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3 years ago

8

Handsaw teeth are very sharp: to avoid being cut by the teeth, keep hands and fingers well away from the

2 answers:

alexdok [17]3 years ago

8 0

Yes, this seems logical

siniylev [52]3 years ago

6 0

Handsaw teeth are very sharp: to avoid being cut by the teeth, keep hands and fingers well away from the
path of the blade

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1. Technician A says an automotive computer can scan its input and output circuits to detect incorrect voltage problems. Technic

Marta_Voda [28]

B is the correct answer

4 0

3 years ago

Air is compressed steadily from 100kPa and 20oC to 1MPa by an adiabatic compressor. If the mass flow rate of the air is 1kg/s an

igomit [66]

Answer:

(a). 575 kJ/kg.

(b). 290kw.

Explanation:

We have the following set of information or parameters from the question above;

Pressure(1) = 100kPa, Pressure (2) = 1MPa, temperature(1) = b1= 12°C = 285K = 285kJ/kg, efficiency = 80% and the mass flow rate of the air = 1kg/s.

At a temperature of 12°C, we have the value from steam table; gx= 1.2, thus gx22 = 1.2 × (1000/100) = 12.

We have that the value for b12 = 517.

Therefore, the value for h2a can be calculated as;

80/100 = (517 - 285)/ (tp at exist) - 285.

0.8 = 232/ (tp at exist) - 285.

232 = 0.8 × (tp at exist) - 285).

232 = 0.8 (tp at exist) - 228 .

(tp at exist) = 575.

Therefore, the temperature 575 kJ/kg.

Thus, the required power input of the compressor = 1kg/s × ( 575 - 285) = 290kw.

6 0

3 years ago

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75

Inessa [10]

Answer:

(a). The value of temperature at the end of heat addition process $T_{3}$ = 2042.56 K

(b). The value of pressure at the end of heat addition process $P_{3}$ = 1555.46 k pa

(c). The thermal efficiency of an Otto cycle $E_{otto}$ = 0.4478

(d). The value of mean effective pressure of the cycle $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

Explanation:

Compression ratio $r_{p}$ = 8

Initial pressure $P_{1}$ = 95 k pa

Initial temperature $T_{1}$ = 278 °c = 551 K

Final pressure $P_{2}$ = 8 × $P_{1}$ = 8 × 95 = 760 k pa

Final temperature $T_{2}$ = $T_{1}$ × $r_{p} ^{\frac{\gamma - 1}{\gamma} }$

Final temperature $T_{2}$ = 551 × $8 ^{\frac{1.4 - 1}{1.4} }$

Final temperature $T_{2}$ = 998 K

Heat transferred at constant volume Q = 750 $\frac{KJ}{kg}$

(a). We know that Heat transferred at constant volume $Q_{S}$ = $m C_{v} ( T_{3} - T_{2} )$

⇒ 1 × 0.718 × ( $T_{3}$ - 998 ) = 750

⇒ $T_{3}$ = 2042.56 K

This is the value of temperature at the end of heat addition process.

Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.

⇒ P ∝ T

⇒ $\frac{P_{3} }{P_{2} }$ = $\frac{T_{3} }{T_{2} }$

⇒ $P_{3}$ = $\frac{2042.56}{998}$ × 760

⇒ $P_{3}$ = 1555.46 k pa

This is the value of pressure at the end of heat addition process.

(b). Heat rejected from the cycle $Q_{R} = m C_{v} ( T_{4} - T_{1} )$

For the compression and expansion process,

⇒ $\frac{T_{3} }{T_{2} } = \frac{T_{4} }{T_{1} }$

⇒ $\frac{2042.56}{998} = \frac{T_{4} }{551}$

⇒ $T_{4}$ = 1127.7 K

Heat rejected $Q_{R}$ = 1 × 0.718 × ( 1127.7 - 551)

⇒ $Q_{R}$ = 414.07 $\frac{KJ}{kg}$

Net heat interaction from the cycle $Q_{net}$ = $Q_{S}$ - $Q_{R}$

Put the values of $Q_{S}$ & $Q_{R}$ we get,

⇒ $Q_{net}$ = 750 - 414.07

⇒ $Q_{net}$ = 335.93 $\frac{KJ}{kg}$

We know that for a cyclic process net heat interaction is equal to net work transfer.

⇒ $Q_{net}$ = $W_{net}$

⇒ $W_{net}$ = 335.93 $\frac{KJ}{kg}$

This is the net work output from the cycle.

(c). Thermal efficiency of an Otto cycle is given by

$E_{otto} = 1- \frac{T_{1} }{T_{2} }$

Put the values of $T_{1}$ & $T_{2}$ in the above formula we get,

$E_{otto} = 1- \frac{551 }{998 }$

⇒ $E_{otto}$ = 0.4478

This is the thermal efficiency of an Otto cycle.

(d). Mean effective pressure $P_{m}$ :-

We know that mean effective pressure of the Otto cycle is given by

$P_{m}$ = $\frac{W_{net} }{V_{s} }$ ---------- (1)

where $V_{s}$ is the swept volume.

$V_{s}$ = $V_{1} - V_{2}$ ---------- ( 2 )

From ideal gas equation $P_{1}$ $V_{1}$ = m × R × $T_{1}$

Put all the values in above formula we get,

⇒ 95 × $V_{1}$ = 1 × 0.287 × 551

⇒ $V_{1}$ = 0.6 $m^{3}$

From the same ideal gas equation

$P_{2}$ $V_{2}$ = m × R × $T_{2}$

⇒ 760 × $V_{2}$ = 1 × 0.287 × 998

⇒ $V_{2}$ = 0.377 $m^{3}$

Thus swept volume $V_{s}$ = 0.6 - 0.377

⇒ $V_{s}$ = 0.223 $m^{3}$

Thus from equation 1 the mean effective pressure

⇒ $P_{m}$ = $\frac{335.93}{0.223}$

⇒ $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

This is the value of mean effective pressure of the cycle.

4 0

3 years ago

An_______<br> employee is always prompt and on time.

Paha777 [63]

Do u won’t to go out

5 0

4 years ago

A diesel engine with CR= 20 has inlet at 520R, a maximum pressure of 920 psia and maximum temperature of 3200 R. With cold air p

Stella [2.4K]

Answer:

Cut-off ratio $\dfrac{V_3}{V_2}=6.15$

Cxpansion ratio $\dfrac{V_4}{V_3}=3.25$

The exhaust temperature $T_4=1997.5R$

Explanation:

Compression ratio CR(r)=20

$\dfrac{V_1}{V_2}=20$

$P_2=P_3=920 psia$

$T_1=520 R ,T_{max}=T_3,T_3=3200 R$

We know that for air γ=1.4

If we assume that in diesel engine all process is adiabatic then

$\dfrac{T_2}{T_1}=r^{\gamma -1}$

$\dfrac{T_2}{520}=20^{1.4 -1}$

$T_2=1723.28R$

$\dfrac{V_3}{V_2}=\dfrac{T_3}{T_2}$

$\dfrac{V_3}{V_2}=\dfrac{3200}{520}$

So cut-off ratio $\dfrac{V_3}{V_2}=6.15$

$\dfrac{V_1}{V_2}=\dfrac{V_4}{V_3}\times\dfrac{V_3}{V_2}$

Now putting the values in above equation

$\dfrac20=\dfrac{V_4}{V_3}\times 6.15$

$\dfrac{V_4}{V_3}=3.25$

So expansion ratio $\dfrac{V_4}{V_3}=3.25$ .

$\dfrac{T_4}{T_3}=(expansion\ ratio)^{\gamma -1}$

$\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}$

$T_4=1997.5R$

So the exhaust temperature $T_4=1997.5R$

3 0

4 years ago