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3 years ago

Giving 100 points for answer plzz

Engineering

2 answers:

34kurt3 years ago

6 0

Answer:

the meaning of life is to die

Explanation:

Volgvan3 years ago

4 0

Answer:

the meaning of life is to have fun boi

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Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat

Alexandra [31]

Answer:

V = 0.30787 m³/s

m = 2.6963 kg/s

v2 = 0.3705 m³/s

v2 = 6.017 m/s

Explanation:

given data

diameter = 28 cm

steadily =200 kPa

temperature = 20°C

velocity = 5 m/s

solution

we know mass flow rate is

m = ρ A v

floe rate V = Av

m = ρ V

flow rate = V = $\frac{m}{\rho}$

V = Av = $\frac{\pi}{4} * d^2 * v1$

V = $\frac{\pi}{4} * 0.28^2 * 5$

V = 0.30787 m³/s

and

mass flow rate of the refrigerant is

m = ρ A v

m = ρ V

m = $\frac{V}{v}$ = $\frac{0.30787}{0.11418}$

m = 2.6963 kg/s

and

velocity and volume flow rate at exit

velocity = mass × v

v2 = 2.6963 × 0.13741 = 0.3705 m³/s

and

v2 = A2×v2

v2 = $\frac{v2}{A2}$

v2 = $\frac{0.3705}{\frac{\pi}{4} * 0.28^2}$

v2 = 6.017 m/s

7 0

4 years ago

A pump of a water distribution system at 25°C is powered by a 15 kW electric motor whose efficiency is 90 percent. The water flo

IRISSAK [1]

The friction loss in the system is 3.480 kilowatts.

<h2>Procedure - Friction loss through a pump</h2><h2 /><h3>Pump model</h3><h3 />

Let suppose that the pump within a distribution system is an open system at steady state, whose mass and energy balances are shown below:

<h3>Mass balance</h3>

$\dot m_{in}-\dot m_{out} = 0$ (1)

$\dot m_{in} = \frac{\dot V_{in}}{\nu_{in}}$ (2)

$\dot m_{out} = \frac{\dot V_{out}}{\nu_{out}}$ (3)

<h3>Energy balance</h3>

$\eta \cdot \dot W_{el} + \dot m_{in}\cdot (h_{in}-h_{out}) - \dot W_{f} = 0$ (4)

Where:

$\dot m_{in}$ - Inlet mass flow, in kilograms per second.
$\dot m_{out}$ - Outlet mass flow, in kilograms per second.
$\dot V_{in}$ - Inlet volume flow, in cubic meters per second.
$\dot V_{out}$ - Outlet volume flow, in cubic meters per second.
$\nu_{in}$ - Inlet specific volume, in cubic meters per kilogram.
$\nu_{out}$ - Outlet specific volume, in cubic meters per kilogram.
$\eta$ - Pump efficiency, no unit.
$\dot W_{el}$ - Electric motor power, in kilowatts.
$h_{in}$ - Inlet specific enthalpy, in kilojoules per kilogram.
$h_{out}$ - Outlet specific enthalpy, in kilojoules per kilogram.
$\dot W$ - Work losses due to friction, in kilowatts.

<h3>Data from steam tables</h3>

From steam tables we get the following water properties at inlet and outlet:

Inlet

$p = 100\,kPa$ , $T = 25\,^{\circ}C$ , $\nu = 0.001003\,\frac{kJ}{kg}$ , $h = 104.927\,\frac{kJ}{kg}$ , Subcooled liquid

Outlet

$p = 300\,kPa$ , $T = 25\,^{\circ}C$ , $\nu = 0.001003\,\frac{kJ}{kg}$ , $h = 105.128\,\frac{kJ}{kg}$ , Subcooled liquid

<h3>Calculation of the friction loss in the system</h3>

If we know that $\dot V_{in} = 0.05\,\frac{m^{3}}{s}$ , $\nu_{in} = 0.001003\,\frac{m^{3}}{kg}$ , $h_{in} = 104.927\,\frac{kJ}{kg}$ , $h_{out} = 105.128\,\frac{kJ}{kg}$ , $\eta = 0.90$ and $\dot W_{el} = 15\,kW$ , then the friction loss in the system is:

$\dot W_{f} = \frac{\dot V_{in}}{\nu_{in}}\cdot (h_{in} - h_{out}) + \eta \cdot \dot W_{el}$

$\dot W_{f} = \left(\frac{0.05\,\frac{m^{3}}{s} }{0.001003\,\frac{m^{3}}{kg} } \right)\cdot \left(104.927\,\frac{kJ}{kg}-105.128\,\frac{kJ}{kg}\right) + (0.90)\cdot (15\,kW)$