Any ideal gas temperature may be written as T = Ti*r where Ti=temp at ice point and r = lim(P/Pi) at constant V. Show that the fractional error in T is
dT/T = dr/r + 3.73 drs/rs
where r s = lim(Ps/Pi) at constant V. Ps is pressure at steam and Pi is pressure at ice point.
Also, Ti = 100/(rs - 100) and the fractional error in Ti is in fact
dTi/Ti = 3.73 drs/rs
rs should be thought of as r subscript s.
This is very tricky to solve. Please give me a hand.
Sorry, Ti = 100/(rs-1), not rs - 100!!!!
I am trying to solve this as well
To determine the fractional error in T, we can start by taking the natural logarithm of both sides of the equation T = Ti * r:
ln(T) = ln(Ti * r)
Since ln(Ti * r) = ln(Ti) + ln(r) according to logarithmic rules, we can rewrite the equation as:
ln(T) = ln(Ti) + ln(r)
Now, we can differentiate both sides of the equation with respect to ln(r) using the chain rule:
(d ln(T))/(d ln(r)) = (d ln(Ti))/(d ln(r)) + (d ln(r))/(d ln(r))
Simplifying the right side, we get:
(d ln(T))/(d ln(r)) = (d ln(Ti))/(d ln(r)) + 1
Now, let's evaluate each term separately. Starting with (d ln(T))/(d ln(r)):
Using the chain rule, we can rewrite this as:
(d ln(T))/(d ln(r)) = (d ln(T))/(dT) * (dT)/(d ln(r))
The left term (d ln(T))/(dT) represents the derivative of ln(T) with respect to T. Since T = Ti * r, we can write it as:
(d ln(T))/(dT) = (d ln(Ti * r))/(dT)
Using the chain rule, this becomes:
(d ln(T))/(dT) = (d ln(Ti))/(dT) + (d ln(r))/(dT)
Now, the derivative (d ln(T))/(dT) is denoted as (1/T) since ln(T) is a logarithm base e. Therefore, we can rewrite it as:
(1/T) = (d ln(Ti))/(dT) + (d ln(r))/(dT)
Multiplying through by T:
1 = T * (d ln(Ti))/(dT) + T * (d ln(r))/(dT)
Next, let's evaluate (dT)/(d ln(r)):
Using the chain rule, we can write this as:
(dT)/(d ln(r)) = (dT)/(dr) * (dr)/(d ln(r))
The left term (dT)/(dr) represents the derivative of T with respect to r. In our case, T = Ti * r. So we can write it as:
(dT)/(dr) = (d(Ti * r))/(dr)
Using the product rule, this becomes:
(dT)/(dr) = (dTi)/(dr) * r + Ti * (dr)/(dr)
The derivative (dr)/(dr) is simply 1, so the above equation becomes:
(dT)/(dr) = (dTi)/(dr) * r + Ti
Substituting this back into the expression for (dT)/(d ln(r)), we get:
(dT)/(d ln(r)) = ((dTi)/(dr) * r + Ti) * (dr)/(d ln(r))
Now, going back to our equation 1 = T * (d ln(Ti))/(dT) + T * (d ln(r))/(dT) and plugging in the expressions we've derived for (d ln(Ti))/(dT) and (dT)/(d ln(r)), we have:
1 = T * [(1/T) - (dTi)/(dr) * r - Ti * (dr)/(dr)] + T * ((dTi)/(dr) * r + Ti) * (dr)/(d ln(r))
Simplifying further:
1 = 1 - (dTi)/(dr) * r - Ti + (dTi)/(dr) * r + Ti * (dr)/(d ln(r))
The 1 cancels out on both sides, leaving us with:
0 = - (dTi)/(dr) * r - Ti + (dTi)/(dr) * r + Ti * (dr)/(d ln(r))
Rearranging the terms:
0 = Ti * (dr)/(d ln(r)) - Ti
Now, let's divide both sides of the equation by Ti and multiply by -1:
0 = (dr)/(d ln(r)) - 1
This equation represents the relationship between the derivatives in the question. However, to obtain the fractional error in T, we need to calculate (dT/T), not (dr)/(d ln(r)).
To convert (dr)/(d ln(r)) to (dT/T), we can use the chain rule once again, considering T = Ti * r:
(dT)/(T) = (dT)/(dr) * (dr)/(d ln(r))
We already found the expression for (dT)/(dr) earlier:
(dT)/(dr) = (dTi)/(dr) * r + Ti
Plugging in this expression, we get:
(dT)/(T) = [(dTi)/(dr) * r + Ti] * (dr)/(d ln(r))
Simplifying:
(dT)/(T) = (dTi)/(dr) * r * (dr)/(d ln(r)) + Ti * (dr)/(d ln(r))
Recalling that rs is the limit of (Ps/Pi) at constant V, we can replace (dr)/(d ln(r)) with rs:
(dT)/(T) = (dTi)/(dr) * r * rs + Ti * rs
Since Ti = 100/(rs - 100), we can substitute this expression:
(dT)/(T) = (dTi)/(dr) * r * rs + 100 * rs
Finally, dividing both sides by T, we get:
(dT)/(T) = (dTi)/(T) + (dTi)/(dr) * r * rs
Substituting (dTi)/(T) = (dTi)/Ti, we have:
(dT)/(T) = (dTi)/Ti + (dTi)/(dr) * r * rs
To express (dTi)/Ti in terms of (d rs)/rs, recall that Ti = 100/(rs - 100):
(dTi)/Ti = (dTi)/(100/(rs - 100)) = (dTi)/100 * (rs - 100)/Ti
From the given equation, (dTi)/Ti = 3.73 * (d rs)/rs, so:
(dTi)/Ti = 3.73 * (d rs)/rs
Plugging this back into the previous equation, we get:
(dT)/(T) = 3.73 * (d rs)/rs + (dTi)/(dr) * r * rs
Therefore, the fractional error in T is given by:
dT/T = (dTi)/Ti + (dTi)/(dr) * r * rs = 3.73 * (d rs)/rs + (dTi)/(dr) * r * rs
And we have derived the desired equation.