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3 years ago

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The enthalpy of the water entering an actual pump is 500 kJ/kg and the enthalpy of the water leaving it is 550 kJ/kg. The pump h

as 98% efficiency, what would have been the enthalpies at the inlet and outlet if the pump was 100% efficient?

1 answer:

n200080 [17]3 years ago

3 0

Answer:500,551.02

Explanation:

Given

Initial enthaly of pump \left ( h_1\right )=500KJ/kg

Final enthaly of pump \left ( h_2\right )=550KJ/kg

Final enthaly of pump when efficiency is 100%= $h_2^{'}$

Now pump efficiency is 98%

$\eta$ = $\frac{h_2-h_1}{h_2^{'}-h_1}$

0.98= $\frac{550-500}{h_2-500}$

$h_2=551.02KJ/kg$

therefore initial and final enthalpy of pump for 100 % efficiency

initial=500KJ/kg

Final=551.02KJ/kg

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Assume a steel pipe of inner radius r1= 20 mm and outer radius r2= 25 mm, which is exposed to natural convection at h = 50 W/m2.

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Answer:

98,614.82 W/m²

Explanation:

$Q = 2\pi hL(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})$

Where;

Q = the amount of heat loss from the pipe

h = the heat transfer coefficient of the pipe = 50 W/m².K

T₁ = the ambient temperature of the pipe = 30⁰C

T₂ = the outside temperature of the pipe = 100⁰C

L= the length of pipe

r₁ = inner radius of the pipe = 20mm

r₂ = outer radius of the pipe = 25mm

To determine the amount of heat loss from the pipe per unit length

From the equation above

$\frac{Q}{L} = 2\pi h(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})$

$\frac{Q}{L} = 2\pi* 50(\frac{100-30}{Ln\frac{25}{20}})$

$\frac{Q}{L} = 314.159(\frac{70}{0.223})$

$\frac{Q}{L} = 314.159(313.901)$ = 98,614.82 W/m²

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4 years ago

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3 years ago

An air-standard dual cycle has a compression ratio of 9.1 and displacement of Vd = 2.2 L. At the beginning of compression, p1 =

jok3333 [9.3K]

Answer:

a) T₂ is 701.479 K

T₃ is 1226.05 K

T₄ is 2350.34 K

T₅ is 1260.56 K

b) The net work of the cycle in kJ is 2.28 kJ

c) The power developed is 114.2 kW

d) The thermal efficiency, $\eta _{dual}$ is 53.78%

e) The mean effective pressure is 1038.25 kPa

Explanation:

a) Here we have;

$\frac{T_{2}}{T_{1}}=\left (\frac{v_{1}}{v_{2}} \right )^{\gamma -1} = \left (r \right )^{\gamma -1} = \left (\frac{p_{2}}{p_{1}} \right )^{\frac{\gamma -1}{\gamma }}$

Where:

p₁ = Initial pressure = 95 kPa

p₂ = Final pressure =

T₁ = Initial temperature = 290 K

T₂ = Final temperature

v₁ = Initial volume

v₂ = Final volume

$v_d$ = Displacement volume =

γ = Ratio of specific heats at constant pressure and constant volume cp/cv = 1.4 for air

r = Compression ratio = 9.1

Total heat added = 4.25 kJ

1/4 × Total heat added = $c_v \times (T_3 - T_2)$

3/4 × Total heat added = $c_p \times (T_4 - T_3)$

$c_v$ = Specific heat at constant volume = 0.718×2.821× 10⁻³

$c_p$ = Specific heat at constant pressure = 1.005×2.821× 10⁻³

v₁ - v₂ = 2.2 L

$\left \frac{v_{1}}{v_{2}} \right =r \right = 9.1$

v₁ = v₂·9.1

∴ 9.1·v₂ - v₂ = 2.2 L = 2.2 × 10⁻³ m³

8.1·v₂ = 2.2 × 10⁻³ m³

v₂ = 2.2 × 10⁻³ m³ ÷ 8.1 = 2.72 × 10⁻⁴ m³

v₁ = v₂×9.1 = 2.72 × 10⁻⁴ m³ × 9.1 = 2.47 × 10⁻³ m³

Plugging in the values, we have;

${T_{2}}= T_{1} \times \left (r \right )^{\gamma -1} = 290 \times 9.1^{1.4 - 1} = 701.479 \, K$

From;

$\left (\frac{p_{2}}{p_{1}} \right )^{\frac{\gamma -1}{\gamma }}= \left (r \right )^{\gamma -1}$ we have;

$p_{2} = p_{1}} \times \left (r \right )^{\gamma } = 95 \times \left (9.1 \right )^{1.4} = 2091.13 \ kPa$

1/4×4.25 = $0.718 \times 2.821 \times 10^{-3}\times (T_3 - 701.479)$

∴ T₃ = 1226.05 K

Also;

3/4 × Total heat added = $c_p \times (T_4 - T_3)$ gives;

3/4 × 4.25 = $1.005 \times 2.821 \times 10^{-3} \times (T_4 - 1226.05)$ gives;

T₄ = 2350.34 K

$\frac{T_{4}}{T_{5}}=\left (\frac{v_{5}}{v_{4}} \right )^{\gamma -1} = \left (\frac{r}{\rho } \right )^{\gamma -1}$

$\rho = \frac{T_4}{T_3} = \frac{2350.34}{1226.04} = 1.92$

$T_{5} = \frac{T_{4}}{\left (\frac{r}{\rho } \right )^{\gamma -1}}= \frac{2350.34 }{\left (\frac{9.1}{1.92 } \right )^{1.4-1}} =1260.56 \ K$

b) Heat rejected = $c_v \times (T_5 - T_1)$

$Therefore \ heat \ rejected = 0.718 \times 2.821 \times 10^{-3}\times (1260.56 - 290) = 1.966 kJ$

The net work done = Heat added - Heat rejected

∴ The net work done = 4.25 - 1.966 = 2.28 kJ

The net work of the cycle in kJ = 2.28 kJ

c) Power = Work done per each cycle × Number of cycles completed each second

Where we have 3000 cycles per minute, we have 3000/60 = 50 cycles per second

Hence, the power developed = 2.28 kJ/cycle × 50 cycle/second = 114.2 kW

d)

$Thermal \ efficiency, \, \eta _{dual} = \frac{Work \ done}{Heat \ supplied} = \frac{2.28}{4.25} \times 100 = 53.74 \%$

The thermal efficiency, $\eta _{dual}$ = 53.78%

e) The mean effective pressure, $p_m$ , is found as follows;

$p_m = \frac{W}{v_1 - v_2} =\frac{2.28}{2.2 \times 10^{-3}} = 1038.25 \ kPa$

The mean effective pressure = 1038.25 kPa.

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4 years ago

Technician A states that if someone is asking a question of you, then he or she is showing interest. Technician B states that si

eimsori [14]

Answer:

They are both correct

Explanation:

When someone is asking a question of you, it means he or she has either been paying attention, and is interested in you, what you're doing, or what you're saying. Also, silence when used properly can be golden in the sense that it can prevent unnecessary problems from arising, and can save one from a lot of unforeseen problem. Whatever is said cannot be taken back again, and some things should never be said at all, especially in a professional setting.

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4 years ago

A vehicle of 1 200 kg is moving at a speed of 40 km/h on an incline of 1 in 50. The total constant rolling and wind resistance i

Readme [11.4K]

Answer:11.602 KW

Explanation:

mass of vehicle $\left ( m\right )=1200 kg$

speed=40Km/h

Resistance=600 N

$\eta =80%$

Gear ratio $\left ( G\right )=4:1$

$D_{effective}=500mm$

Net force to overcome by engine is

F=Resistance + sin component of weight

F=600+ $mgsin\theta$

Where $tan\theta =[tex]\frac{1}{50}$

$\theta =1.1457^{\circ}$

$F=600+1200\times 9.81\times sin\left ( 1.1457\right )$

F=600+235.38=835.38 N

power= $F.v=835.38\frac{100}{9}$

Engine Power= $\frac{835.\frac{100}{9}}{\eta }$ =11.602 KW

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4 years ago