Refer to Problem 10.32 (on page 442 ). Provide a 95 % Cl for...

Key Concepts

Odds Ratio

The odds ratio (OR) is a measure used in statistics, particularly in epidemiology and logistic regression, to quantify the strength of association between two events or characteristics. It compares the odds of an outcome occurring in one group relative to the odds of it occurring in another, providing insight into the relative risk or effect size.

Confidence Interval

A confidence interval (CI) is a statistical range, computed from sample data, that is likely to contain the true value of an unknown population parameter. A 95% CI implies that there is a 95% probability that the calculated interval includes the true parameter value, thereby conveying the precision and reliability of the estimate.

Logarithmic Transformation for Confidence Intervals

When calculating confidence intervals for an odds ratio, it is common to work with the natural logarithm of the OR. This transformation stabilizes the variance and makes the distribution of the estimate more symmetric, allowing the use of normal theory methods to compute the CI which is then exponentiated back to the original scale.

Transcript

00:01 In this question we have a carnot cycle heat pump in which refrigerant r410a is used as a working fluid.

00:09 The heat rejection takes place at 35 degrees centigrade.

00:15 So this is higher temperature and during this process, refrigerant changes from saturated vapor to saturated liquid.

00:23 Now it is given that heat addition takes place at 0 degree centigrade.

00:29 We are required to draw this cycle on temperature entropy diagram then we need to find the dryness fraction at the beginning and at the end of the isothermal heat addition process at 0 degree centigrade and finally we need to evaluate the coefficient of performance of this cycle so let's see how to solve this question the temperature entropy curve for the given heat pump is shown below so this is the temperature entropy curve.

01:05 This temperature is higher temperature t .h that is equals to 35 degree centigrade and this temperature is lower temperature or we can say pl is equal to 0 degree centigrade.

01:19 So this is the final answer for part a.

01:23 Now let's move to part b.

01:27 So refer table b.

01:34 So from this table entropy at 35 degree centigrade it means s3 is equals to 1 .1673 kilojoule per kilogramme kelvin and since this is a isentropic line therefore s4 will be equals to s3 and this will be equals to 1 .1673 kilojoule per kilogram kelvin.

02:11 Now let's apply the standard formula to calculate entropy at 0 .4.

02:17 So we can write s4 is equal to sf plus x4 into sfg.

02:28 This will be equals to s4 that means 1 .1673 is equal to sf and from this table we get sf corresponding to 0 degree centigrade is equal to 1 .0 plus x4 multiplied by sfg that means 0 .7 to 62 so when we further calculate we get x4 is equals to 0 .2303 similarly now we will find the value of x1 so so from the same table b .5 .1, the value of entropy at point two, s2 will be equals to sg and this will be equal to 1 .7139 kilojoules per kilogram kelvin.

03:29 Now s2 will be equals to s1, since s1 and s2 is a isentropic line, hence s1 will be equals to s2 and this will be equals to 1 .7139 kilojoule per kilogram kelvin...

Please give Ace some feedback