

T_H= 334.3^{o} K

We are given:

T_H=1.2T_L

T_L=\frac{T_H}{1.2}

T_L=\frac{334.3}{1.2}

T_L=278.58^{o} K

Converting T_L from Kelvin to Celsius:

278.58^{o} K-273 = 5.58^{o} C

T_L= 5.58^{o} C

From Refrigerant 134-a tables P_m_i_n at T_L=5.58^{o} C is 356.9 KPa. (We calculated this by interpolation).

P_m_i_n= 356.9 KPa

8 0
3 years ago
A chemistry student accidentally drops a large mercury thermometer and it breaks. The thermometer contained 2 grams of mercury (
loris [4]

Answer:

23.1 35f

Explanation:

not dumb

4 0
3 years ago
GMA MIG weiding is a
Dvinal [7]

Gas metal arc welding (GMAW), sometimes referred to by its subtypes metal inert gas (MIG) welding or metal active gas (MAG) welding, is a welding process in which an electric arc forms between a consumable MIG wire electrode and the workpiece metal(s), which heats the workpiece metal(s), causing them to melt and join.

6 0
3 years ago
Introduction to gear and gear ratios: If you have two gears of the same size does the output speed increase, decrease, or remain
creativ13 [48]

Answer:

remain the same constant

Explanation:

if a small gear is making 5 rpm turning a big gear thats making 1 rpm the bug gear is making lots of torque but if its going the opposite way its much harder to turn the big gear 1 rpm just to make the little gear do 5 rpm and it makes way less torque if thre both the same size there ging to stay the same speed

4 0
3 years ago
A cast-iron tube is used to support a compressive load. Knowing that E 5 10 3 106 psi and that the maximum allowable change in l
Papessa [141]

Answer:

(a) 2.5 ksi

(b) 0.1075 in

Explanation:

(a)

E=\frac {\sigma}{\epsilon}

Making \sigma the subject then

\sigma=E\epsilon

where \sigma is the stress and \epsilon is the strain

Since strain is given as 0.025% of the length then strain is \frac {0.025}{100}=0.00025

Now substituting E for 10\times 10^{6} psi then

\sigma=(10\times 10^{6} psi)\times 0.00025=2500 si= 2.5 ksi

(b)

Stress, \sigma= \frac {F}{A} making A the subject then

A=\frac {F}{\sigma}

A=\frac {\pi(d_o^{2}-d_i^{2})}{4}

where d is the diameter and subscripts o and i denote outer and inner respectively.

We know that 2t=d_o - d_i where t is thickness

Now substituting

\frac {\pi(d_o^{2}-d_i^{2})}{4}=\frac {1600}{2500}

\pi(d_o^{2}-d_i^{2})=\frac {1600}{2500}\times 4

(d_o^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4

But the outer diameter is given as 2 in hence

(2^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4

2^{2}-(\frac {1600}{2500\times \pi}\times 4)=d_i^{2}

d_i=\sqrt {2^{2}-(\frac {1600}{2500\times \pi}\times 4)}=1.784692324 in\approx 1.785 in

As already mentioned, 2t=d_o - d_i hence t=0.5(d_o - d_i)

t=0.5(2-1.785)=0.1075 in

3 0
3 years ago
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