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ELEN [110]
3 years ago
10

Item110pointseBook HintPrintReferences Check my work Check My Work button is now disabled5Item 1Item 1 10 pointsAn ideal Diesel

cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the maximum air temperature and the rate of heat addition to this cycle when it produces 150 hp of power, the cycle is repeated 1200 times per minute, and the state of the air at the beginning of the compression is 95 kPa and 17°C. Use constant specific heats at room temperature. The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4.
Engineering
1 answer:
konstantin123 [22]3 years ago
5 0

Answer:

The answer is given below

Explanation:

Given that:

Compression ratio (r) = 18, cut off ratio (r_c) = 1.5, k = 1.4, W = 150 hp, initial temperature (Ti) = 17°C = 290 K

The rate of heat addition (Qin) is given by the equation:

Q_{in}=\frac{W*r^{k-1}*k(r_c-1)}{r^{k-1}*k(r_c-1)-(r_c^k-1)} \\\\substituting\ values\ into\ the \ equation\ gives:\\\\Q_{in}=\frac{150*18^{1.4-1}*1.4(1.5-1)}{18^{1.4-1}*1.4(1.5-1)-(1.5^{1.4}-1)}=\frac{333.6555}{2.2244-0.7641} =\frac{333.6555}{1.4603} =228.5\ hp

Therefore, The rate of heat addition is 228.5 hp

The maximum air temperature is determined using the Compression ratio, cut off ratio and the initial temperature. The maximum air temperature (T_{max}) is given by the formula:

T_{max}=T_{in}*r^{k-1}*r_c=290*18^{1.4-1}*1.5=1382.3\ K=1109.3^oC

The maximum air temperature is 1109.3°C

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2 years ago
An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capaci
Serggg [28]

Answer:

Explanation:

f = 50.0 Hz, L = 0.650 H, π = 3.14

C = 4.80 μF, R = 301 Ω resistor. V = 120volts

XL = wL = 2πfL

= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

√ 301630.9844

= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

= 0.256A ×204.1

=52.2496

= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

1 = 0.000030144×f×4.082×f

1 = 0.000123f^2

f^2 = 1/0.000123048

f^2 = 8126.922

f =√8126.922

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8 0
3 years ago
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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