Lockheed Martin Skunk Works designs and produces aircraft for defense using rapid prototyping tools

Determine the following parameters for the water having quality x=0.7 at 200 kPa:

ra1l [238]

Solution :

Given :

Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa

The phase diagram is provided below.

a). The phase is a standard mixture.

b). At pressure, p = 200 kPa, T = $T_{saturated}$

Temperature = 120.21°C

c). Specific volume

$v_{f}= 0.001061, \ \ v_g=0.88578 \ m^3/kg$

$v_x=v_f+x(v_g-v_f)$

$=0.001061+0.7(0.88578-0.001061)$

$=0.62036 \ m^3/kg$

d). Specific energy ( $u_x$ )

$u_f=504.5 \ kJ/kg, \ \ u_{fg}=2024.6 \ kJ/kg$

$u_x=504.5 + 0.7(2024.6)$

$=1921.72 \ kJ/kg$

e). Specific enthalpy $(h_x)$

At $h_f = 504.71, \ \ h_{fg} = 2201.6$

$h_x=504.71+(0.7\times 2201.6)$

$= 2045.83 \ kJ/kg$

f). Enthalpy at m = 0.5 kg

$H=mh_x$

$= 0.5 \times 2045.83$

= 1022.91 kJ

7 0

3 years ago

To assist in completing this question, you may reference the Animated Technique Video - MALDI-TOF Mass Spectroscopy. Complete th

a_sh-v [17]

Complete Question

The complete question is shown on the first uploaded image.

Answer:

The answer is shown on the second uploaded image

Explanation:

The explanation is also shown on the second uploaded image

3 0

3 years ago

There is an electric field near the Earth's surface whose magnitude is about 145 V/m . How much energy is stored per cubic meter

weqwewe [10]

Answer:

u_e = 9.3 * 10^-8 J / m^3 ( 2 sig. fig)

Explanation:

Given:

- Electric Field strength near earth's surface E = 145 V / m

- permittivity of free space (electric constant) e_o = 8.854 *10^-12 s^4 A^2 / m^3 kg

Find:

- How much energy is stored per cubic meter in this field?

Solution:

- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:

u_e = 0.5*e_o * E^2

- Plug in the values given:

u_e = 0.5*8.854 *10^-12 *145^2

u_e = 9.30777 * 10^-8 J/m^3

5 0

3 years ago

No question but thx<br> reeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

jeka94

Answer:

why you doin this

Explanation:

is this so we get free points?

5 0

3 years ago

Read 2 more answers

The structure of a house is such that it loses heat at a rate of 4500kJ/h per °C difference between the indoors and outdoors. A

adelina 88 [10]

Answer:

15.24°C

Explanation:

The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as

$COP=\frac{Q_{in}}{W}$

Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat

You can think of this quantity as similar to heat engine's efficiency

In our case, the COP of our heater is

$COP_{heater} = \frac{\frac{4500\ kJ}{3600\ s} *(T_{house}-T_{out})}{4\ kW}$

Where T_{house} = 24°C and T_{out} is temperature outside

To achieve maximum heating, we will have to use the most efficient heat pump, and, according to the second law of thermodynamics, nothing is more efficient that Carnot Heat Pump

Which has COP of:

$COP_{carnot}=\frac{T_{house}}{T_{house}-T_{out}}$

So we equate the COP of our heater with COP of Carnot heater

$\frac{1.25 *(T_{house}-T_{out})}{4}=\frac{T_{house}}{T_{house}-T_{out}}$

Rearrange the equation

$\frac{1.25}{4}(24-T_{out})^2-24=0$

Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be

15.24°C

4 0

3 years ago