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2 years ago

Order of Design Process steps ?

Engineering

1 answer:

gregori [183]2 years ago

5 0

Answer:

The five stages of Design Thinking, according to d.school, are as follows: Empathise, Define , Ideate, Prototype, and Test. Let's take a closer look at the five different stages of Design Thinking

Explanation:

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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

Harmonic excitation of motion is represent as

Gennadij [26K]

Harmonic excitation refers to a sinusoidal external force of a certain frequency applied to a system. ... Resonance occurs when the external excitation has the same frequency as the natural frequency of the system. It leads to large displacements and can cause a system to exceed its elastic range and fail structurally.

6 0

3 years ago

At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r

ludmilkaskok [199]

Answer:

a) $COP_{R} = 25.014$ , b) $T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

$COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}$

$COP_{R} = 25.014$

b) The respective coefficient of performance is determined:

$COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}$

$COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}$

$COP_{R} = 5$

But:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

The temperature at hot reservoir is found with some algebraic help:

$COP_{R} \cdot (T_{H}-T_{L})=T_{L}$

$T_{H}-T_{L} = \frac{T_{L}}{COP_{R}}$

$T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}} \right)$

$T_{H} = 273.15\,K \cdot \left(1+\frac{1}{5} \right)$

$T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

8 0

3 years ago

Read 2 more answers

Water vapor at 100 psi, 500 F and a velocity of 100 ft./sec enters a nozzle operating at steady sate and expands adiabatically t

almond37 [142]

Answer:

a)exit velocity of the steam, V2 = 2016.8 ft/s

b) the amount of entropy produced is 0.006 Btu/Ibm.R

Explanation:

Given:

P1 = 100 psi

V1 = 100 ft./sec

T1 = 500f

P2 = 40 psi

n = 95% = 0.95

a) for nozzle:

Let's apply steady gas equation.

$h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2}$

h1 and h2 = inlet and exit enthalpy respectively.

At T1 = 500f and P1 = 100 psi,

h1 = 1278.8 Btu/Ibm

s1 = 1.708 Btu/Ibm.R

At P2 = 40psi and s1 = 1.708 Btu/Ibm.R

1193.5 Btu/Ibm

Let's find the actual h2 using the formula :

$n = \frac{h_1 - h_2*}{h_1 - h_2}$

$n = \frac{1278.8 - h_2*}{1278.8 - 1193.5}$

solving for h2, we have

$h_2 = 1197.77 Btu/Ibm$

Take Btu/Ibm = 25037 ft²/s²

Using the first equation, exit velocity of the steam =

$(1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}$

Solving for V2, we have

V2 = 2016.8 ft/s

b) The amount of entropy produced in BTU/ lbm R will be calculated using :

Δs = s2 - s1

Where s1 = 1.708 Btu/Ibm.R

At h2 = 1197.77 Btu/Ibm and P2 =40 psi,

S2 = 1.714 Btu/Ibm.R

Therefore, amount of entropy produced will be:

Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R

= 0.006 Btu/Ibm.R

3 0

3 years ago

Which reference source may be consulted to answer questions regarding the Professional Engineers Act?

9966 [12]

Answer: c. The Professional Engineers Act and Board Rules

Explanation:

The reference source may be consulted to answer questions regarding the Professional Engineers Act is the The Professional Engineers Act and Board Rules.

The Professional Engineers Act and Board Rules is an Act that was established in order to regulate the qualifications for professional engineered, register them and also make sure that their conducts and behavior are looked into.

6 0

3 years ago