1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
chubhunter [2.5K]
4 years ago
5

. (Use equations not the psychrometric chart) The dry- and wet-bulb temperatures of atmospheric air at 95 kPa are 25 and 17oC, r

espectively. Determine (a) the specific humidity, (b) the relative humidity, and (c) the dew-point temperature
Physics
1 answer:
Fantom [35]4 years ago
4 0

Answer:

a) The specific humidity of air is 9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}.

b) The specific humidity of air is 0.464.

c) The dew-point temperature is 12.665 ºC.

Explanation:

a) The temperature of atmospheric air is considered the dry-bulb temperature, whereas the temperature of entirely saturated air is the the wet-bulb temperature. Dry bulb pressure is the atmospheric air. First we need to find the specific humidity at wet bulb temperature (\omega_{wb}), measured in kilograms of water per kilogram of dry air:

\omega_{wb} = \frac{0.622\cdot P_{wb}}{P_{db}-P_{wb}} (Eq. 1)

Where:

P_{wb} - Wet bulb pressure, measured in kilopascals.

P_{db} - Dry bulb pressure, measured in kilopascals.

Wet bulb pressure is the saturation pressure of water evaluated at wet bulb temperature, while dry bulb pressure in the pressure presented on statement. If P_{db} = 95\,kPa and P_{wb} = 1.9591\,kPa, then the specific humidity at wet bulb temperature is:

\omega_{wb} = \frac{0.622\cdot (1.9591\,kPa)}{95\,kPa-1.9591\,kPa}

\omega_{wb} = 0.0131\,\frac{kg\,H_{2}O}{kg\,DA}

Now we use the following equation to determine the dry bulb specific humidity (\omega_{db}), measured in kilograms of water per kilogram of dry air:

\omega_{db} = \frac{c_{p,a}\cdot (T_{wb}-T_{db})+\omega_{wb}\cdot h_{fg,wb}}{h_{g,db}-h_{f,wb}} (Eq. 2)

Where:

c_{p,a} - Isobaric specific heat of air, measured in kilojoules per kilogram-Celsius.

T_{wb} - Wet-bulb temperature, measured in Celsius.

T_{db} - Dry-bulb temperature, measured in Celsius.

\omega_{wb} - Wet-bulb specific humidity, measured in kilograms of water per kilogram of dry air.

h_{fg,wb} - Wet-bulb specific enthalpy of vaporization of water, measured in kilojoules per kilogram.

h_{g,db} - Dry-bulb specific enthalpy of saturated vapor, measured in kilojoules per kilogram.

h_{f,wb} - Wet-bulb specific enthalpy of liquid vapor, measured in kilojoules per kilogram.

If we know that T_{wb} = 17\,^{\circ}C, T_{db} = 25\,^{\circ}C, c_{p,a} = 1.005\,\frac{kJ}{kg\cdot ^{\circ}C}, \omega_{wb} = 0.0131\,\frac{kg\,H_{2}O}{kg\,DA}, h_{fg, wb} = 2460.6\,\frac{kJ}{kg}, h_{g,db} = 2546.5\,\frac{kJ}{kg} and h_{f,wb} = 71.355\,\frac{kJ}{kg}, the dry bulb specific humidity is:

\omega_{db} = \frac{\left(1.005\,\frac{kJ}{kg\cdot ^{\circ}C} \right)\cdot (17\,^{\circ}C-25\,^{\circ}C)+\left(0.0131\,\frac{kg\,H_{2}O}{kg\,DA} \right)\cdot \left(2460.6\,\frac{kJ}{kg} \right)}{2546.5\,\frac{kJ}{kg}-71.355\,\frac{kJ}{kg}  }

\omega_{db} = 9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}

The specific humidity of air is 9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}.

b) Then, the relative humidity of air (\phi_{db}), dimensionless, is obtained from this expression:

\phi_{db} = \frac{\omega_{db}\cdot P_{db}}{(0.622+\omega_{db})\cdot P_{sat, db}} (Eq. 3)

Where P_{sat, db} is the saturation pressure at dry-bulb temperature, measured in kilopascals.

If we know that \omega_{db} = 9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}, P_{db} = 95\,kPa and P_{sat, db} = 3.1698\,kPa, the relative humidity of air is:

\phi_{db} = \frac{\left(9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA} \right)\cdot (95\,kPa)}{\left(0.622+9.774\times 10^{-3}\,\frac{kg\,H_{2}O}{kg\,DA}\right)\cdot 3.1698\,kPa}

\phi_{db} = 0.464

The specific humidity of air is 0.464.

c) The dew point temperature is the temperature at which water is condensated when air is cooled at constant pressure. That temperature is equivalent to the saturation temperature at vapor pressure (P_{v}), measured in kilopascals:

P_{v} = \phi_{db} \cdot P_{sat, db} (Eq. 4)

(\phi_{db} = 0.464, P_{sat, db} = 3.1698\,kPa)

P_{v} = 0.464\cdot (3.1698\,kPa)

P_{v} = 1.4707\,kPa

The saturation temperature at given vapor pressure is:

T_{dp} = 12.665\,^{\circ}C

The dew-point temperature is 12.665 ºC.

You might be interested in
The temperature of a smelting furnace is found to be 2000 degree Celsius.find the temperature on Fahrenheit scale​
Dima020 [189]

Answer:

3632.

Explanation:

Based on the formula:

2000×(9/5)+32=3632

6 0
3 years ago
A ray of light parallel to the principal axis of spherical mirror
lesya692 [45]

Answer:

This question is incomplete but the completed question is below

A ray of light parallel to the principal axis of a spherical mirror passes through its _____ after reflection.

(a) focus (b) pole (c) centre of curvature (d) imaginary centre of curvature

The correct option is (a)

Explanation:

After reflection, a ray of light will usually appear to diverge after they hit a spherical mirror, extending behind the mirror to meet at a point (called the focus or focal point) between the pole and the centre of curvature of this mirror.  Hence, option a is the correct option.

6 0
3 years ago
James is planning a science fair project on sound waves. He places an alarm inside a jar which he can remove the
otez555 [7]

Answer: c

Explanation:

Sound waves cannot travel through a medium

3 0
3 years ago
50 points Answer fast please
erma4kov [3.2K]

Answer:

22 hours

Explanation:

6 0
3 years ago
A steel piano string of mass per unit lenght 4×10-⁴kgm-¹ was struck, if the speed of the sound on the string is 300ms-¹. Calcula
Snezhnost [94]

Answer:

36 N

Explanation:

Velocity of a standing wave in a stretched string is:

v = √(T/ρ),

where T is the tension and ρ is the mass per unit length.

300 m/s = √(T / 4×10⁻⁴ kg/m)

T = 36 N

7 0
3 years ago
Other questions:
  • A boy throws a ball and accidentally breaks a window. The momentum of the ball and all the pieces of glass taken together after
    7·2 answers
  • A 4.80-kg watermelon is dropped from rest from the roof of a 23.0-m -tall building and feels no appreciable air resistance Just
    14·1 answer
  • given the perimeter of a rectangle and either the length or width, find the unknown measurement using the appropriate formula. T
    9·1 answer
  • A four-resistor circuit that has three parallel resistors is considered to be what type of circuit?
    8·1 answer
  • What occurs when the surface of a large body of water is heated by the sun and the water at the depths remains cool because of l
    10·2 answers
  • Suppose a zero-resistance rod slides to the right on two zero-res a distance of 0.330 m. The rails are connected by a 16 2 resis
    8·2 answers
  • A penny is dropped from the top of a building 290 m high. Ignoring air resistance, if
    14·1 answer
  • 90. An uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment
    6·1 answer
  • What type of fault is shown in the image below?
    14·1 answer
  • You and you friend are sitting on the floor in an empty room a distance, d, apart, and you both have an equal mass. Assume the o
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!