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grandymaker [24]

3 years ago

7

In using the drag coefficient care needs to be taken to use the correct area when determining the drag force. What is a typical

example of the appropriate area to use?

1 answer:

stealth61 [152]3 years ago

6 0

Answer:

Explanation:

We know that Drag force $F_D$

$F_D=\dfrac{1}{2}C_D\rho AV^2$

Where

$C_D$ is the drag force constant.

A is the projected area.

V is the velocity.

ρ is the density of fluid.

Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.

Example:

When we place our hand out of the window in a moving car ,we feel a force in the opposite direction and feel like some one trying to pull our hand .This pulling force is nothing but it is drag force.

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Water is the working fluid in an ideal Rankine cycle. Superheatedvapor enters the turbine at 10MPa, 480°C, and the condenser pre

choli [55]

Answer:

Explanation:

Given that:

Superheated vapor enters the turbine at 10 MPa, 480°C,

From the tables of superheated steam tables; the following values are obtained

$h_1 = 3322.02 \ kJ/kg\\\\ s_1 = 6.52846 \ kJ/kg.K$

Also; from the system, the isentropic line is 1-2 in which s_2 is in wet state

$s_2 = s_{f \ 6 kpa} +xs_{fg \ 6 kpa}$

$s_2 =0.51624 + x(7.82)$

$s_2 =0.51624 + 7.82x$

From the values obtained;

$s_1 =s_2= 6.52846 \ kJ/kg.K$

Therefore;

6.52846 = 0.51624+7.82x

6.52846 - 0.51624 = 7.82 x

6.01222 = 7.82 x

x = 6.01222/7.82

x = 0.7688

The enthalpy for this process at state (s_2) can be determined as follows:

$h_2 = h _f +xh_{fg} \\ \\ h_2 = 150.15 +(0.77 \times 2415.92) \\ \\ h_2 =150.15 +( 1629.2584 ) \\ \\ h_2 =2010.4084 \ kJ/kg$

The actual enthalpy at s_2 by using the isentropic efficiency of the turbine can determined by using the expression:

$n_T = \dfrac{h_1-h_{2a}}{h_1-h_2}$

$0.8 = \dfrac{3322.02-h_{2a}}{3322.02-2010.4084}$

$0.8 = \dfrac{3322.02-h_{2a}}{1311.6116}$

$0.8 * {1311.6116}= {3322.02-h_{2a}$

$1049.28928= {3322.02-h_{2a}$

$h_{2a}= {3322.02- 1049.28928$

$h_{2a}= 2272.73072$ kJ/kg

The work pump is calculated by applying the formula:

$w_p = v_{f \ 6 kpa} (p_4-p_3)$

$w_p = 0.0010062 * (10000-6)$

$w_p = 0.0010062 *9994$

$w_p = 10.0559628 \ kJ/kg$

However;

$w_p = h_4 -h_3$

From the process;

$h_3 = h_{f(6 kpa)} = 150.15 \ kJ/kg$

$10.0559628 = h_4 - 150.15$

$10.0559628+ 150.15 = h_4$

$160.2059628= h_4$

$h_4= 160.2059628 \ kJ/kg$

The actual enthalpy at s_4 by using the isentropic efficiency of the turbine can determined by using the expression:

$n_P = \dfrac{h_4-h_{3}}{h_{4a}-h_3}$

6 0

2 years ago

Identify renewable energy sources you will propose. Explain the key elements to your solution and the basic technical principles

MArishka [77]

Answer:

A renewable electricity generation technology harnesses a naturally existing energy. But they have other features that a few fringe customers value.

Explanation:

3 0

2 years ago

A heat pump with refrigerant-134a as the working uid is used to keep a space at 25C by absorbing heat from geothermal water that

Snezhnost [94]

Answer:

A) 0.03382 kg/s

B) 7.0372 Kw

C) 4.3982

D) 0.7396 kw

Explanation:

Given data:

Evaporator at 60 C

Space temperature = 25 C

power consumed by compressor = 1.6 kw

T1( evaporator temperature ) = 12°C

attached below is the detailed solution

8 0

2 years ago

Select the correct answer.

andre [41]

Answer:

A. energy transformations

Explanation:

8 0

1 year ago

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

8 0

2 years ago