Answer:
Length = 129.55m, 129.55m
Explanation:
Given:
cp of water = 4180 J/kg·°C
Diameter, D = 2.5 cm
Temperature of water in = 17°C
Temperature of water out = 80°C
mass rate of water =1.8 kg/s.
Steam condensing at 120°C
Temperature at saturation = 120°C
hfg of steam at 120°C = 2203 kJ/kg
overall heat transfer coefficient of the heat exchanger = 700 W/m2 ·°C
U = 700 W/m2 ·°C
Since Temperature of steam is at saturation,
temperature of steam going in = temperature of steam out = 120°C
Energy balance:
Heat gained by water = Heat loss by steam
Let specific capacity of steam = 2010kJ/Kg .°C
Find attached the full solution to the question.
Answer:
Q = 125.538 W
Explanation:
Given data:
D = 30 cm
Temperature 
degree celcius
Heat coefficient = 12 W/m^2 K
Efficiency 80% = 0.8
Q = 125.538 W
Answer:
knowledge of animal behavior and anatomy
Explanation:
the qualification that will make Milton successful in his research is a knowledge of animal behaviour and also their anatomy. the knowledge of whales behaviour has opened his eyes into their world so he knows to a great deal about them. it is through his knowledge of the behaviour of whales that he's able to get used to their migrating patterns to know where and when to find them. Also, through the body anatomy of whales he knows what their movement is like.
Answer:
Flow-rate = 0.0025 m^3/s
Explanation:
We need to assume that the flow-rate of pure water entering the pond is the same as the flow-rate of brine leaving the pond, in other words, the volume of liquid in the pond stays constant at 20,000 m^3. Using the previous assumption we can calculate the flow rate entering or leaving the tank (they are the same) building a separable differential equation dQ/dt, where Q is the milligrams (mg) of salt in a given time t, to find a solution to our problem we build a differential equation as follow:
dQ/dt = -(Q/20,000)*r where r is the flow rate in m^3/s
what we pose with this equation is that the variable rate at which the salt leaves the pond (salt leaving over time) is equal to the concentration (amount of salt per unit of volume of liquid at a given time) times the constant rate at which the liquid leaves the tank, the minus sign in the equation is because this is the rate at which salt leaves the pond.
Rearranging the equation we get dQ/Q = -(r/20000) dt then integrating in both sides ∫dQ/Q = -∫(r/20000) dt and solving ln(Q) = -(r/20000)*t + C where C is a constant (initial value) result of solving the integrals. Please note that the integral of dQ/Q is ln(Q) and r/20000 is a constant, therefore, the integral of dt is t.
To find the initial value (C) we evaluate the integrated equation for t = 0, therefore, ln(Q) = C, because at time zero we have a concentration of 25000 mg/L = 250000000 mg/m^3 and Q is equal to the concentration of salt (mg/m^3) by the amount of liquid (always 20000 m^3) -> Q = 250000000 mg/m^3 * 20000 m^3 = 5*10^11 mg -> C = ln(5*10^11) = 26.9378. Now the equation is ln(Q) = -(r/20000)*t + 26.9378, the only thing missing is to find the constant flow rate (r) required to reduce the salt concentration in the pond to 500 mg/L = 500000 mg/m^3 within one year (equivalent to 31536000 seconds), to do so we need to find the Q we want in one year, that is Q = 500000 mg/m^3 * 20000 m^3 = 1*10^10 mg, therefore, ln(1*10^10) = -(r/20000)*31536000 + 26.9378 solving for r -> r = 0.002481 m^3/s that is approximately 0.0025 m^3/s.
Note:
- ln() refer to natural logarithm
- The amount of liquid in the tank never changes because the flow-rate-in is the same as the flow-rate-out
- When solving the differential equation we calculated the flow-rate-out and we were asked for the flow-rate-in but because they are the same we could solve the problem
- During the solving process, we always converted units to m^3 and seconds because we were asked to give the answer in m^3/seg
Answer:
The barometer reading will be 29.43 in
Explanation:
Using the formula of pressure variation
p2 - p1 = -yair * H
= 7.65 * 
= 38.5 
According to the relationship between the pressure and the height of the mercury column
p = yHg * h --> where yHg and h is the barometer reading
yHg 
- yHg * h1 = 38.5
h1 = (
) - 
![[(\frac{29.97}{12} ft) - 0.0455 ft] - 12 \frac{in}{ft} \\\\h1 = 29.43 in](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B29.97%7D%7B12%7D%20ft%29%20-%200.0455%20ft%5D%20-%2012%20%5Cfrac%7Bin%7D%7Bft%7D%20%5C%5C%5C%5Ch1%20%3D%2029.43%20in)
