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anastassius [24]
2 years ago
13

HAPPINESS DISCUSSION

Engineering
2 answers:
RideAnS [48]2 years ago
8 0

Answer:

uh because life sucks o_<

KengaRu [80]2 years ago
7 0
I personally have more bad days then good days because I don’t cherish the good days, on the good days I’m happy that I can feel something like sadness, or anger on my bad days, because without my bad days there would be no good days. But on my bad days I don’t care about anything except for what is making my day bad.
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A flow of 12 m/s passes through a 6 m wide, 2 m deep rectangular channel with a bed slope of 0. 001. If the mean velocity of flo
prohojiy [21]

Answer:

manning's coefficient is 0.0357

Explanation:

Given:

Velocity of flow, v = 12 m/s

Width of the channel, b = 6 m

Depth of the channel, d = 2 m

bed slope, s = 0.001

mean velocity of flow, V = 1 m/s

now, the velocity is given as:

V= \frac{1}{n}R^{\frac{2}{3}}S^{\frac{1}{2}}

where,

n is the manning's coefficient

R is hydraulic mean depth

R = (Area of the channel) / (wetted Perimeter of the channel)

now,

R = (2 × 6) / ((2 × 2) + 6)

or

R = 12 / 10 = 1.2 m

now, on substituting the values in the equation for velocity, we get

1= \frac{1}{n}1.2^{\frac{2}{3}}(0.001)^{\frac{1}{2}}

or

n= 1.2^{\frac{2}{3}}(0.001)^{\frac{1}{2}}

or

n = 0.0357

hence, the value of manning's coefficient is 0.0357

8 0
3 years ago
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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

8 0
3 years ago
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