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2 years ago

Tech A says that failure of a battery hold down can cause battery plate damage. Tech B says that a bungee

Engineering

1 answer:

Pavel [41]2 years ago

7 0

Answer:

Technician A

Explanation:

1- It's was easy to understand.

2- Holding down a battery could actually cause battery plate damage. Most of the batteries I interact with only use a plastic cover and a small spring to hold the battery in place.

-Hope this helped :) -

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An Otto cycle engine is analyzed using the air standard method. Given the conditions at state 1, compression ratio (r), and pres

My name is Ann [436]

Answer:

A) 222.58 kJ / kg

B) 0.8897 M^3/ kg

c) 0.7737 m^3/kg

D) 746.542 k

E) 536.017 kj/kg

efficiency = 58% ( approximately )

Explanation:

Given Data :

Gas constant (R) = 0.287 kJ/ kg.K

T1 = 310 k

P1 ( Kpa ) = 100

r = 11.5 ( compression ratio )

rp = 1.95 ( pressure ratio )

A ) specific internal energy at state 1

= Cv*T1 = 0.718 * 310 = 222.58 kJ / kg

B) Relative specific volume at state 1

= P1*V1 = R*T1 ( ideal gas equation )

V1 = R*T1 / P1 = (0.287* 10^3*310 ) / 100 * 10^3

V1 = 88.97 / 100 = 0.8897 M^3/ kg

C ) relative specific volume at state 2

Applying r ( compression ratio) = V1 / V2

11.5 = 0.8897 / V2

V2 = 0.8897 / 11.5 = 0.7737 m^3/kg

D) The temperature (k) at state 2

since the process is an Isentropic process we will apply the p-v-t relation

$\frac{T1}{T2} = (\frac{V1}{V2}^{n-1} ) = (\frac{P2}{P1} )^{\frac{n-1}{n} }$

hence T2 = $9^{1.4-1} * 310$ = 2.4082 * 310 = 746.542 k

e) specific internal energy at state 2

= Cv*T2 = 0.718 * 746.542 = 536.017 kj/kg

efficiency = output /input = 390.3511 / 667.5448 ≈ 58%

attached is a free hand diagram of an Otto cycle is attached below

3 0

3 years ago

How would you expect an increase in the austenite grain size to affect the hardenability of a steel alloy? Why?

seraphim [82]

Answer:

The hardenability increases with increasing austenite grain size, because the grain boundary area is decreasing. This means that the sites for the nucleation of ferrite and pearlite are being reduced in number, with the result that these transformations are slowed down, and the hardenability is therefore increased.

3 0

3 years ago

A 600-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 54 percent. Determine the rate of hea

Gennadij [26K]

Answer:

$\dot Q _{L} = 511.111 MW$ . Heat transfer can be higher if themal efficiency is lower.

Explanation:

The heat transfer rate to the river water is calculated by this expression:

$\dot Q_{L} = \dot Q_{H} - \dot W$

$\dot Q_{L} = (\frac{1}{\eta_{th}}-1 )\cdot \dot W\\\dot Q_{L} = (\frac{1}{0.54}-1)\cdot (600 MW)\\\dot Q _{L} = 511.111 MW$

The actual heat transfer can be higher if the steam power plant reports an thermal efficiency lower than expected.

8 0

3 years ago

Assume the impedance of a circuit element is Z = (3 + j4) Ω. Determine the circuit element’s conductance and susceptance.

djyliett [7]

Answer:

B. G = 333 mS, B = j250 mS

Explanation:

impedance of a circuit element is Z = (3 + j4) Ω

The general equation for impedance

Z = (R + jX) Ω

where

R = resistance in ohm

X = reactance

R = 3Ω X = 4Ω

Conductance = 1/R while Susceptance = 1/X

Conductance = 1/3 = 0.333S

= 333 mS

Susceptance = 1/4 = 0.25S

= 250mS

The right option is B. G = 333 mS, B = j250 mS

8 0

3 years ago

The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200K and 400K, re

Anni [7]

Answer:

a) W_cycle = 200 KW , n_th = 33.33 % , Irreversible

b) W_cycle = 600 KW , n_th = 100 % , Impossible

c) W_cycle = 400 KW , n_th = 66.67 % , Reversible

Explanation:

Given:

- The temperatures for hot and cold reservoirs are as follows:

TL = 400 K

TH = 1200 K

Find:

For each case W_cycle , n_th ( Thermal Efficiency ) :

(a) QH = 600 kW, QC = 400 kW

(b) QH = 600 kW, QC = 0 kW

- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

Solution:

- The work done by the cycle is given by first law of thermodynamics:

W_cycle = QH - QC

- For categorization of cycle is given by second law of thermodynamics which states that:

n_th < n_max ...... irreversible

n_th = n_max ...... reversible

n_th > n_max ...... impossible

- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:

n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %

And, n_th = W_cycle / QH

a) QH = 600 kW, QC = 400 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 400 = 200 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 200 / 600 = 33.33 %

- The type of process according to second Law of thermodynamics:

n_th = 33.333 % n_max = 66.67 %

n_th < n_max

Hence, Irreversible Process

b) QH = 600 kW, QC = 0 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 0 = 600 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 600 / 600 = 100 %

- The type of process according to second Law of thermodynamics:

n_th = 100 % n_max = 66.67 %

n_th > n_max

Hence, Impossible Process

c) QH = 600 kW, QC = 200 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 200 = 400 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 400 / 600 = 66.67 %

- The type of process according to second Law of thermodynamics:

n_th = 66.67 % n_max = 66.67 %

n_th = n_max

Hence, Reversible Process

7 0

3 years ago