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3 years ago

What is coefficient of thermal conductivity

Engineering

1 answer:

True [87]3 years ago

7 0

Answer:

<em>The coefficient of thermal conductivity -- is the quantity of heat that is transported through a unit cube of two surfaces of perpendicular distance to each other in a given unit of time when the difference in temperature of the two surfaces is 1°C and its unit is W/m∗K</em>

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______ are an idication that your vehicle may be developing a cooling system problem.

iris [78.8K]

Answer:

The temperature gauge showing that the vehicle has been running warmer or has recently began to have issues from overheating is an idication that your vehicle may be developing a cooling system problem.

Explanation:

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3 years ago

A debugging process where you, the programmer, pretend you are a computer and step through each statement while recording the va

Ipatiy [6.2K]

Answer:

hand tracing

Explanation:

as a programmer when we pretend computer in the debugging process by the step of each statement in recording

then there value of variable is hand tracing because as The hand tracking feature is the use of hands as an input method

so while recording value of each variable each step is hand tracing

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3 years ago

A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this

tresset_1 [31]

Answer:

a)COP=5.01

b) $W_{in}=2.998$ KW

c)COP=6.01

d) $Q_R=17.99 KW$

Explanation:

Given

$T_L$ = -12°C, $T_H$ =40°C

For refrigeration

We know that Carnot cycle is an ideal cycle that have all reversible process.

So COP of refrigeration is given as follows

$COP=\dfrac{T_L}{T_H-T_L}$ ,T in Kelvin.

$COP=\dfrac{261}{313-261}$

a)COP=5.01

Given that refrigeration effect= 15 KW

We know that $COP=\dfrac{RE}{W_{in}}$

RE is the refrigeration effect

5.01= $\dfrac{15}{W_{in}}$

b) $W_{in}=2.998$ KW

For heat pump

So COP of heat pump is given as follows

$COP=\dfrac{T_h}{T_H-T_L}$ ,T in Kelvin.

$COP=\dfrac{313}{313-261}$

c)COP=6.01

In heat pump

Heat rejection at high temperature=heat absorb at low temperature+work in put

$Q_R=Q_A+W_{in}$

Given that $Q_A=15$ KW

We know that $COP=\dfrac{Q_R}{W_{in}}$

$COP=\dfrac{Q_R}{Q_R-Q_A}$

$6.01=\dfrac{Q_R}{Q_R-15}$

d) $Q_R=17.99 KW$

5 0

4 years ago

H2O enters a conical nozzle, operates at a steady state, at 2 MPa, 300 oC, with the inlet velocity 30 m/s and the mass flow rate

Colt1911 [192]

Answer:

The flow velocity at outlet is approximately 37.823 meters per second.

The inlet radius of the nozzle is approximately 0.258 meters.

Explanation:

A conical nozzle is a steady state device used to increase the velocity of a fluid at the expense of pressure. By First Law of Thermodynamics, we have the energy balance of the nozzle:

Energy Balance

$\dot m \cdot \left[\left(h_{in}+\frac{v_{in}^{2}}{2} \right)-\left(h_{out}+\frac{v_{out}^{2}}{2} \right)\right]= 0$ (1)

Where:

$\dot m$ - Mass flow, in kilograms per second.

$h_{in}$ , $h_{out}$ - Specific enthalpies at inlet and outlet, in kilojoules per second.

$v_{in}, v_{out}$ - Flow speed at inlet and outlet, in meters per second.

It is recommended to use water in the form of superheated steam to avoid the appearing of corrosion issues on the nozzle. From Property Charts of water we find the missing specific enthalpies:

Inlet (Superheated steam)

$p = 2000\,kPa$

$T = 300\,^{\circ}C$

$h_{in} = 3024.2\,\frac{kJ}{kg}$

$\nu_{in} = 0.12551\,\frac{m^{3}}{kg}$

Where $\nu_{in}$ is the specific volume of water at inlet, in cubic meters per kilogram.

Outlet (Superheated steam)

$p = 600\,kPa$

$T = 160\,^{\circ}C$

$h_{out} = 2758.9\,\frac{kJ}{kg}$

If we know that $\dot m = 50\,\frac{kJ}{kg}$ , $h_{in} = 3024.2\,\frac{kJ}{kg}$ , $h_{out} = 2758.9\,\frac{kJ}{kg}$ and $v_{in} = 30\,\frac{m}{s}$ , then the flow speed at outlet is:

$35765-25\cdot v_{out}^{2} = 0$ (2)

$v_{out} \approx 37.823\,\frac{m}{s}$

The flow velocity at outlet is approximately 37.823 meters per second.

The mass flow is related to the inlet radius ( $r_{in}$ ), in meters, by this expression:

$\dot m = \frac{\pi \cdot v_{in}\cdot r_{in}^{2} }{\nu_{in}}$ (3)

If we know that $\dot m = 50\,\frac{kJ}{kg}$ , $v_{in} = 30\,\frac{m}{s}$ and $\nu_{in} = 0.12551\,\frac{m^{3}}{kg}$ , then the inlet radius is:

$r_{in} = \sqrt{\frac{\dot m\cdot \nu_{in}}{\pi\cdot v_{in}}}$

$r_{in}\approx 0.258\,m$

The inlet radius of the nozzle is approximately 0.258 meters.

7 0

3 years ago

You plan to install an active, liquid-based solar heating system for hot water. There are four candidate collector systems. Your

olchik [2.2K]

Solution:

The given formula,

$x=F_{R} U_{L} \times \frac{P l}{F R_{1}} \times\left(T_{r e f}-\bar{T}_{a}\right) \Delta t \times \frac{A_{c}}{L}$

$y=F_{R}(\tau \alpha)_{n} x \frac{F_{R}^{\prime}}{F_{R}} \times \frac{(\bar{\tau} d)}{(T d)_{n}} \times \bar{H}_{T} N \times \frac{A C}{L}$

$\frac{x}{y}=\frac{ u_{L} \times\left(T_{x t}-\bar{T}_{a}\right) \times \Delta t}{\left(\tau_{x}\right)_{h} \times\left(\frac{\bar{\tau}_{d}}{\left.| \tau_{d}\right)_{n}}\right) \times \bar{H}+N}$

From the table,

$1) $\quad x=2 \cdot 87, \quad y=0.96$\\$\frac{x}{y}=\frac{2187}{0.96}$22895\\\\2) $x=3 \cdot 466 \cdot y=6 \cdot 998$\\$\frac{x}{y}=\frac{3 \cdot 466}{0.898}$$=3 \cdot 4729$$

$3$x=3 \cdot 229, y=1 \cdot 08$\\$\frac{x}{x}=\frac{3 \cdot 229}{1 \cdot 08}$\\=2.9898\)\\\\4) $x=6.525, y=1.094$\\$\frac{x}{y}=\frac{5.625}{1.094}$\\=5.0502$

8 0

4 years ago

What is coefficient of thermal conductivity​

What is coefficient of thermal conductivity