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Alexus [3.1K]
3 years ago
13

What is rainfall intensity

Engineering
2 answers:
natulia [17]3 years ago
7 0

Answer:

the ratio of the total amount of rain (rainfall depth) falling during a given period to the duration of the period

bearhunter [10]3 years ago
3 0

Answer:

Rainfall intensity is defined as the ratio of the total amount of rain (rainfall depth) falling during a given period to the duration of the period It is expressed in depth units per unit time, usually as mm per hour (mm/h).

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Explanation:

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Pick a subjectarea/field/topic that you are interested in. For each of the following Bonham- Carver uses of GIS give an example
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How many coordinates are required for a 3-dimensional force?
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The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is
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The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is hollow and has an outer diameter of 25 mm; it is made of a brass for which the allowable shearing stress is 50 MPa.

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3 years ago
Carbon dioxide (CO2) is compressed in a piston-cylinder assembly from p1 = 0.7 bar, T1 = 320 K to p2 = 11 bar. The initial volum
tekilochka [14]

Answer:

W_{12}=-53.9056KJ

Part A:

Q=-7.03734 KJ/Kg (-ve sign shows heat is getting out)

Part B:

Q=1.5265KJ/Kg (Heat getting in)

The value of Q at constant specific heat is approximately 361% in difference with variable specific heat and at constant specific heat Q has opposite direction (going in) than Q which is calculated in Part B from table A-23. So taking constant specific heat is not a good idea and is questionable.

Explanation:

Assumptions:

  1. Gas is ideal
  2. System is closed system.
  3. K.E and P.E is neglected
  4. Process is polytropic

Since Process is polytropic so  W_{12} =\frac{P_{2}V_{2}-P_{1}V_{1}}{1-n}

Where n=1.25

Since Process is polytropic :

\frac{V_{2}}{V_{1}}=(\frac{P_{1}}{P_{2}})^{\frac{1}{1.25}} \\V_{2}= (\frac{P_{1}}{P_{2}})^{\frac{1}{1.25}} *V_{1}

V_{2}= (\frac{0.7}{11})^{\frac{1}{1.25}} *0.262\\V_{2}=0.028924 m^3

Now,W_{12} =\frac{P_{2}V_{2}-P_{1}V_{1}}{1-n}

W_{12} =\frac{11*0.028924-0.7*0.262}{1-1.25}(\frac{10^{5}N/m^2}{1 bar})(\frac{1  KJ}{10^{3}Nm})

W_{12}=-53.9056KJ

We will now calculate mass (m) and Temperature T_2.

m=\frac{P_{1}V_{1}}{RT_{1}}\\ m=\frac{0.7*0.262}{\frac{8.314KJ}{44.01Kg.K}*320}(\frac{10^{5}N/m^2}{1 bar})(\frac{1  KJ}{10^{3}Nm})\\m=0.30338Kg

T_{2} =\frac{P_{2}V_{2}}{Rm}\\ m=\frac{11*0.028924}{\frac{8.314KJ}{44.01Kg.K}*0.30338}(\frac{10^{5}N/m^2}{1 bar})(\frac{1  KJ}{10^{3}Nm})\\T_{2} =555.14K

Part A:

According to energy balance::

Q=mc_{v}(T_{2}-T_{1})+W_{12}

From A-20, C_v for Carbon dioxide at 300 K is 0.657 KJ/Kg.k

Q=0.30338*0.657(555.14-320)+(-53.9056)

Q=-7.03734 KJ/Kg (-ve sign shows heat is getting out)

Part B:

From Table A-23:

u_{1} at 320K = 7526 KJ/Kg

u_{2} at 555.14K = 15567.292 (By interpolation)

Q=m(\frac{u(T_{2})-u(T_{1})}{M} )+W_{12}

Q=0.30338(\frac{15567.292-7526}{44.01} )+(-53.9056)

Q=1.5265KJ/Kg (Heat getting in)

The value of Q at constant specific heat is approximately 361% in difference with variable specific heat and at constant specific heat Q has opposite direction (going in) than Q which is calculated in Part B from table A-23. So taking constant specific heat is not a good idea and is questionable.

7 0
4 years ago
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