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NemiM [27]
3 years ago
15

A controlled process is described by the closed-loop transfer function G(s).

Engineering
1 answer:
MissTica3 years ago
3 0

Answer:

The answer is "Option B".

Explanation:

Given equation:

G(s) =\frac{K(s + 1)}{2s^2 + (K-1)s + (K-1)}\\\\

if

\to 2s^2 + (K-1)s + (K-1)=0

Calculating by the Routh's Hurwitz table:

\to s^2  \ \ \ \ \    2  \ \ \ \ \ \  K-1 \\\\\to s^2  \ \ \ \ \    K-1  \ \ \ \ \ \   \\\\\to s^0 \ \  ( \frac{(K-1)(K-1)(-2) (0)}{K-1}  \\\\    \ \ \ \  = (K-1) )

Form the above table:

\to K-1 > 0 \\\\ \to K > 1

In the above, the value of k is greater than 1.

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Engine oil (unused) flows at 1.81 x 10^-3 kg/s inside a 1-cm diameter tube that is heated electrically at a rate of 76 W/m. At a
Ann [662]

Answer:

(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K

Explanation:

Solution

Given that:

The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s

Diameter of the tube, D = 1cm =0.01 m

Electrical heat rate, q =76 W/m

Wall Temperature, Ts = 370 K

Now,

From the properties table of engine oil we can deduce as follows:

thermal conductivity, k =0.139 W/m .K

Density, ρ = 854 kg/m³

Specific heat, cp = 2120 J/kg.K

(a) Thus

The wall heat flux is given as follows:

qs = q/πD

=76/π *0.01

= 2419.16 W/m²

Now

The oil mean temperature is given as follows:

Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)

Tb =370 - 11/24 * (2419.16 * 0.005/0.139)

Tb = 330.12 K

(b) The center line temperature is given below:

Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)

Tc =304.73 K

(c) The flow velocity is given as follows:

V = m/ρ (πR²)

Now,

The The axial gradient of the mean temperature is given below:

dTb/dx = 2 *qs/ρ *V*cp * R

=2 *qs/ρ*[m/ρ (πR²) *cp * R

=2 *qs/[m/(πR)*cp

dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120

dTb/dx = 19.81 K/m

(d) The heat transfer coefficient is given below:

h =48/11 (k/D)

=48/11 (0.139/0.01)

h =60.65 W/m². K

8 0
3 years ago
A heat pump cycle whose coefficient of performance is 2.5 delivers energy by heat transfer to a dwelling at a rate of 20kW.
12345 [234]

Answer:

a) 8kW

b) $128

Explanation:

Given the coefficient of performance of the heat pump cycle to be 2.5

Energy delivered by the heat pump = 20kW

a) net power required to operate the heat pump = Energy delivered / coefficient of performance

Net power required = 20/2.5

= 8kW

b) Given the cost of electricity is $0.08 for 1kWhour

Since net power required to operate heat pump = 8kW

If the heat pump operate for 200hours, total power required for a month = 8kW×200hours = 1600kWhour

since 1kWh of electricity costs $0.08, cost of electricity used in a month when the pump operates for 200hour will be 1600kWh×$0.08 which is equivalent to $128

8 0
3 years ago
Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.
nikitadnepr [17]

Answer:

a)  m_e= 3.05 Kg

b)  \rho=1072.3kg/m^3

c)  m_e= 3.05 Kg

Explanation:

From the question we are told that:

Beaker Mass m_b=1.20

Liquid Mass m_l=1.85

Balance D:

Mass m_d=3.10

Balance E:

Mass m_e=7.50

Volume v=4.15*10^{-3}m^3

a)

Generally the equation for Liquid's density is mathematically given by

m_e=m_b+m_l+(\rho*v)

\rho=\frac{7.50-(1.2+1.85)}{4.15*10^{-3}}

\rho=1072.3kg/m^3

b)

Generally the equation for D's Reading at A pulled is mathematically given by

m_d = mass of block - mass of liquid displaced

m_d=m- (\rho *v )

m=3.10+ (1072.30 *4.15*10^{-3}m^3 )

m=18.10kg

c)

Generally the equation for E's Reading at A pulled is mathematically given by

m_e=m_b+m_l

m_e = 1.20 + 1.85

m_e= 3.05 Kg

6 0
3 years ago
How long...you...novels? A. Have/write B. Do/write C. Have/written D. Did/go
soldi70 [24.7K]

Answer:

Answer is B.

Explanation:

Subject verb agreement

3 0
3 years ago
Read 2 more answers
An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 3 m. If the height of the tank sides is 4 m. What
VARVARA [1.3K]

Answer:

ay max = 4.91 m/s²  

so here acceleration would be either right or left

Explanation:

given data

wide b = 1 m

long l  = 2 m

depth d = 3 m

height of  tank sides h = 4 m

solution

here for prevent spilling condition is

\frac{dz}{dy} ≤ - \frac{1.5 - 1 }{1}  ..........1

\frac{dz}{dy}  ≤ - 0.50

and when here

\frac{dz}{dy} =  -   \frac{ay}{g + az}    ......2  

when az is 0 ay will be

ay = -  \frac{dz}{dy} g    

and ay max will be

ay max = -( -0.50) (9.81 )

ay max = 4.91 m/s²  

so here acceleration would be either right or left

8 0
3 years ago
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