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2 years ago

9

Precast concrete curtain wall panels: Group of answer choices are typically manufactured on site and then hoisted into place. of

ten are unreinforced.

1 answer:

amid [387]2 years ago

5 0

Answer:

c) can be made with a variety of surface finishes.

Explanation:

The missing options are;

When it comes to concrete work in construction, the concrete can be cast either in-situ or in form of pre-cast concrete.

Now in-situ concrete means concrete done on the construction site being built while pre cast concrete simply means concrete cast outside in a factory or yard and brought to site to mount.

These pre cast concrete could have different surface finishes as required as this is one of it's advantages over in situ because there is a lot of space and room to have the desired concrete finish.

a) are typically manufactured on site and then hoisted into place.

b) cannot be fiber-reinforced.

c) can be made with a variety of surface finishes.

d) never include insulation.

e) often are unreinforced.

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Consider a modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each repl

Ulleksa [173]

Answer:

The answers to the question are

(1) Process 1 to 2

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

(2) Process 2 to 3

W = 0

Q = 1135.376 kJ/kg

(3) Process 3 to 4

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

(4) Process 4 to 3

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency = 49.9 %

(c) The mean effective pressure is 9.44 bar

Explanation:

(a) Volume compression ratio $\frac{v_1}{v_2} = 10$

Initial pressure p₁ = 1 bar

Initial temperature, T₁ = 310 K

cp = 1.005 kJ/kg⋅K

Temperature T₃ = 2200 K from the isentropic chart of the Otto cycle

For a polytropic process we have

$\frac{p_1}{p_2} = (\frac{v_2}{v_1} )^n$ Therefore p₂ = p₁ ÷ $(\frac{v_2}{v_1} )^n$ = (1 bar) ÷ $(\frac{1}{10} )^{1.3}$ = 19.953 bar

Similarly for a polytropic process we have

$\frac{T_1}{T_2} = (\frac{v_2}{v_1} )^{n-1}$ or T₂ = T₁ ÷ $(\frac{v_2}{v_1} )^{n-1}$ = $\frac{310}{0.1^{0.3}}$ = 618.531 K

The molar mass of air is 28.9628 g/mol.

Therefore R = $\frac{8.3145}{28.9628}$ = 0.287 kJ/kg⋅K

cp = 1.005 kJ/kg⋅K Therefore cv = cp - R = 1.005- 0.287 = 0.718 kJ/kg⋅K

1). For process 1 to 2 which is polytropic process we have

W = $\frac{R(T_2-T_1)}{n-1}$ = $\frac{0.287(618.531-310)}{1.3 - 1}$ = 295.16 kJ/kg

Q = $(\frac{n-\gamma}{\gamma - 1} )W$ = $(\frac{1.3-1.4}{1.4-1} ) 295.16 kJ/kg$ = -73.79 kJ/kg

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

2). For process 2 to 3 which is reversible constant volume heating we have

W = 0 and Q = cv×(T₃ - T₂) = 0.718× (2200-618.531) = 1135.376 kJ/kg

W = 0

Q = 1135.376 kJ/kg

3). For process 3 to 4 which is polytropic process we have

W = $\frac{R(T_4-T_3)}{n-1}$ = Where T₄ is given by $\frac{T_4}{T_3} = (\frac{v_3}{v_4} )^{n-1}$ or T₄ = T₃ × $0.1^{0.3}$

= 2200 × $0.1^{0.3}$ T₄ = 1102.611 K

W = $\frac{0.287(1102.611-2200)}{1.3 - 1}$ = -1049.835 kJ/kg

and Q = 262.459 kJ/kg

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

4). For process 4 to 1 which is reversible constant volume cooling we have

W = 0 and Q = cv×(T₁ - T₄) = 0.718×(310 - 1102.611) = -569.09 kJ/kg

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency is given by

$\eta = 1-\frac{T_4-T_1}{T_3-T_2}$ = $1-\frac{1102.611-310}{2200-618.531}$ = 0.499 or 49.9 % Efficient

(c) The mean effective pressure is given by

$p_{m} = \frac{p_1r[(r^{n-1}-1)(r_p-1)]}{ (n-1)(r-1)}$ where r = compression ratio and $r_p$ = $\frac{p_3}{p_2}$

However p₃ = $\frac{p_2T_3}{T_2}$ = $\frac{(19.953)(2200)}{618.531}$ =70.97 atm

$r_p$ = $\frac{p_3}{p_2}$ = $\frac{70.97}{19.953} = 3.56$

Therefore $p_m =\frac{1*10*[(10^{0.3}-1)(3.56-1)]}{0.3*9}$ = 9.44 bar

Please find attached generalized diagrams of the Otto cycle

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3 years ago

Which one of the following activities is not an example of incident coordination

Lady bird [3.3K]

Directing, ordering, or controlling

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3 years ago

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EleoNora [17]

Answer: why?

Explanation:

6 0

3 years ago

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Let be a real-valued signal for which when . Amplitude modulation is preformed to produce the signal . A proposed demodulation t

densk [106]

Answer:

hello your question is incomplete attached below is the complete question

answer : attached below

Explanation:

let ; x(t) be a real value signal for x ( jw ) = 0 , |w| > 200 $\pi$

g(t) = x ( t ) sin ( 2000 $\pi t )$

$x_{1} (t) = \frac{1}{2} x(t) sin ( 4000\pi t )$

next we apply Fourier transform

attached below is the remaining part of the solution

6 0

3 years ago

A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60

Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

T_2 = T_1 * r^(k-1/k)

T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

T_7 = T_6 * r^(-k-1/k)

T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

flow(m) = 7.941 lbm/s

- The heat input is as follows:

Q_in = c_p*(T_6 - T_5)

Q_in = 0.24*(1400 - 1022.84)

Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

Q_out = c_p*(T_10 - T_1)

Q_out = 0.24*(711.744 - 520)

Q_out = 56.01856 Btu/lbm

5 0

3 years ago