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3 years ago

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Using the tables for water, determine the specified property data at the indicated states. In each case, locate the state on ske

tches of the p–υ and T–υ diagrams.a. At p = 3 bar, υ = 0.5 m3 /kg, find T in °C and u in kJ/kg.b. At T = 320°C, υ = 0.03 m3 /kg, find p in MPa and u in kJ/kg.c. At p = 28 MPa, T = 520°C, find υ in m3 /kg and h in kJ/kg.d. At T = 10°C, υ = 100 m3 /kg, find p in kPa and h in kJ/kg.e. At p = 4 MPa, T = 160°C, find υ in m3 /kg and u in kJ/kg.

1 answer:

Anarel [89]3 years ago

5 0

heres a link! ;)

http://www.econsulat.ro/

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telo118 [61]

Explanation:

dbdhsndvhsbzbxbxjsnd

8 0

3 years ago

Charging the system is being discussed. Technician A says that the refrigerant must be weighed (or metered) to obtain a proper c

andrey2020 [161]

Charging the system is being discussed. Technician A is correct and Technician B is incorrect because overcharging can cause the system to reduce it capacity, but it does not reduce its cooling efficiency.

<h3>What is a coolant?</h3>

The coolants are the liquid that are present in machines to prevent the heating of the system or machines. Because machine and cars run they generate lots of heat that why there are coolants present.

Thus, the correct option is Technician A.

Learn more about coolant

brainly.com/question/11912897

#SPJ1

7 0

2 years ago

Intravenous infusions are usually driven by gravity by hanging the bottle at a sufficient height to counteract the blood pressur

Bingel [31]

Answer:

(a) BP = 11.99 KPa

(b) h = 2 m

Explanation:

(a)

Since, the fluid pressure and blood pressure balance each other. Therefore:

BP = ρgh

where,

BP = Blood Pressure

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = 1.2 m

Therefore,

BP = (1020 kg/m³)(9.8 m/s²)(1.2 m)

<u>BP = 11995.2 Pa = 11.99 KPa</u>

(b)

Again using the equation:

P = ρgh

with data:

P = Gauge Pressure = 20 KPa = 20000 Pa

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = ?

Therefore,

20000 Pa = (1020 kg/m³)(9.8 m/s²)h

<u>h = 2 m</u>

7 0

3 years ago

adell [148]

Answer 1:

Like most electrical mechanical devices, electric motors have mechanical bearings that eventually wear out.

The graphite brushes and commutator of older DC motors can also wear out over time

Answer2:

Question two is not specific enough!!
Are we talking DC?? Because if so, then yes, nothing will happen. Assuming you are insulted by ground.
Ac is a different story, as the two would light up dangerously and etc.

-your welcome!!

5 0

4 years ago

Consider the following hypothetical scenario for Jordan Lake, NC. In a given year, the average watershed inflow to the lake is 9

dybincka [34]

Answer:

The lake can withdraw a maximum of $1.464\times 10^{10}$ cubic feet per year to provide water supply for the Triangle area.

Explanation:

The maximum amount of water that can be withdrawn from the lake is represented by the following formula:

$V = V_{in}+V_{p}-V_{e}-V_{out}$ (Eq. 1)

Where:

$V$ - Available amount of water for water supply in the Triangle area, measured in cubic feet per year.

$V_{in}$ - Inflow amount of water, measured in cubic feet per year.

$V_{out}$ - Amount of water released for the benefit of fish and downstream water users, measured in cubic feet per year.

$V_{p}$ - Amount of water due to precipitation, measured in cubic feet per year.

$V_{e}$ - Amount of evaporated water, measured in cubic feet per year.

Then, we can expand this expression as follows:

$V = f_{in}\cdot \Delta t+h_{p}\cdot A_{l}-h_{e}\cdot A_{l}-f_{out}\cdot \Delta t$

$V = (f_{in}-f_{out})\cdot \Delta t +(h_{p}-h_{e})\cdot A_{l}$ (Eq. 2)

Where:

$f_{in}$ - Average watershed inflow, measured in cubic feet per second.

$f_{out}$ - Average flow to be released, measured in cubic feet per second.

$\Delta t$ - Yearly time, measured in seconds per year.

$h_{p}$ - Change in lake height due to precipitation, measured in feet per year.

$h_{e}$ - Change in lake height due to evaporation, measured in feet per year.

$A_{l}$ - Surface area of the lake, measured in square feet.

If we know that $f_{in} = 900\,\frac{ft^{3}}{s}$ , $f_{out} = 300\,\frac{ft^{3}}{s}$ , $\Delta t = 31,536,000\,\frac{second}{yr}$ , $h_{p} = 32\,\frac{in}{yr}$ , $h_{e} = 55\,\frac{in}{yr}$ and $A_{l} = 47,000\,acres$ , the available amount of water for supply purposes in the Triangle area is:

$V = \left(900\,\frac{ft^{2}}{s}-300\,\frac{ft^{3}}{s} \right)\cdot \left(31,536,000\,\frac{s}{yr} \right) +\left(32\,\frac{in}{yr}-55\,\frac{in}{yr} \right)\cdot \left(\frac{1}{12}\,\frac{ft}{in}\right)\cdot (47000\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)$ $V = 1.464\times 10^{10}\,\frac{ft^{3}}{yr}$

The lake can withdraw a maximum of $1.464\times 10^{10}$ cubic feet per year to provide water supply for the Triangle area.

5 0

3 years ago