A double-pipe heat exchanger is used to cool a hot fluid (cp = 3800 J/kg·K) entering the heat exchanger at 200°C with a flow rat
e of 0.4 kg/s. In the cold side, cooling fluid (cp = 4200 J/kg·K) enters the heat exchanger at 10°C with a mass flow rate of 0.5 kg/s. The double-pipe heat exchanger has a thin-walled inner tube, with convection heat transfer coefficients inside and outside of the inner tube estimated to be 1400 W/m2 ·K and 1100 W/m2 ·K, respectively. The heat exchanger has a heat transfer surface area of 2.5 m2 , and the estimated fouling factor caused by the accumulation of deposit on the surfaces is 0.0002 m2 ·K/W. (a) Determine the effectiveness values for the parallel- and counter-flow configurations. (b) Determine outlet temperatures of the hot fluid for the parallel- and counter-flow configurations. (c) Determine outlet temperatures of the cold fluid for the parallel- and counter-flow configurations.
1 answer:
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Answer:
0.14% probability of observing more than 4 errors in the carpet
Explanation:
When we only have the mean, we use the Poisson distribution.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8.
This means that
What is the probability of observing more than 4 errors in the carpet
Either we observe 4 or less errors, or we observe more than 4. The sum of the probabilities of these outcomes is 1. So
We want P(X > 4). Then
In which
0.14% probability of observing more than 4 errors in the carpet
Solution :
Using the data table for refrigerant-134a at P = 120 psia
∴
For pressure, P = 20 psia
Change in temperature,
Now we find the quality,
The final energy,
Change in internal energy
= 38.09297-40.5485
= -2.4556
Answer:
attached below
Explanation:
