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nalin [4]
3 years ago
14

Problem 1: A catchment has the following Horton’s infiltration parameters: f0=280 mm/hr, fc=25 mm/hr and k = 2.5 hr-1 . For the

rainfall distribution is given below Interval (mins) 0-10 10-20 20-30 30-40 40-50 Rainfall (mm/hr) 25 50 80 205 170 (a) What is the cumulative infiltration rate (F), infiltration rate (fp), depression storage (S) and runoff (R) at each time step (b) What time does ponding occur?

Engineering
1 answer:
harkovskaia [24]3 years ago
6 0

Answer:

Ponding will occur in 40mins

Explanation:

We say that the infiltration rate is the velocity or speed at which water enters into the soil. This often times is measured by the depth (in mm) of the water layer that can enter the soil in one hour. An infiltration rate of 15 mm/hour means that a water layer of 15 mm on the soil surface, will take one hour to infiltrate.

Consider checking attachment for the step by step solution.

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At the instant shown car A is travelling with a velocity of 24 m/s and which is decreasing at 4 m/s2 along the highway. At the s
SVEN [57.7K]

(a) V(A/B) = (14 i - 17.32 J) m/s

(b) acc(A/B) = ( 5.11 i + 5.13 j ) m/s²

<u>Explanation:</u>

We will solve with respect to Cartesian vector form.

So,

V(A)= (24i) m/s

acc(A) = (4i) m/s²

There are two components of Car B, cos 60⁰ and sin 60⁰

V(B) = 20 cos 60° i + 20 sin 60° j

V(B) = (10 i + 17.32 j ) m/s

The car B moves along a curve, so it will have a tangential acceleration and a normal acceleration.

The tangential acceleration, a(t) = 5 m/s²

Normal acceleration, a(n) = \frac{v^2}{p} \\\\

So,

a(n) = \frac{(20)^2}{250}\\ \\a(n) = 1.6 m/s^2

For the tangential acceleration, the acceleration is slowing down. So,

a(t) = (-5 cos 60° i - 5 sin 60° j ) m/s²

a(t) = ( -2.5 i - 4.33 j) m/s²

For normal acceleration, it towards center. So,

a(n) = (1.6 sin 60° i - 1.6 cos 60° j) m/s²

a(n) = (1.39 i - 0.8 j ) m/s²

Total acceleration of Car B:

acc(B) = a(t) + a(n)

acc(B) = ( -2.5 i - 4.33 j) m/s² + (1.39 i - 0.8 j ) m/s²

acc(B) = (-1.11i - 5.13 j ) m/s²

(a) V(A/B) = ?

V(A) = V(B) + V(A/B)

(24i) m/s = (10 i + 17.32 j ) m/s + V(A/B)

V(A/B) = (14 i - 17.32 J) m/s

(b) acc(A/B) = ?

acc(A) = acc(B) + acc(A/B)

(4i) m/s² = (-1.11i - 5.13 j ) m/s² + acc(A/B)

acc(A/B) = ( 5.11 i + 5.13 j ) m/s²

3 0
2 years ago
Find the thickness of the material that will allow heat transfer of 6706.8 *10^6 kcal during the 5 months through the rectangle
Vinvika [58]

Answer:

The thickness of the material is 6.23 cm

Explanation:

Given;

quantity of heat, Q = 6706.8 *10⁶ kcal  

duration of the heat transfer, t = 5 months

thermal conductivity of copper, k = 385 W/mk

outside temperature of the heater, T₁ = 30° C

inside  temperature of the heater, T₂ = 50° C

dimension of the rectangular heater = 450 cm by 384 cm

1 kcal = 1.163000 Watt-hour

6706.8 *10⁶ kcal  = 7800008400 watt-hour

I month = 730 hours

5 months = 3650 hours

Rate of heat transfer, P = \frac{7800008400 \ Watt-Hour}{3650 \ Hours}  = 2136988.6 \ W

Rate of heat transfer, P = \frac{K*A *\delta T}{L}

where;

P is the rate of heat transfer (W)

k si the thermal conductivity (W/mk)

ΔT is change in temperature (K)

A is area of the heater (m²)

L is thickness of the heater (m)

P = \frac{KA(T_2-T_1)}{L} \\\\L =  \frac{KA(T_2-T_1)}{P}\\\\L =  \frac{385(4.5*3.84)(50-30)}{2136988.6}\\\\L = 0.0623 \ m

L = 6.23 cm

Therefore, the thickness of the material is 6.23 cm

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2 years ago
Inspection with considering a variable uses gages to determine if the product is good or bad. True or False?
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Answer:A. 40% B.50% C. 60% Od 70%

Explanation:A. True B. False

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