Which ONE of the following options is CORRECT in relation to the standard drill pipe?
Show Hint
- Nominal weight is equal to the actual weight.
- Nominal weight is less than the actual weight.
- Nominal weight is greater than the actual weight.
- Nominal weight is twice the actual weight.
The Correct Option is B
Solution and Explanation
For standard drill pipe, the \emph{nominal weight} (lb/ft or kg/m) is the linear weight of the \emph{pipe body only} - it excludes the heavier tool joints and upsets. Step 2: What is "actual weight"?
The \emph{actual weight} accounts for the complete joint: pipe body \emph{plus} tool joints (box and pin) and upsets. Tool joints are much thicker and denser, so they add significant extra weight per joint. Step 3: Inequality relation.
Let \(W_n\) be nominal weight (pipe body/ft). For a joint of length \(L\) (ft) with tool-joint assembly weight \(W_{tj}\) (lb) then \[ W_{\text{actual per ft}} \;=\; \frac{W_{tj} + W_n \cdot L}{L} \;=\; W_n \;+\; \frac{W_{tj}}{L} \;\;>\; W_n. \] Hence, \(\text{Actual weight} > \text{Nominal weight}\). Step 4: Concrete illustration (typical).
A common 5 in DP may have \(W_n \approx 19.5\) lb/ft (body). Including tool joints, actual linear weight often rises by \(\sim 1\!-\!3\) lb/ft, confirming \(W_{\text{actual}} > W_n\). Final Answer:\; \[ \boxed{\text{Nominal weight } < \text{ actual weight}} \]
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