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2 years ago

Which of the following is an example of a hardwood? A maple B spruce C pine D fir

Engineering

1 answer:

bearhunter [10]2 years ago

7 0

Answer:

A. Maple

Explanation:

Maple is a hardwood.

Hope that helps!

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What are the two main what are the two main concerns in the research of fluid power efficiency?

Galina-37 [17]

Answer:

The correct option is;

Materials and Components

Explanation:

The efficiency of fluid power is influenced by the components and the materials used to deliver the power of the fluid as such fluid power control are focused on

1) Advances in fluid power

2) Making use of the advantages

3) Making use of the other externally available technological advantages

4) Giving allowance for disadvantages

Areas of interest in advances in fluid power are;

a. Computer optimized flow

b. The use of new and improved materials/coatings

c. The use of components that save energy, such as intelligent supply pressure adapting systems

3 0

2 years ago

Read 2 more answers

Plz solve the problem

julsineya [31]

I attached a photo that explains and gives the answer to your questions. Had to add a border because the whole picture didn’t fit.

6 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

Which of the following applies to a module?

Ipatiy [6.2K]

Answer:

D all of the above

Explanation:

because I said so

3 0

2 years ago

Refrigerant 134a enters an air conditioner compressor at 4 bar, 20 C, and is compressed at steady state to 12 bar, 80 C. The vol

sleet_krkn [62]

Answer:

$Q=15.7Kw$

Explanation:

From the question we are told that:

Initial Pressure $P_1=4bar$

Initial Temperature $T_1=20 C$

Final Pressure $P_2=12 bar$

Final Temperature $T_2=80C$

Work Output $W= 60 kJ/kg$

Generally Specific Energy from table is

At initial state

$P_1=4bar \& T_1=20 C$

$E_1=262.96KJ/Kg$

With

Specific Volume $V'=0.05397m^3/kg$

At Final state

$P_2=12 bar \& P_2=80C$

$E_1=310.24KJ/Kg$

Generally the equation for The Process is mathematically given by

$m_1E_1+w=m_2E_2+Q$

Assuming Mass to be Equal

$m_1=m_1$

Where

$m=\frac{V}{V'}$

$m=frac{0.06666}{V'=0.05397m^3/kg}$

$m=1.24$

Therefore

$1.24*262.96+60)=1.24*310.24+Q$

$Q=15.7Kw$

4 0

3 years ago