Given asphalt content test data:
1 answer:
Answer:
hello your question is incomplete attached below is the complete question
A) overall mean = 5.535, standard deviation ≈ 0.3239
B ) upper limit = 5.85, lower limit = 5.0
C) Not all the samples meet the contract specifications
D) fluctuation ( unstable Asphalt content )
Explanation:
B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday
The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85
The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0
attached below is the required plot
C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :
15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20
D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed
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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 =
Rate of heat transfer,
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)
L = 6.23 cm
Therefore, the thickness of the material is 6.23 cm
The rate of heat transfer by the air conditioner using constant specific heat of 1.004kj/kg.K is 15.06 kW.
<h3>What is the rate of heat transfer?</h3>Rate of heat transfer is the power rating of the machine.
Work done and changes in potential and kinetic energy are neglected since it is a steady state process.
The specific heat in terms of specific heat capacity and temperature change is given as:
The rate of heat transfer, is then determined as follows:
- Qout = flow rate × specific heat
Qout = 0.75 × 20.08 = 15.06 kW
Therefore, the rate of heat transfer by the air conditioner is 15.06 kW.
Learn more about rate of heat transfer at: brainly.com/question/17152804
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