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babunello [35]
3 years ago
12

To be able to write an ss-domain equation for a circuit, use partial fraction decomposition to separate the terms in this equati

on, and then inverse transform the equation back into the time-domain. Using the mesh-current method to analyze the following circuit yields a set of two integro-differential equations which would be difficult to solve analytically. However, applying a Laplace transform allows for these equations to be rewritten as polynomial equations which can be manipulated algebraically.

Engineering
1 answer:
Ann [662]3 years ago
4 0

Answer:

Assuming it is for the circuit attached below. To construct a domain equivalent for t > 0

Explanation:

See hand written solution

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Answer:

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3 years ago
Balanced forces will result in which of the following
tia_tia [17]
D is the answer. Hope this helped
8 0
3 years ago
A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy by heat
Hoochie [10]

Answer:

a) Ql=33120000 kJ

b) COP = 5.6

c) COPreversible= 29.3

Explanation:

a) of the attached figure we have:

HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:

W=Qh-Ql

Ql=Qh-W

where W=2000 kWh

Qh=120000 kJ/h

Q_{l}=14days(\frac{24 h}{1 day})(\frac{120000 kJ}{1 h})-2000 kWh(\frac{3600 s}{1 h})=33120000 kJ

b) The coefficient of performance is:

COP=\frac{Q_{h} }{W}=\frac{120000 kJ/h*14(\frac{24 h}{1 day}) }{2000 kWh(\frac{3600 s}{1 h}) } = 5.6

c) The coefficient of performance of a reversible heat pump is:

COP_{reversible}=\frac{T_{h} }{T_{h}-T_{l}  }

Th=20+273=293 K

Tl=10+273=283K

Replacing:

COP_{reversible}=\frac{293}{293-283}=29.3

4 0
3 years ago
What are some common ways of converting fuel to useful energy? Check all that apply.
Rainbow [258]

Answer: Burning it.

Explanation: When your car moves its due to the burning of fuel, works like a turbine the molecules evaporate and turn a turbine like object.

3 0
3 years ago
Read 2 more answers
An experiment to determine the convection coefficient associated with airflow over the surface of a thick stainless steel castin
Maksim231197 [3]

Answer:

h = 375 KW/m^2K

Explanation:

Given:

Thermo-couple distances: L_1 = 10 mm , L_2 = 20 mm

steel thermal conductivity k = 15 W / mK

Thermo-couple temperature measurements: T_1 = 50 C , T_2 = 40 C

Air Temp T_∞ = 100 C

Assuming there are no other energy sources, energy balance equation is:

                                               E_in = E_out

                                        q"_cond = q"_conv

Since, its a case 1-D steady state conduction, the total heat transfer rate can be found from Fourier's Law for surfaces 1 and 2

q"_cond = k * (T_1 - T_2) / (L_2 - L_1) = 15 * (50 - 40) / (0.02 - 0.01)

=15KW/m^2

Assuming SS is solid, temperature at the surface exposed to air will be 60 C since its gradient is linear in the case of conduction, and there are two temperatures given in the problem. Convection coefficient can be found from Newton's Law of cooling:

q"_conv = h * ( T_∞  - T_s ) ----> h = q"_conv / ( T_∞  - T_s )

                                                   h = 15000 W / (100 - 60 ) C = 375 KW/m^2K

4 0
3 years ago
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