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KatRina [158]
3 years ago
12

The steam requirements of a manufacturing facility are being met by a boiler whose rated heat input is 5.5 x 106 Btu/h. The comb

ustion efficiency of the boiler is measured to be 0.65. After tuning up the boiler, the combustion efficiency rises to 0.75. The boiler operates 4000 hours a year intermittently. Take the unit cost of energy to be $4.35/106 Btu. Assuming the same heat output for before and after the tunning:________.
Engineering
1 answer:
Veronika [31]3 years ago
7 0

Answer:

Explanation:uhhhhhh all?

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BlackZzzverrR [31]

Answer:

the answer is c

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How do you put air knight in slingshock mode.
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3 0
3 years ago
The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet
alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

T_{h_i}=200^{\circ}C

T_{h_o}=120^{\circ}C

T_{c_i}=100^{\circ}C

T_{c_o}=125^{\circ}C

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]

\frac{m_hc_{ph}}{m_cc_{pc}}=\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )

we can see that heat capacity of hot fluid is minimum

Also from energy balance

Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )

NTU=\frac{UA}{\left ( mc_p\right )_{h}}=\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}

T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}

T_m=41.63^{\circ}C

NTU=1.921

And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}

\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}

\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}

\epsilon =\frac{1-0.2669}{1-0.0834}

\epsilon =0.799

5 0
3 years ago
Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w
Dafna11 [192]

Answer:

\mathbf{h_2 =1021.9 \  mm}

Explanation:

Given that :

The initial volume of water V_1 = 1000.00 mL = 1000000 mm³

The initial temperature of the water  T_1 = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water T_2 = 70° C

The coefficient of thermal expansion for the glass is  ∝ = 3.8*10^{-6 } mm/mm  \ per ^oC

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature T_1 = 10 ^ 0C  the density of the water is \rho = 999.7 \ kg/m^3

At temperature T_2 = 70^0 C  the density of the water is \rho = 977.8 \ kg/m^3

The mass of the water is  \rho V = \rho _1 V_1 = \rho _2 V_2

Thus; we can say \rho _1 V_1 = \rho _2 V_2;

⇒ 999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2

V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }

V_2 = 1022.40 \ mL

v_2 = 1022400 \ mm^3

Thus, the volume of the water after heating to a required temperature of  70^0C is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

V_1 = A_1 *h_1

The area of the water before heating is:

A_1 = \dfrac{V_1}{h_1}

A_1 = \dfrac{1000000}{1000}

A_1 = 1000 \ mm^2

The area of the heated water is :

A_2 = A_1 (1  + \Delta t  \alpha )^2

A_2 = A_1 (1  + (T_2-T_1) \alpha )^2

A_2 = 1000 (1  + (70-10) 3.8*10^{-6} )^2

A_2 = 1000.5 \ mm^2

Finally, the depth of the heated hot water is:

h_2 = \dfrac{V_2}{A_2}

h_2 = \dfrac{1022400}{1000.5}

\mathbf{h_2 =1021.9 \  mm}

Hence the depth of the heated hot  water is \mathbf{h_2 =1021.9 \  mm}

4 0
3 years ago
Earth completes one full ____ on its axis every 24 hours
mars1129 [50]

Answer:

rotation

Explanation:

I just answered this!

7 0
2 years ago
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