Question

A chemical plant uses process steam that is supplied from a remote boiler house. The steam pipe is uninsulated and has an exposed length of 250 meters. It is currently estimated that the energy losses amount to $3750 annually. It is estimated that the application of 25 mm of insulation will reduce energy losses by 25%, cost of the installation is $25 per meter. If 50 mm insulation is applied, the cost will be $35 per meter and will result in a 33% reduction in energy losses. The expected life of the insulation is 15 years with no salvage value. Compare the present worths of the 25 mm insulation, the 50 mm insulation and no insulation if the minimum rate of return is 109 and fuel costs are expected to increase by 8% annually

Answer 1

The chemical plant has three alternative scenarios to choose from:

1) No change

2) Apply 25 mm insulation on the steam pipe, or

3) Apply 50 mm insulation on the steam pipe

To recommend the best alternative, we need to find the present value of scenarios 2 and 3.

The relevant data to be considered is as follows:

Sr. No.	Particulars	Label	25 mm insulation	50 mm insulation
1	Insulation cost per meter ($)		25	35
2	Total length (meters)		250	250
3	Total cost of insulation ($) - (1*2)		6250	8750
4	Energy loss per year ($)		3750	3750
5	Energy loss saving (%)		25%	33%
6	Energy loss saved in year 1 ($) - (4*5)	P	937.50	1237.50
7	Fuel cost increase p.a. (%)	g	8%	8%
8	Expected rate of return (%)	r	10%	10%
9	Expected life of insulation (years)	n	15	15

The formula for calculating the present value of an annuity which increases at a specific rate is as follows:

PV= P/(r-g) * [1-{(1+g)/(1+r)}^n] where PV stands for Present Value, P stands for the annual amount in year 1, r stands for expected rate of return, g stands for the annual growth rate in P, and n stands for the number of years.

Using this formula, and the labels as identified in row nos. 6,7,8 and 9 of the table above, we calculate the Present value of scenario 2 and scenario 3 as below:

Scenario 2 - (application of 25 mm insulation)

PV= P/(r-g) * [1-{(1+g)/(1+r)}^n]

= 937.5/ (0.10-0.08) * [1-{(1+0.08)/(1+0.10)}^15

= 937.5/0.02 *[1-{1.08/1.10}^15]

= 46875 * [1-0.98181818^15]

= 46875* [1-0.759392044]

= 46875 *0.240607956

= $ 11279

Scenario 3 - (application of 50 mm insulation)

PV = P/(r-g) * [1-{(1+g)/(1+r)}^n]

= 1237.5/ (0.10-0.08) * [1-{(1+0.08)/(1+0.10)}^15

= 1237.5/0.02 *[1-{1.08/1.10}^15]

= 61875 * [1-0.98181818^15]

= 61875* [1-0.759392044]

= 61875 *0.240607956

= $ 14888

Now, we compare the present values of the 2 scenarios in the table below.

Sr. No.	Particulars	25 mm insulation	50 mm insulation
1	Present value of future cash flows ($)	11279	14888
2	Investment required ($)	6250	8750
3	NPV ($) - (1-2)	5029	6138
4	Multiple - (1/2)	1.8x	1.7x

Thus, both the alternatives - 25 mm insulation as well as 50 mm insulation stand favorable as compared to the current situation as they are expected to return more than 10% on capital invested. The 50 mm insulation works better compared to the 25 mm insulation in terms of the return in absolute numbers, though the return on capital is higher in case of 25 mm insulation as indicated by the higher multiple.